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## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C.

Other Exercises

Question 1.
In the given circle with diameter AB, find the value of x. (2003)

Solution:

∠ABD = ∠ACD = 30° (Angle in the same segment)
But ∠ADB = 90° (Angle in a semi-circle)
∴ x + 90° + 30° = 180°
⇒ x + 120° = 180°
∴ x = 180°- 120° = 60°

Question 2.
In the given figure, O is the centre of the circle with radius 5 cm, OP and OQ are perpendiculars to AB and CD respectively. AB = 8cm and CD = 6cm. Determine the length of PQ.

Solution:

Radius of the circle whose centre is O = 5 cm
OP ⊥ AB and OQ ⊥ CD, AB = 8 cm and CD = 6 cm.
Join OA and OC, then OA = OC=5cm.
∴ OP ⊥ AB
∴ P is the mid-point of AB.
Similarly, Q is the mid-point of CD.
In right ∆OAP,
OA² = OP² + AP² (Pythagorous theoram)
⇒ (5)² =OP² +(4)² ( ∵ AP = AB = $$\frac { 1 }{ 2 }$$ x 8 = 4cm)
⇒ 25 = OP² + 16
⇒ OP3 = 25 – 16 = 9 = (3)²
∵ OP = 3 cm
Similarly, in right ∆ OCQ,
OC2 = OQ2 + CQ2
⇒ (5)2 =OQ2+(3)2 (∵ CQ = CD = $$\frac { 1 }{ 2 }$$ x 6 = 3cm)
⇒ 25 = OQ2 + 9
⇒ OQ3 = 25 – 9 = 16 = (4)2
∴ OQ = 4 cm
Hence, PQ = OP + OQ = 3-4 = 7 cm.

Question 3.
The given figure shows two circles with centres A and B; and radii 5cm and 3cm respectively, touching each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q, find the length of PQ.

Solution:

Join AP and produce AB to meet the bigger circle at C.
AB = AC – BC = 5 cm – 3 cm = 2 cm.
But, M is the mid-point of AB
∴ AM = $$\frac { 2 }{ 2 }$$ = 1cm.
Now in right ∆APM,
AP2 = MP2 + AM2 (Pythagorous theorem)
⇒ (5)2 = MP2 -1- (1 )2
⇒ 25 = MP: + 1
⇒ MP: = 25 – 1 = 24

Question 4.
In the given figure, ABC is a triangle in which ∠BAC = 30°. Show that BC is equal to the radius of the circumcircle of the triangle ABC, whose centre is O.

Solution:

Given: In the figure ABC is a triangle in winch ∠A = 30°
To Prove: BC is the radius of circumcircle of ∆ABC whose O is the centre.
Const: Join OB and OC.
Proof: ∠BOC is at the centre and ∠BAC is at the remaining part of the circle
∴ ∠BOC = 2 ∠BAC = 2 x 30° = 60°
Now in ∆OBC,
OB = OC (Radii of the same circle)
∴ ∠OBC = ∠OCB
But ∠OBC + ∠OCB + ∠BOC – 180°
∠OBC + ∠OBC + 60°- 180°
⇒ 2 ∠OBC = 180°- 60° = 120°
⇒ ∠OBC = $$\frac { { 120 }^{ circ } }{ 2 }$$ = 60°
∴ ∆OBC is an equilateral triangle.
∴ BC = OB = OC
But OB and OC are the radii of the circumcircle
∴ BC is also the radius of the circumcircle.

Question 5.
Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
Solution:

Given: In ∆ABC, AB-AC and a circle with AB as diameter is drawn which intersects the side BC and D.
To Prove: D is the mid point of BC.
Proof: ∠ 1 = 90° (Angle in a semi-circle)
But ∠ 1 + ∠ 2 – 180° (Linear pair)
∴ ∠ 2 = 90°
Now, in right ∆ ABD and ∆ ACD,
Hyp. AB – Hyp. AC (Given)
∴ ∆ABD = ∆ACD (RHS criterion of congruency)
∴ BD = DC (c.p.c.t.)
Hence D is he mid point of BC. Q.E.D.

Question 6.
In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠CBE = 65°, calculate ∠DEC.

Solution:
Join OE,
Arc EC subtends ∠EOC at the centre and ∠EBC at the remaining part of the circle.
∴ ∠EOC = 2 ∠EBC = 2 x 65° = 130°
Now, in ∆OEC, OE = OC (radii of the same circle)
∴ ∠OEC = ∠OCE
But ∠ OEC + ∠ OCE + ∠ EOC = 180°
⇒ ∠ OCE + ∠ OCE + ∠ EOC = 180°

Question 7.
Chords AB and CD of a circle intersect each other at point P, such that AP = CP. S.how that AB = CD.

Solution:
Given: Two chords AB and CD intersect each other at P inside the circle with centre O and AP = CP.

To Prove: AB = CD.
Proof: ∵ Two chords AB and CD intersect each other inside the circle at P.
∴ AP x PB = CP x PD ⇒ $$\frac { AP}{ CP }$$ = $$\frac { AD }{ PB }$$
But AP = CP ….(i) (given)
∴ PD = PB or PB = PD …,(ii)
AP + PB = CP + PD
⇒ AB = CD Q.E.D.

Question 8.
The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.
Solution:
Given: ABCD is a cyclic quadrilateral and PRQS is a quadrilateral formed by the angle bisectors of angle ∠A, ∠B, ∠C and ∠D.
To Prove: PRQS is a cyclic quadrilateral.
Proof: In ∆APD,

∠1 +∠2 + ∠P= 180° ,…(i)
Similarly, in ∆BQC.
∠3 + ∠4 + ∠Q= 180° ….(ii)
Adding (i) and (ii), we get:
∠1 + ∠2 + ∠P + ∠3 + ∠4 ∠Q = 180° + 180° = 360°
⇒ ∠1 + ∠2 + ∠3 + ∠4 + ∠P + ∠Q = 360° ..(iii)
But ∠1+∠2+∠3+∠4 = $$\frac { 1 }{ 2 }$$ (∠A+∠B+∠C+∠D)
= $$\frac { 1 }{ 2 }$$ x 360°= 180°
∴ ∠P + ∠Q = 360° – 180° = 180° [From (iii)]
But these are the sum of opposite angles of quadrilateral PRQS

Question 9.
In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate:
(I) ∠BDC
(ii) ∠BEC
(iii) ∠BAC (2014)

Solution:
∠DBC = 58°
BD is diameter
∴ ∠DCB=90° (Angle in semi circle)

(i) In ∆BDC
∠BDC + ∠DCB + ∠CBD = 180°
∠BDC = 180°- 90° – 58° = 32°
(ii) ∠BEC =180°-32°
= 148°
(iii) ∠BAC = ∠BDC = 32°
(Angles in same segment)

Question 10.
D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Prove that the points B, C, E and D are concyclic.
Solution:

Given: In ∆ABC, AB = AC and D and E are points on AB and AC such that AD = AE, DE is joined.
To Prove: B,C,E,D. are concyclic.
Proof: In ∆ABC, AB = AC
∴ ∠B = ∠C (Angles opposite to equal sides)
In ∆ABC,
∵ $$\frac { AP }{ AB }$$ = $$\frac { AE }{ AC }$$
∴ DE || BC.
∴ ∠ADE = ∠B (Corresponding angles)
But ∠B = ∠C (Proved)
∴ Ext. ∠ADE = its interior opposite ∠C.
∴ BCED is a cyclic quadrilateral.
Hence B,C, E and D are concyclic.

Question 11.
In the given figure, ABCD is a cyclic quadrilateral. AF is drawn parallel to CB and DA is produced to point E. If ∠ADC = 92°, ∠FAE’= 20°; determine ∠BCD. Give reason in support of your answer.

Solution:

AF || CB and DA is produced to E such that
∠ADC = 92° and ∠FAE – 20°
Now, we have to find the measure of ∠BCD In cyclic quad. ABCD,
∠B – ∠D = 180° ⇒ ∠B + 92° = 180″
⇒∠B = 180°-92° = 88°
∵ AF || CB.
∴∠FAB = ∠B = 88°
But ∠FAE – 20° (Given)
Ext. ∠BAE – ∠BAF + ∠FAE
= 88° + 20° = 108°
But Ext. ∠BAE – ∠BCD
∴ ∠BCD = 108°

Question 12.
If I is the incentre of triangle ABC and AI when produced meets the circumcircle of triangle ABC in point D. If ∠BAC = 66° and ∠ABC – 80°, calculate
(i) ∠DBC,
(ii) ∠IBC,
(iii) ∠BIC.

Solution:

Join DB and DC, IB and IC.
∠BAC = 66°, ∠ABC = 80°. I is the incentre of the ∆ABC.
(i) ∵ ∠DBC and ∠DAC are in the same segment
∴ ∠DBC – ∠DAC.
But ∠DAC = $$\frac { 1 }{ 2 }$$∠BAC = $$\frac { 1 }{ 2 }$$ x 66° = 33°
∴ ∠DBC = 33°.
(ii) ∵ I is the incentre of ∆ABC.
∴ IB bisect ∠ABC
∴ ∠ IBC = $$\frac { 1 }{ 2 }$$ ∠ABC = $$\frac { 1 }{ 2 }$$ x 80° = 40°.
(iii) ∴∠BAC = 66° ∠ABC = 80°
∴ In ∆ABC,
∠ACB = 180° – (∠ABC + ∠CAB)
= 180°-(80°+ 66°)= 180°- 156° = 34°
∵ IC bisects the ∠C
∴ ∠ ICB = $$\frac { 1 }{ 2 }$$ ∠C = $$\frac { 1 }{ 2 }$$ x 34° = 17°.
Now in ∆IBC,
∠ IBC + ∠ ICB + ∠ BIC = 180°
⇒ 40° + 17° + ∠BIC = 180°
⇒ ∠ BIC = 180° – (40° + 17°) = 180° – 57°
= 123°

Question 13.
In the given figure, AB = AD = DC= PB and ∠DBC = x°. Determine in terms of x :
(i) ∠ABD
(ii) ∠APB

Hence or otherwise prove that AP is parallel to DB.
Solution:
Given: In figure, AB = AD = DC = PB.
∠DBC = x. Join AC and BD.
To Find : the measure of ∠ABD and ∠APB.
Proof: ∠DAC=∠DBC= x(angles in the same segment)
But ∠DCA = ∠DAC (∵ AD = DC)
= x
But ∠ABD = ∠DAC (Angles in the same segment)
In ∆ABP, ext. ∠ABC = ∠BAP + ∠APB
But ∠ BAP = ∠APB (∵ AB = BP)
2 x x = ∠APB + ∠APB = 2∠APB
∴ 2∠APB = 2x
⇒ ∠APB = x
∵ ∠APB = ∠DBC = x
But these are corresponding angles
∴ AP || DB. Q.E.D.

Question 14.
In the given figure, ABC, AEQ and CEP are straight lines. Show that ∠APE and ∠CQE are supplementary.

Solution:
Given: In the figure, ABC, AEQ and CEP are straight lines.
To prove: ∠APE + ∠CQE = 180°.
Const: Join EB.
∠APE+ ∠ABE= 180° ….(i)

∠CQE +∠CBE = 180° ….(ii)
∠APE +∠ABE + ∠CQE +∠CBE = 180° + 180° = 360°
⇒ ∠APE + ∠CQE + ∠ABE + ∠CBE = 360°
But ∠ABE + ∠CBE = 180° (Linear pair)
∴ ∠APE + ∠CQE + 180° = 360°
⇒ ∠APE + ∠CQE = 360° – 180° = 180°
Hence ∠APE and ∠CQE are supplementary. Q.E.D.

Question 15.
In the given figure, AB is the diameter of the circle with centre O.

If ∠ADC = 32°, find angle BOC.
Solution:
Arc AC subtends ∠AOC at the centre and ∠ ADC at the remaining part of the circle

= 2 x 32° = 64°
∵ ∠AOC + ∠ BOC = 180° (Linear pair)
⇒ 64° + ∠ BOC = 180°
⇒ ∠ BOC=180° – 64° =116° .

Question 16.
In a cyclic-quadrilateral PQRS, angle PQR = 135°. Sides SP and RQ produced meet at point A : whereas sides PQ and SR produced meet at point B.
If ∠ A : ∠ B = 2 : 1 ; find angles A and B.
Solution:
PQRS is a cyclic-quadrilateral in which ∠ PQR =135°
Sides SP and RQ are produced to meet at A and sides PQ and SR are produced to meet at B.

Question 17.
In the following figure, AB is the diameter of a circle with centre O and CD is the chord with length equal to radius OA. If AC produced and BD produced meet at point P ; show that ∠ APB = 60°.

Solution:
Given : In the figure, AB is the diameter of the circle with centre O.
CD is the chord with length equal to the radius OA.
AC and BD are produced to meet at P.
To prove : ∠ APB = 60°
Const : Join OC and OD
Proof : ∵ CD = OC = OD (Given)
∴ ∆OCD is an equilateral triangle
∴ ∠ OCD = ∠ ODC = ∠ COD = 60°
In ∆ AOC, OA = OC (Radii of the same circle)
∴ ∠ A = ∠ 1
Similarly, in ∆ BOD,
OB = OD
∴∠2= ∠B
∠ A CD + ∠B = 180°

∠60°+ ∠ 1 + ∠B = 180°
= ∠ 1 + ∠B = 180° – 60°
⇒∠ 1 + ∠B = 120°
But ∠ 1 = ∠ A
∴ ∠ A + ∠B = 120° …(i)
Now, in ∆ APB,
∠ P + ∠ A + ∠ B = 180° (Sum of angles of a triangle)
⇒ ∠P+120°=180° [From (i)]
⇒ ∠P = 180°- 120°= 60°
Hence ∠ P = 60° or ∠ APB = 60° Hence proved.

Question 18.
In the following figure,

ABCD is a cyclic quadrilateral in which AD is parallel to BC.
If the bisector of angle A meets BC at point E and the given circle at point F, prove that :
(i) EF = FC
(ii) BF = DF
Solution:

Given : ABCD is a cyclic quadrilateral in which AD || BC.
Bisector of ∠ A meets BC at E and the given circle at F. DF and BF are joined.
To prove :
(i) EF = FC
(ii) BF = DF
Proof : ∵ ABCD is a cyclic -quadrilateral and AD || BC
∵ AF is the bisector of ∠ A
∴ ∠ BAF = ∠ DAF
∴ Arc BF = Arc DF (equal arcs subtends equal angles)
⇒ BF = DF(equal arcs have equal chords)
Hence proved

Question 19.
ABCD is a cyclic quadrilateral. Sides AB and DC produced meet at point E ; whereas sides BC and AD produced meet at point F. If ∠ DCF : ∠ F : ∠ E = 3 : 5 : 4, find the angles of the cyclic quadrilateral ABCD.
Solution:
Given : In a circle, ABCD is a cyclic quadrilateral AB and DC are produce to meet at E and BC and AD are produced to meet at F.

Question 20.
The following figure shows a circle with PR as its diameter.

If PQ = 7 cm, QR = 3 cm, RS = 6 cm. find the perimeter of the cyclic quadrilateral PQRS. (1992)
Solution:
In the figure, PQRS is a cyclic quadrilateral in which PR is a diameter
PQ = 7 cm,
QR = 3 RS = 6cm
∴ 3 RS = 6cm
and RS = $$\frac { 6 }{ 3 }$$ = 2cm
Now in ∆ PQR,
∠ Q = 90° (Angle in a semi-circle)
∴ PR2 = PQ2 + QR2 (Pythagoras theorem)
= (7)2 + (6)2 = 49 + 36 = 85
Again, in right ∆ PSQ, PR2 = PS2 + RS2
⇒ 85 = PS2 + (2)2
⇒ 85 = PS2 + 4
⇒ PS2 = 85 – 4 = 81 = (9)2
∴ PS = 9cm
Now, perimeter of quad. PQRS = PQ + QR + RS + SP = (7 + 9 + 2 + 6) cm = 24cm

Question 21.
In the following figure, AB is the diameter of a circle with centre O.

If chord AC = chord AD, prove that :
(i)arc BC = arc DB
(ii) AB is bisector of ∠ CAD. Further, if the length of arc AC is twice the length of arc BC, find :
(a) ∠ BAC
(b) ∠ ABC
Solution:
Given : In a circle with centre O, AB is the diameter and AC and AD are two chords such that AC = AD.
To prove : (i) arc BC = arc DB
(ii) AB is the bisector of ∠CAD
(iii) If arc AC = 2 arc BC, then find
(a) ∠BAC (b) ∠ABC

Construction: Join BC and BD.
Proof : In right angled A ABC and A ABD
Hyp. AB = AB (Common)
∴ ∆ ABC ≅ ∆ ABD (R.H.S. axiom)
(i) ∴ BC = BD (C.P.C.T)
∴ Arc BC = Arc BD (equal chords have equal arcs)
(ii) ∠ BAC = ∠ BAD (C.P.C.T)
∴ AB is the bisector of ∠CAD.
(iii) If arc AC = 2 arc B
Then ∠ABC = 2 ∠BAC
But ∠ABC – 2 ∠BAC = 90°
∴ 2 ∠BAC + ∠BAC = 90°
⇒ 3 ∠BAC = 90° ⇒ ∠BAC = 30°
and ∠ABC = 2 ∠BAC = 2 * 30° = 60°

Question 22.
∠ BAC = 30° and ∠ CBD = 70°, find:
(i) ∠ BCD
(ii) ∠ BCA
(iii) ∠ ABC
Solution:
∠ BAC = 30°, ∠ CBD = 70°
∠ DAC = ∠ CBD , (Angles in the same segment)

But ∠ CBD = 70°
∴ ∠ DAC = 70°
⇒ ∠ BAD = ∠ BAC + ∠ DAC = 30° + 70° = 100°
But ∠ BAD + ∠ BCD = 180°
(Sum of opposite angles of a cyclic quad.)
⇒100°+ ∠ BCD=180° ⇒ ∠ BCD=180° – 100° = 80°
∴ ∠ BCD = 80°
∴ ∠ ACD = ∠ BDC
(Equal chords subtends equal angles)
But ∠ ACB = ∠ ADB
(Angles in the same segment)
∴ ∠ ACD + ∠ ACB = ∠ BDC + ∠ ADB
⇒ ∠ BCD = ∠ ADC = 80° (∵ ∠ BCD = 80°)
But in ∆ BCD,
∠ CBD + ∠ BCD + ∠ BDC = 180° (Angles of a triangle)
⇒ 70° + 80° + ∠ BDC = ∠ 180°
⇒ 150°+ ∠ BDC = 180°
∴ ∠ BDC = 180° – 150° = 30°
⇒ ∠ ACD = 30° (∵ ∠ ACD = ∠ BDC)
∴ ∠ BCA = ∠ BCD – ∠ ACD = 80° – 30° = 50°
∠ ADC + ∠ABC = 180°
(Sum of opp. angles of a cyclic quadrilateral)
⇒ 80°+ABC = 180°
⇒ ∠ ABC=180°-80° = 100°

Question 23.
In the given figure, if ∠ ACE = 43° and ∠CAF = 62°. Find the values of a, b and c.

Solution:
Now, ∠ ACE = 43° and ∠ CAF = 62° (given)
In ∆ AEC,

∠ ACE + ∠ CAE + ∠ AEC = 180°
∴ 43° + 62° + ∠ AEC = 180°
105° + ∠ AEC = 180°
⇒ ∠ AEC = 180°- 105° = 75°
Now, ∠ ABD + ∠AED=180°
(Opposite ∠ s of a cyclic quad, and ∠ AED = ∠ AEC)
⇒ 0 + 75°= 180° a = 180° – 75° = 105°
∠ EDF = ∠ BAE (Angles in the alternate segments)
∴ c = 62°
In ∆BAF, ∠a + 62° + ∠b = 180°
⇒ 105°+ 62°+ ∠b= 180°
⇒ 167° + ∠6 = 180°
⇒ ∠b= 180°-167°= 13°
Hence, a= 105°, 6=13° and c = 62°.

Question 24.
In the given figure, AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25° .

Find:
(ii)∠CBD
Solution:

In the given figure,
ABCD is a cyclic quad, in which AB || DC
∴ ABCD is an isosceles trapezium AD = BC
(i) Join BD
{ Ext. angle of a cyclic quad, is equal to interior opposite angle}
∴ ∠BAD = 80° (∵ ∠BCE = 80°)
But ∠BAC = 25°
∴ ∠CAD = ∠BAD – ∠BAC = 80° – 25° = 55°
(ii) ∠CBD = ∠CAD (Angles in the same segment)
= 55°
(iii) ∠ADC = ∠BCD (Angles of the isosceles trapezium)
= 180°- ∠BCE =180°- 80° = 100°

Question 25.
ABCD is a cyclic quadrilateral of a circle with centre O such that AB is a diameter of this circle and the length of the chord CD is equal to the radius of the circle. If AJD and BC produced meet at P, show that ∠APB = 60°.
Solution:

Given : In a circle, ABCD is a cyclic quadrilateral in which AB is the diameter and chord CD is equal to the radius of the circle
To prove: ∠APB = 60°
Construction : Join OC and OD
Proof: ∵ chord CD = CO = DO (radii of the circle)
∴ ∆DOC is an equilateral triangle
∠DOC = ∠ODC = ∠OCD – 60°
Let ∠A = x and ∠B = y
∵ OA = OD = OC = OB (radii of the same circle)
∴ ∠ODA = ∠OAD = x and ∠OCB = ∠OBC =y
∴∠AOD = 180° – 2x and ∠BOC = 180° – 2y
But AOB is a straight line
∴ ∠AOD + ∠BOC + ∠COD = 180°
180°- 2x + 180° -2y + 60° = 180°
⇒ 2x + 2y = 240°
⇒ x + y = 120°
But ∠A + ∠B + ∠P = 180° (Angles of a triangle)
⇒ 120° + ∠P = 180°
⇒ ∠P = 180° – 120° = 60°
Hence ∠APB = 60°

Question 26.
In the figure, given alongside, CP bisects angle ACB. Show that DP bisects angle ADB.

Solution:
Given : In the figure,
CP is the bisector of ∠ACB
To prove : DP is the bisector of ∠ADB
Proof: ∵ CP is the bisector of
∴ ∠ACB ∠ACP = ∠BCP
But ∠ACP = ∠ADP {Angles in the same segment of the circle}
and ∠BCP = ∠BDP
But ∠ACP = ∠BCP
∴ DP is the bisector of ∠ADB

Question 27.
In the figure, given below, AD = BC, ∠BAC = 30° and ∠CBD = 70°. Find :
(i) ∠BCD
(ii) ∠BCA
(iii) ∠ABC

Solution:
In the figure,
AC and BD are its diagonals
∠BAC = 30° and ∠CBD = 70°
Now we have to find the measures of ∠BCD, ∠BCA, ∠ABC and ∠ADB
(Angles in the same segment)
Similarly ∠BAC = ∠BDC = 30°
∴ ∠BAD = ∠BAC + ∠CAD = 30° + 70° = 100°
(i) Now ∠BCD + ∠BAD = 180° (opposite angles of cyclic quad.)
⇒ ∠BCD + 100°= 180°
⇒ ∠BCD = 180°-100° = 80°
(ii) ∵ AD = BC (given)
∴ ∠ABCD is an isosceles trapezium
and AB || DC
∴ ∠BAC = ∠DCA (alternate angles)
⇒ ∠DCA = 30°
∠ABD = ∠D AC = 30° (Angles in the same segment)
∴ ∠BCA = ∠BCD – ∠DAC = 80° – 30° = 50°
(iii) ∠ABC = ∠ABD + ∠CBD = 30° + 70° = 100°
(iv) ∠ADB = ∠BCA = 50° (Angles in the same segment)

P.Q.
In the given below figure AB and CD are parallel chords and O is the centre.
If the radius of the circle is 15 cm, find the distance MN between the two chords of length 24 cm and 18 cm respectively.

Solution:
Given : AB = 24 cm, CD = 18 cm
⇒AM = 12 cm, CN = 9 cm
Also, OA = OC = 15 cm
Let MO = y cm, and ON = x cm
In right angled ∆AMO

(OA)2 = (AM)2 + (OM)2
(15)2 = (12)2 + (y2
⇒ (15)2-(12)2
⇒ y2 = 225-144
⇒ y2 = 81 = 9 cm
In right angled ∆CON
(OC)2 = (ON)2 + (CN)2
⇒(15 )2= x2 + (9)2
⇒ x2 = 225-81
⇒x2= 144
⇒ x = 12 cm
Now, MN = MO + ON =y + x = 9 cm + 12 cm = 21 cm

Question 28.
In the figure given below, AD is a diameter. O is the centre of the circle. AD is parallel to BC and ∠CBD = 32°. Find :
(i) ∠OBD
(ii) ∠AOB
(iii) ∠BED (2016)
Solution:

(i) AD is parallel to BC, that is, OD is parallel to BC and BD is transversal.
∴ ∠ODB = ∠CBD = 32° (Alternate angles)
In ∆OBD,
OD = OB (Radii of the same circle)
⇒ ∠ODB = ∠OBD = 32°
(ii) AD is parallel to BC, that is, AO is parallel to BC and OB is transversal.
∴∠AOB = ∠OBC (Alternate angles)
∠OBC = ∠OBD + ∠DBC
⇒ ∠OBC = 32° + 32°
⇒ ∠OBC = 64°
∴ ∠AOB = 64°
(iii) In AOAB,
OA = OB(Radii of the same circle)
∴ ∠OAB = ∠OBA = x (say)
∠OAB + ∠OBA + ∠AOB = 180°
⇒ x + x + 64° = 180°
⇒ 2x = 180° – 64°
⇒ 2x= 116°
⇒ x = 58°
∴ ∠OAB = 58°
That is ∠DAB = 58°
∴ ∠DAB = ∠BED = 58°
(Angles inscribed in the same arc are equal)

Question 29.
In the given figure PQRS is a cyclic quadrilateral PQ and SR produced meet at T.

(i) Prove ∆TPS ~ ∆TRQ
(ii) Find SP if TP = 18 cm, RQ = 4 cm and TR = 6 cm.
(iii) Find area of quadrilateral PQRS if area of ∆PTS = 27 cm2.(2016)
Solution:
(i) Since PQRS is a cyclic quadrilateral ∠RSP + ∠RQP = 180°
(Since sum of the opposite angles of a cyclic quadrilateral is 180°)
⇒ ∠RQP = 180° – ∠RSP …(i)
∠RQT + ∠RQP = 180°
(Since angles from a linear pair)
⇒ ∠RQP = 180° – ∠RQT …(ii)
From (i) and (ii),
180° – ∠RSP = 180° – ∠RQT
⇒ ∠RSP = ∠RQT …(iii)
In ∆TPS and ∆TRQ,
∠PTS = ∠RTQ (common angle)
∠RSP = ∠RQT [From (iii)]
∴ ATPS ~ ATRQ (AA similarity criterion)
(ii) Since ∆TPS ~ ∆TRQ implies that corresponding sides are proportional that

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.


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# Articles from the Institute Letter

Additional articles from new and past issues of the Institute Letter will continue to be posted over time and as they become available.

## Economic, Political, and Social Outcomes of Community-Driven Development

### by Ruben Enikolopov

How has Afghanistan's largest development program affected democratic processes, counterinsurgency, and the position of women?

Each year, billions of dollars in foreign aid are directed to the developing world. Assistance comes in a variety of forms, but one particular method of delivery—community-driven development (CDD)—which came about as a response to large-scale top-down initiatives that were criticized for failing to empower aid recipients, has become especially popular. This approach emphasizes involvement of local communities in planning decisions and controlling investment of resources. Beyond benefiting communities with their involvement in planning decisions and the investment of resources, CDD is intended to encourage sustained participation through local representative institutions, thus improving social capital and local governance.

The CDD approach is particularly popular in the context of weak or fragile states, in which government bureaucracy often fails to provide public goods and services. From 1996 to 2003, World Bank lending alone for such projects rose from $325 million to$2 billion per year, reaching $30 billion in total as of 2012, toward the support of four hundred programs with CDD components in ninety-four countries. Yet, rigorous empirical evidence of CDD value remains limited. My work with Andrew Beath from the World Bank and Fotini Christia from the Massachusetts Institute of Technology contributes to understanding the effectiveness of this approach by assessing the impact of a large-scale CDD program in Afghanistan known as the National Solidarity Program (NSP). NSP is the country’s largest development program, implemented in over thirty-two thousand out of Afghanistan’s thirty-nine thousand villages at a cost of over$1 billion. Funded by a consortium of international donors and administered by the Afghan national government, its aim is to both improve the access of rural villagers to critical services and to create a structure for village governance centered on democratic processes and female participation.

## Painting Time: Impressionism and the Modern Temporal Order

### By André Dombrowski

How quickening brushwork arose from the industrialization of time

It is often said that impressionism sought to make represented time and the time of representation coterminous. With its seemingly quick and unpolished touch, it gave the modern cultures of speed their first appropriately modernist forms. But art historians have rarely if ever interrogated the concrete histories and technologies of time (and time keeping) that underwrote this seismic stylistic shift or to inquire into the links between quickening brushwork and the nineteenth century’s industrialization of time. This is especially remarkable given the fact that two key scientific events in the measuring of modern time—the advent of quantifiable nervous reaction time in France around 1865 and the standardization of universal time in 1884—overlap so precisely with the history of impressionism, its rise in the mid-1860s, and the turn toward postimpressionism around the mid-1880s.

## Understanding Networks and Defining Freedom

From capturing interactions and inferring the structure of data to determining the infringement of freedom

In November, the Association of Members of the Institute for Advanced Study (AMIAS) sponsored two lectures by Jennifer Chayes, Member (1994–95, 97) in the School of Mathematics, and Quentin Skinner, Member in the Schools of Historical Studies (1974–75) and Social Science (1976–79). All current and former Institute Members and Visitors are members of AMIAS, which includes some 6,000 scholars in more than fifty countries. To learn more about the organization, upcoming events, and opportunities to support the mission of the Institute, please visit www.ias.edu/people/amias/. Following are brief summaries of the lectures by Chayes and Skinner; full videos are available at http://video.ias.edu/2013-amias-chayes and http://video.ias.edu/2013-amias-skinner/.

AGE OF NETWORKS

Jennifer Chayes, Distinguished Scientist and Managing Director of Microsoft Research New England and New York City:

Everywhere we turn, networks can be used to describe relevant interactions. In the high-tech world, we see the internet, the world wide web, mobile phone networks, and online social networks. In economics, we are increasingly experiencing both the positive and negative effects of a globally networked economy. In epidemiology, we find disease spreading over ever-growing social networks, complicated by mutation of the disease agents. In problems of world health, distribution of limited resources, such as water resources, quickly becomes a problem of finding the optimal network for resource allocation. In biomedical research, we are beginning to understand the structure of gene-regulatory networks, with the prospect of using this knowledge to manage many human diseases.

## How Open-Source Ideas Can Help Us Study Exoplanets

### By Hanno Rein

Pluto, the ninth planet in our solar system1 was discovered in 1930, the same year the Institute was founded. While the Institute hosted more than five thousand members in the following sixty-five years, not a single new planet was discovered during the same time.

Finally, in 1995, astronomers spotted an object they called 51 Pegasi b. It was the first discovery of a planet in over half a century. Not only that, it was also the first planet around a Sun-like star outside our own solar system. We now call these planets extrasolar planets, or in short, exoplanets.

As it turns out, 51 Pegasi b is a pretty weird object. It is almost as massive as Jupiter, but it orbits its host star in only four days. Jupiter, as a comparison, needs twelve years to go around the Sun once. Because 51 Pegasi b is very close to the star, its equilibrium temperature is very high. These types of planets are often referred to as “hot Jupiters.”

Since the first exoplanet was discovered, the technology has improved dramatically, and worldwide efforts by astronomers to detect exoplanets now yield a large number of planet detections each year. In 2011, 189 planets were discovered, approximately the number of visiting Members at the Institute every year. In 2012, 130 new planets were found. As of May 20 of this year, the total number of confirmed exoplanets was 892 in 691 different planetary systems.

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Distance away from earth to see it as a full disk [duplicate]

This question is more space-related than physics-related, but here goes...

How far away the earth would I have to be in order to see the earth as a full disk? What I'm looking for is a distance in kilometers or miles. For example, when I fly in an airplane at 40,000 feet (about 12000 m), I can begin to see the curvature of the earth, but the view I have of the earth is a tiny piece of the total. Also, I know if I'm on the moon I can see the earth as a full disk. But what is the minimum distance away the earth I'd have to be to see the full disk?

The main reason I ask this question is because I am interested to know how much of the earth you can see from the International Space Station (ISS). I've seen various photo collections that supposedly show views of the earth from the ISS and some indicate that you can see the earth as a full disk. However, I am skeptical that you can see the earth as a full disk from the ISS. Below I've done some calculations to try and determine how far away from the earth you would have to be to see the earth as a full disk.

The human visual field of view is approximately 120 degrees in both the horizontal and vertical directions. If we construct a right triangle where one angle is 60 degrees (half of 120 degrees), "d" is the distance to the earth, and "r" is the radius of the earth, then d = r/tan(60) = 6371 km/1.732 = 3678 km = 2285 miles. This says that you would have to be 3678 km (2285 miles) away from the earth to see it as a full disk. Since the ISS is orbiting at an altitude of 347 km (216 miles) perigee and 360 km (224 miles) apogee (the mean is about 353 km (219 miles)), I believe that you will not be able to see the earth as a full disk from the space station.

Is this analysis correct? If not, what is the correct analysis?

Mike

-
Duplicate of this one from 2012: physics.stackexchange.com/q/25509/2498 –  Rory Alsop May 12 at 11:03

marked as duplicate by Manishearth♦May 12 at 11:31

How far away from earth's surface do we need to go to see its full hemispherical area? The strict mathematical answer is: no matter how far you go away from earth, you will never see its full hemisphere.

So a better question is the one that generalizes your remark "I am interested to know how much of the earth you can see from the International Space Station (ISS)":

At height $h$ above earth's surface, what fraction of earth's hemispherical area will remain invisible?

Some straightforward math (let me know via a comment in case more detail is needed) reveals an elegant answer to this question. The fraction of the hemispherical area missing equates to $\frac{r}{r+h}$, where $r$ denotes earth's radius, and $h$ observer's height above earth's surface.

So, when you observe earth ($r$ = 6.3 megameter) from ISS ($h$ = 0.35 megameter), some 95% of earth's hemispherical area will be missing from your view. Even when observing from the moon ($h$ = 360 megameter), about 1.7% of earth's hemispherical area will remain hidden from your view.

-

It seems to me that you have calculated the distance to the horizon rather than the altitude.

The altitude would be $\frac{r}{\sin(60^\circ)} - r = r \times 0.1547 = 986\text{ km}$.

Update:

From within spacecraft the windows would restrict field of view, but in 1966 on gemini-11 mission Richard Gordon did a spacewalk at altitude of 1369km where the Earth would be 110 degrees across. The field of view of his helmet may have been sufficient to include the full width but probably not the full height of the disc. Astronauts on Apollo 8 photographed Earthrise from first lunar orbit in 1968. The Earth is 2 degrees across at that distance. In theory they could have seen Earth disk on outward journey through a window if it pointed in right direction, but 3 of 5 windows were fogged and they did not see it until they came round from far side of moon. Picture is from Gemini-11.

-
Which indeed proves that you would need very-wide-angle eyes/camera to see the full disc from ISS. I believe the first sight of earth like that, was from moon orbit: independent.co.uk/news/science/… –  Kris Van Bael May 12 at 7:46
I think Mike is right that field of view is about 120 degrees including peripheral vision. From 1000km outwards it will look disk-like. From moon it is 2 degrees across. Highest shuttle orbit was about 620km –  Philip Gibbs May 12 at 8:18
Gemini-11 reached 1369km altitude in 1966 with 2 crew. Astronaut Richard Gordon did a space-walk at that altitude. Windows on these craft would reduce angle of view to much less than 120 degrees. Even a spacesuit helmet would reduce field of view during space-walks so not sure if Gordon could get the full disk effect. I will add info to answer. –  Philip Gibbs May 12 at 8:34


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# tricky error with updating an external data source, and 'data - sort - copy output to'.

hello all, :-)

i have built me a sheet to download and evaluate the corona data from JHU,

one special thing goes wrong:

if i update the data in the sheet (by allowing 'update' after opening the file or with 'edit - links to external files - update') - note the red marked cells JF1:JG2 before, the references run 'smoothly through' to column JZ and more columns are downloaded for the sheet 'cases_raw', then the columns on the sheet 'cases_sort' change from JG on, and the formulas reference other cells, e.g. 'cases_sort.JG1' changes from '=cases_raw.JG1' to '=cases_raw.JK1', this is not intended and messes up all downstream work ...

the two sheets are connected in two different ways, in row 1 and 2 of 'cases_sort' there are hard coded references to cells in the first two rows of 'cases_raw', below that there is an area which was filled with data by selecting 'data_1' on sheet cases_raw and then doing 'data - sort - criteria col_b ascending, col_a ascending - options - copy output to cases_sort.A3',

somehow the new download now has an effect on 'cases_sort' that i did not intend and do not understand ... may this be due to some cryptic options? or is it a bug?

tia fah 'thanks in advance' - 'for any help'

reg.

b.

edit retag close merge delete

Sort by » oldest newest most voted

My assumption is that at some point in time 5 columns were inserted on sheet cases_raw before column JG so the references on sheet cases_sort were adjusted. If you want the references to always point to some fixed but relative position without being adjusted upon insert/delete/move then use a named range expression, i.e. on cell cases_sort.A1 define a name headercell as $cases_raw.A1 (note relative column and row reference pointing to the same position on the other sheet) and use the name as cell formula. You can simply select cells A1:JZ2 and type in =headercell and close with Alt+Enter, which writes the input into all selected cells. more ## Comments @erAck: :-) it works, 1. don't expect me to understand how and why, 2. 'at some point in time 5 columns were inserted on sheet cases_raw before column JG' - you mean by the download process? that could match as - perhaps - the download routine tries to not to damage work done beside the old download area, and as a precaution inserts new space, then updating or not the references is wrong in the one or other case (they are used to access space beside the download or prepared to be used with later download ... can't say if i'd call it a bug, 3. how the heck should a 'normal user' get the idea to make special use of named cells, up to now i only understood assigning a name to a range ... mysteries over mysteries ... thanks for your help and profound knowledge ... ( 2020-10-08 20:45:04 +0100 )edit '1. don't expect me to understand how and why Relative cell references in named expressions are not and can not be adjusted because they are expected to always be relative to the cell where they are used; it's a property of named expressions. '2. [...] As I don't know what you did there I can't tell. If columns were inserted then adjusting cell references is correct and not a bug. '3. how the heck should a 'normal user' get the idea By gaining expertise. Spreadsheets are not easy, only everyone is assuming that (and hence failing). ( 2020-10-08 21:45:23 +0100 )edit Hi and thanks to newbie-02 for having redirecting me to his thread. I have the same issue, columns changing row by theirselves after refreshing the htlm external link sheets. I tried erAck suggestion but didn't worked for me, some columns letters have still changed row from the inital lettter i gave them. Don't know how to solve this issue as it's really annoying.. Note : When define a name for cells i can't add any character like$, it says it's not allowed. So i added a name without the absolute symbol, could this have an effect ?

( 2020-10-16 16:24:31 +0100 )edit

with @erAck's solution you shouldn't have column-letters or row-numbers anymore, but a 'named range expression' - in his sample 'headercell' ...
it was ... not easy to understand ... i just followed his script and it worked ...
either try that, preferably first for another sheet, e.g. the example from this thread, and then transfer it to your task,
or provide your sheet, preferably remove private data first, perhaps someone will fix it for you ... or get's a better understanding of your problem ...

( 2020-10-16 16:35:09 +0100 )edit

Hello, yes i did like erAck suggest, for example i have =NB(Sheet2_D) for second sheet column D90:D150.

I saw the change when ctrl F3, you can see all the named cells and their assigned range and from there i notice some have changed again.

( 2020-10-16 16:43:57 +0100 )edit

What (defined to what) named range changed how, and when doing what?

( 2020-10-16 19:57:32 +0100 )edit

Hello erAck, to explain it a little better i have many sheets with html external data. Those sheet contains only html data, no cells with formula or whatever and they all have various refresh time set.I have one main sheet that contains all the formulas and collect all the data from the html data sheets.

So for example in one of my html data sheet ( the 2nd sheet ) and for the range $Sheet2.$D$90:$D$150 i defined a name Sheet2_D In my main sheet i then have a formula like this : =COUNT(Sheet2_D) Now the data i import contain some tables, sometimes the site doesn't have the table available and the cells are empty and sometimes later the tables are available again and data reappear on the cells if refreshed. What i noticed is when it's refreshing and tables aren't available anymore it's ...(plus) ( 2020-10-16 20:27:06 +0100 )edit All reference points in $Sheet2.$D$90:$D$150 are absolute references by having a $ prefix. Absolute references are adjusted upon move/insert/delete in all formula expressions. In a named expression use relative references by not specifying the$ in that dimension, but note that defining (and displaying) the expression in the Name Manager is always relative to the current cell cursor position where the Name Manager was invoked.

For a relative reference define $Sheet2.D90:D150 for both column-relative and row-relative, so when used in a name in a formula expression in the same cell where the Name Manager was invoked will always address the range D90:D150 on Sheet2. Using it in other cells will shift accordingly. If the rows should be adjusted when inserting rows then define absolute row references but keep the column references relative, so $Sheet2.D$90:D$150. That name then can be ...(more)

( 2020-10-16 20:55:53 +0100 )edit

Thanks for the clarifications erAck, i'll try without the absolute symbol and report.

( 2020-10-17 13:54:23 +0100 )edit


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# Importance of integrity of cell-cell junctions for the mechanics of confluent MDCK II cells

## Abstract

Intercellular junctions are important mechanical couplers between cells in epithelial layers providing adhesion and intercellular communication. Regulation of the junctions occurs in cellular processes such as layer formation, epithelial-to-mesenchymal transition, embryogenesis, and cancer progression. Many studies addressed the role of force generation in cells for establishing lateral cell-cell junctions and the role of cellular force transmission in tissue formation and maintenance. Our atomic force microscopy- (AFM) based study shed light on the role of both, tight junctions and adherens junctions for the mechanical properties of individual epithelial cells that are part of a confluent monolayer. We found that tight junctions are important for the establishment of a functional barrier-forming layer but impairing them does not reduce the mechanical integrity of cells. Depletion of ZO-1 results in a weak increase in cortical tension. An opposite effect was observed for disruption of E-cadherin-mediated adherens junctions using DTT. Opening of adherens junctions leads to substantial alterations of cellular mechanics such as reduced overall stiffness, but these changes turned out to be reversible after re-establishing disulfide bridges in E-cadherin by removal of DTT. We found that regulatory mechanisms exist that preserve mechanical integrity during recovery of disrupted adherens junctions.

## Introduction

Epithelial cells form a dense, stable cell layer lining the outer surface of tissue and organs. Mechanical strength and communication between the cells within a layer is provided by different cell adhesion sites including tight junctions (TJs)1, adherens junctions (AJs)2, desmosomes3, and gap junctions4. Establishment of these intercellular junctions divides polarised cells in apical and basolateral divisions. Owing to their motility and dynamic conditions epithelia are capable of monitoring development5, tissue healing6, and cancer invasion7. To fulfil their role in force transmission between cells lateral cell-cell junctions are connected to the actin cytoskeleton.

Tight junctions are the uppermost cell-cell connection at the lateral cell membrane of polarised epithelia. They form a physical barrier to control the lateral flux of ions, macromolecules, pathogens, and other solutes within the paracellular pathway8. In addition, tight junctions are responsible for the separation of apical and basolateral membrane lipids and proteins9,10. TJs consist of transmembrane proteins such as occludin, claudins, or JAMs and peripheral membrane proteins from the zonula occludens (ZO) or cingulin family8. The actin cytoskeleton is connected to this junctional complex via these zonula occludens proteins, including ZO-1, ZO-2, and ZO-3. As a consequence, loss of ZO proteins influences the actomyosin cortex structure at the tight junctions, for instance, ZO-1/-2 double knock down led to accumulating actin structures as well as an enhanced myosin IIB level at the adherens junctions11. Actin becomes more clustered at the apical cell side11. It was also recently found that epithelial tension and effective viscosity are increased in ZO-1/-2 lacking cells12.

While the molecular details of cell-cell junctions are well-known, their role for cellular mechanics on the mesoscale is largely unexplored. Trepat and co-workers recently demonstrated that the formation of epithelial cracks during cell sheet dilatation is caused by tissue stretching independent of epithelial tension resulting from a transient pressure build-up in the substrate during stretching and compression17. Essentially, the authors found that the cracks do not originate from detachment of the cortex from the membrane but are due to dissolution of adherens junctions. Even tight junctions remain mostly intact in these cracks.

Espinosa and co-workers used micropillar arrays and AFM indentation experiments on single cells and cell pairs to investigate the impact of the desmoplakin–intermediate filament linkage on cell mechanics18. They found that desmosome–intermediate filament networks are important contributors to cell stiffness in cell pairs and sheets underlining the importance of cell-cell contacts for cellular mechanics. Therefore, depending on the cell type different mechanisms might exist to regulate tension.

In the present study, we examined how epithelial cells regulate their mechanical behaviour when cell-cell junctions are impaired. We employed atomic force microscopy (AFM) which is proven to be a versatile tool not only for imaging but also to assess local mechanical properties of adherent cells19,20. Local mechanical properties are an important readout how cells respond to environmental changes21,22,23,24. We stopped tight junction formation by ZO-1 silencing and interfered with the integrity of adherens junctions by a chemical stimulus. Mechanical measurements were analysed using a minimal mechanical model based on the assumption that the restoring force to deformation originates mainly from cortical tension. Our data indicate that the impact of tight junctions is of minor importance for the mechanics of epithelial cells from a monolayer, while their morphology is changed considerably. Loss of adherens junctions, however, changes tension and area compressibility significantly, while the morphology of the cell layer is only slightly affected. Therefore, our study suggests that TJs and AJs impact cellular mechanics in different, opposite ways highlighting their role in establishing a tight but also flexible barrier tissue.

## Results

### Tight junctions are responsible for a complete, dense cell monolayer

Tight junctions form the uppermost cell-cell connections in epithelial cells. This junctional complex consists of the transmembrane proteins occludin and various claudins. On the inner cellular site, it binds to zonula occludens, which establishes the contact to the F-actin cytoskeleton8. We used short interference RNA (siRNA) against zonula occudens-1 (ZO-1) in order to examine the role of tight junctions for the mechanical integrity of epithelial cells. We chose to knock out ZO-1 since it is known to bind to nearly all relevant proteins of the tight junctional complex, especially the perijunctional actomyosin network. It was also shown by Anderson and co-workers that depletion of ZO-1 preserves the transepithelial resistance but generates appreciable changes in cell morphology and organisation of the actomyosin cortex9.

Phase contrast images of confluent MDCK II cells show a tight, densely packed cell monolayer (Fig. 1A1). Fluorescence images of ZO-1 stained cells show that tight junctions built up a gapless mesh surrounding each cell (Fig. 1A2). After applying siRNA against ZO-1 the cell layer still forms but is less uniform. Cells appear to be separated from each other and cell borders are brighter in phase contrast images (Fig. 1B1). ZO-1 is blocked with high efficiency by siRNA as nearly all cells lose their fluorescence signal (Fig. 1B2). This spatial separation between cells is further confirmed by AFM-imaging of ZO-1 deficient cells (Fig. 2). A continuous connection appears in AFM images for a fully established cell monolayer of untreated cells (Fig. 2A–E). In contrast, for ZO-1 deficient cells we found that adjacent cells are no longer connected to each other and only a few finger-like connections are visible (Fig. 2F–J).

Since ZO-1 connects the tight junctional complex with the actin cytoskeleton, the actin distribution of cells after ZO-1 depletion was also analysed by fluorescence microscopy (Fig. 3). On the apical side small point-like structures are found for untreated cells (Fig. 3B1), reflecting the actin-filled microvilli of polarized MDCK II cells25. At the basal side stress fibres are visible (Fig. 3D1). For successfully ZO-1 silenced cells the actin distribution is more clustered on the apical side (Fig. 3B2), while on the basal side stress fibres disappear (Fig. 3D2). At the central focal plane actin fences untreated MDCK II cells, whereas a distinct cell-cell separation becomes visible for ZO-1 deficient cells and the actin signal appears blurry (Fig. 3C).

Mechanical properties are inferred from AFM-based indentation experiments that provide information about cortical strength, membrane tension and excess membrane area. We indented siRNA treated cells with an AFM tip in order to analyse the impact of ZO-1 on these properties. Interestingly, although the cells are visibly separated from each other, the mechanical parameters are only slightly altered by ZO-1 depletion (Fig. 4). The biggest effect is an increased cortical strength after weakening of tight junctions. The average overall tension of untreated cells is found to be $${T}_{0}=(0.275\pm 0.010)\,\text{mN}/{\rm{m}}$$ $$({\rm{median}}\pm {\rm{SEM}})$$. Cells with compromised tight junctions show a significantly increased overall tension of $${T}_{0}=(0.334\pm 0.009)\,\mathrm{mN}/{\rm{m}}$$ (Fig. 4A). The apparent area compressibility modulus is also higher for cells lacking ZO-1 ($${\hat{K}}_{{\rm{A}}}=(0.228\pm 0.007)\,{\mathrm{nN}/\mu {\rm{m}}}^{{\rm{3}}}$$, control: $${\hat{K}}_{{\rm{A}}}=(0.153\pm 0.008)\,{\mathrm{nN}/\mu {\rm{m}}}^{{\rm{3}}}$$) (Fig. 4B). The membrane tension Tt, which is a measure for the strength of membrane-cytoskeleton attachment sites, is found to be $$(0.161\pm 0.004)\,\mathrm{mN}/{\rm{m}}$$ for untreated cells. Nearly the same value is found for ZO-1 lacking cells ($${T}_{{\rm{t}}}=(0.164\pm 0.004)\,\mathrm{mN}/{\rm{m}}$$) (Fig. 4C).

In order to exclude mechanical alterations due to the transfection procedure control measurements using non-targeting siRNA are shown in the supplementary information (suppl. Fig. S1)26. Applying the transfection protocol shifts mechanical parameters to slightly lower values, i.e. in the opposite direction.

In summary, ZO-1 is extremely important for the formation of a dense cell layer, which shields underlying cells. The morphology of ZO-1 lacking cells is changed tremendously but the overall stiffness of the cells increases only slightly compared to well-connected cells.

### Adherens junctions influence topography and mechanics of cells

Phase contrast imaging shows no changes in the cellular structure during the entire observation period (Fig. 5 row A). In contrast to ZO-1 silenced cells, the cell monolayer stays intact after incubation with DTT-containing medium, also tight junctions are not affected by DTT treatment (Fig. 5 row C). The distribution of actin on the apical cell side remains unchanged (Fig. 5 row D), small point-like structures remain. Interestingly, in the central focal plane (Fig. 5 row E) (at the same z-position, where E-cadherin imaging is carried out) the actin fence, found as a solid ring in control cells (Fig. 5E1), is perturbed after 3 h of DTT treatment (Fig. 5E2). 3 h after removal of the drug the cells do not restore the original actin distribution (Fig. 5E3). Only after keeping the cells free from DTT for altogether 6 h, the actin ring reshapes and appears as homogeneous as in untreated cells (Fig. 5E4). In contrast, stress fibre formation at the basal cell side is not significantly affected by disruption of the adherens junctions (Fig. 5 row F).

The intracellular E-cadherin is known to be connected to the actin cytoskeleton via catenins and we found that the actin distribution at the lateral cell side is severely perturbed by DTT treatment (Fig. 5E1, E2). Therefore, we also investigated the distribution of β-catenin (Fig. 6). Control cells display a ring of β-catenin in the same focal plane, where E-cadherin is located (Fig. 6B1). After dissolution of the E-cadherin-E-cadherin bonds the β-catenin ring is also destroyed (Fig. 6A2, B2). The recovery of the β-catenin structure occurs within the same time as the E-cadherin reformation. 3 h after DTT removal the distributions of both, E-cadherin and β-catenin are still irregular (Fig. 6A3, B3). However, after keeping the cells free from DTT for 6 h, they finally manage to restore the distribution of E-cadherin (Fig. 6A4), β-catenin (Fig. 6B4) and F-actin (Fig. 5E4).

AFM imaging confirms the presence of an intact cell monolayer during the whole experimental procedure (Fig. 7). However, a detailed inspection of the topography revealed that the morphology of the connected cells is altered by DTT incubation. Control cells display a typical height of the apical cap of $$(2.0\pm 0.1)\,\mu {\rm{m}}$$ (Fig. 7A). After exposure of the cells to 10 mm DTT for 3 h, a substantially elevated cap of $$(2.8\pm 0.2)\,\mu {\rm{m}}$$ is found (Fig. 7B). After removal of DTT and a recovery time of 3 h cells start to restore their initial topography (cap height: $$(2.5\pm 0.2)\,\mu {\rm{m}}$$, Fig. 7C). 6 h after DTT removal the height of the apical cap is completely restored to $$(2.0\pm 0.2)\,\mu {\rm{m}}$$ (Fig. 7D). Considering that the adherens junctions ‘outstretch’ the cells by pulling at adjacent cells, bulging of the cells after compromising the junctions and therefore removing the tension between them is expected and a strong hint to an altered cellular mechanics.

Next, we asked for the impact of DTT treatment and recovery from the treatment on the mechanical properties of the cells. Fundamental changes of the mechanical parameters are observed during the exposure to DTT and recovery from the drug treatment. Destroyed adherens junctions lead to a substantial drop in the overall tension from $${T}_{0}=(0.275\pm 0.010)\,\text{mN}/{\rm{m}}$$ down to $${T}_{0}=(0.164\pm 0.005)\,\mathrm{mN}/{\rm{m}}$$ after 3 h of DTT treatment. 3 h after DTT removal T0 is found to be $$(0.315\pm 0.010)\,\mathrm{mN}/{\rm{m}}$$, even higher than the initial value. Later, after a recovery time of 6 h the initial value is eventually restored ($${T}_{0}=(0.265\pm 0.013)\,\mathrm{mN}/{\rm{m}}$$) (Fig. 4A). DTT treatment also causes a halving of the apparent area compressibility modulus ($${\hat{K}}_{{\rm{A}},{\rm{c}}{\rm{o}}{\rm{n}}{\rm{t}}{\rm{r}}{\rm{o}}{\rm{l}}}=(0.153\pm 0.008)\,{{\rm{n}}{\rm{N}}/\mu {\rm{m}}}^{3}$$, $${\hat{K}}_{{\rm{A}},3{\rm{h}}{\rm{D}}{\rm{T}}{\rm{T}}}=(0.083\pm 0.006)$$ $${\mathrm{nN}/\mu {\rm{m}}}^{{\rm{3}}}$$. When the drug is removed, cells manage to restore this value within 3 h also with a slight overshoot. It is found to be $${\hat{K}}_{{\rm{A}}}=(0.194\pm 0.024)\,{\mathrm{nN}/\mu {\rm{m}}}^{{\rm{3}}}$$. After 6 h of recovery from the DTT treatment, an apparent area compressibility modulus comparable to untreated cells is detected $$({\hat{K}}_{{\rm{A}}}=(0.187\pm 0.078)\,{\mathrm{nN}/\mu {\rm{m}}}^{{\rm{3}}})$$ (Fig. 4B).

DTT treated cells show a lowered membrane tension ($${T}_{{\rm{t}}}=(0.121\pm 0.036)\,\mathrm{mN}/{\rm{m}}$$) compared to untreated cells $$({{\rm{T}}}_{{\rm{t}},{\rm{c}}{\rm{o}}{\rm{n}}{\rm{t}}{\rm{r}}{\rm{o}}{\rm{l}}}=(0.161\pm 0.003)\,{\rm{m}}{\rm{N}}/{\rm{m}})$$. When the drug is removed for 3 h, Tt is found to be $$(0.227\pm 0.010)\,\mathrm{mN}/{\rm{m}}$$. After 6 h of recovery, a homeostasis occurs. The membrane tension drops again down to $${T}_{{\rm{t}}}=(0.188\pm 0.006)\,\mathrm{mN}/{\rm{m}}$$ (Fig. 4C) close to the initial value.

In summary, we found that opening of adherens junctions by DTT leads to a considerable softening of the individual cells. Both, cortical and membrane tension drop and either the bending modulus of the composite shell is reduced or the excess membrane area is enlarged depending on the interpretation of the apparent area compressibility modulus. Interestingly, the cells entirely recover from this treatment after keeping them for 6 h in the absence of DTT. This holds true both for structure and mechanics. Cells reach their initial values by going through a state of higher stiffness as initially found for control cells (overshoot).

## Discussion

In the present study we examined how different cell-cell junctions contribute to the mechanical properties of individual epithelial cells and how the cells maintain tension homeostasis after selectively destroying the junctions.

It is clear that cell-cell junctions are important for providing efficient and selective barriers against the environment. This also includes a stable mechanical connection between adjacent cells that permits the cell layer to respond to external mechanical stress without losing its integrity. There are seminal studies addressing questions like, which forces are necessary within cells to establish strong cell-cell binding complexes29,30, how different types of junctions influence each other31,32, and how forces are transmitted within a cell layer33,34,35,36. However, it is still not entirely clear if and to what extent cell-cell junctions impact the mechanical behaviour of individual cells of a confluent monolayer.

When one of the major players of tight junctions, ZO-1, is depleted, we found that cells are visibly separated from each other. AFM imaging reveals that only a few finger-like connections between the cells remain (Fig. 2). Surprisingly, however, these significant alterations of the cell morphology have only a small effect on the mechanical behaviour of the individual cells that are still connected via adherens junctions. Slightly elevated overall tension and area compressibility moduli (Fig. 4) can be explained by an altered actin cytoskeleton (Fig. 3B2) with coarser structures on the apical side. This is in good agreement with results from Fanning et al. They also found larger aggregates of actin and accumulating myosin at the cell apex in ZO-1/-2 depleted MDCK II cells11. An enhanced myosin II level and higher actomyosin contractility may also contribute to higher tension values. Contrary to Fanning et al., however, we did not find a stronger actin belt at the lateral apical junctional complex region in response to ZO-1 depletion (Fig. 3C2). This may be due to the knock down of only ZO-1 and thereby explain the only moderately enhanced tension in our case. Compared to their work, we also found less pronounced stress fibre formation, when the cells are separated from each other (Fig. 3D2). Ikenouchi et al. demonstrated that polarisation of EpH4 epithelial cells is slowed down when ZO-1/-2 is lacking. Actin on the lateral cell side is separated between individual cells37. We found the same actin distribution in our experiments with MDCK II cells.

In our experiments cells undergo substantial morphological changes upon ZO-1 depletion (vide supra). The membrane-cytoskeleton attachment describing membrane tension Tt, however, remains largely at the same level that we found for control cells. This supports the work from Fanning et al. who found that cell polarity and ezrin distribution is unaffected in ZO-1/-2 double knockdown cells.

In contrast to ZO-1 depletion, E-cadherin interference has dramatic consequences for the mechanical properties of MDCK II cells, while the impact on the morphology of the cell monolayer is small. Capaldo and Macara reported a swelling of the apical cap in E-cadherin silenced MDCK T23 cells38. Our AFM studies confirm this result, the removal of adherens junctions results in a substantial rounding of the cells indicative of an interfacial tension release. Considering that contractile cortical tension is balanced by the presence of cell-cell contacts it is expected that cells bulge into a more roundish shape after losing the cell-cell counter force. Interestingly, Capaldo and Macara report that β-catenin remains at the cell-cell-boundaries, whereas we found that the protein spreads within the whole cell after DTT treatment (Fig. 6B2). Changes in this protein distribution provoke a reduction of the overall tension and the area compressibility modulus since actin is no longer connected to the lateral cell membrane. A tension generating actin belt is partially degraded, as shown by confocal fluorescence microscopy at the central focal plane (Fig. 5E2). Studies on Xenopus revealed that cadherins influence the actin architecture39. Thus, the mechanical stability provided by the actin belt cannot be maintained after E-cadherin disruption. Additionally, we were able to show that recovery of the original E-cadherin distribution after DTT removal is accompanied by a recovery of the initial cell topography (Fig. 7). Interestingly, after only 3 h of recovery both mechanical parameters, $${T}_{0}$$ and $${\hat{K}}_{{\rm{A}}}$$, return to even higher values as found for untreated cells, which might be due to an altered geometry. At the same time β-catenin is still distributed within the whole cell interior. This is a strong hint that other regulatory mechanisms can rescue the mechanical integrity without participation of β-catenin. At this time point E-cadherin is already partly located at the cell-cell junctions. Other studies put the cadherin-catenin-actin complex as the only cadherin-cytoskeleton connection way into question40. However, there is evidence in the literature that force generation within the cell is necessary to establish cadherin mediated cell-cell contacts41,42,43. In our experiments cells start to rebuild adherens junctions after DTT removal through a temporary increase (overshoot) of the measured mechanical parameters, essentially forming stiffer cells than initially present. During the next 3 h keeping the cells in the absence of DTT, this effect is balanced out through a tug-of-war mechanism. Eventually the cells restore E-cadherin mediated cell-cell junctions (Figs 5B4 and 6A4). β-catenin is again concentrated at the E-cadherin enriched adherens junctions in accordance with junction formation described in textbooks40.

After AJ degradation we found a small decrease in membrane tension followed by an enhanced Tt level during E-cadherin reassembly. It is indisputable that membrane tension is largely governed by the attachment of the cytoskeleton to the plasma membrane44. In studies with C. elegans it was shown that the apical membrane-cytoskeleton linker ezrin homologue ERM-1 is necessary for apical junction formation45. In order to re-establish adherens junctions after DTT removal MDCK II cells might therefore enhance their ezrin level. This linker protein is then available for enhancement of the apical membrane-cytoskeleton connection leading to higher Tt values26. More pinning points to the membrane also increase the level of cross-linking of the actomyosin network that also allows the cells to adopt a higher cortical tension through motor activity. After 6 h, the cells eventually manage to return to a normal AJ formation reflected in a recovery of all three mechanical parameters (Fig. 4).

Tension homeostasis of epithelial cells depends on many complex mechanisms in single cells and the whole tissue. Our study shows that intact lateral cell junctions are not solely responsible for control of mechanical integrity. Tight junctions enable a sealed epithelial monolayer with a controlled perijunctional flux. The impact on mechanical stability of single cells of cell sheets is, however, small. The presence of intact tight junctions is synonymous with softer cells, maybe by decoupling the apical from the basolateral part of the polarised cell. In contrast, loss of functional adherens junctions results in a severe softening of cells that is, however, fully reversible.

## Material and Methods

### Cell culture

Madin-Darby canine kidney cells (strain II, MDCK II; Health Protection Agency, Salisbury, UK) were maintained in minimum essential medium (MEM) with Earle’s salts and 2.2 g/L NaHCO3 supplemented with 4 mm l-glutamine and 10% foetal calf serum at 37 °C in a 5% CO2 humidified incubator. Cells were grown to 70–90% confluency, released from culture flasks using trypsin/EDTA (0.05%/0.02%) and subcultured three times a week. Medium additionally contained penicillin (0.2 mg/mL), streptomycin (0.2 mg/mL), Amphotericin B (0.5 mg/mL) and HEPES (15 μm) during experimentation.

Cells were seeded on Petri dishes (μ-Dish 35 mm, low Grid-500; ibidi, Martinsried, Germany) and grown to confluency. 1,4-dithiothreitol was dissolved in antibiotics and fungicide containing cell culture medium with a drug concentration of 10 mm. Cells were incubated with the drug containing medium for the indicated time at 37 °C.

### ZO-1 silencing

MDCK II cells were seeded on Petri dishes (μ-Dish 35 mm, low Grid-500; ibidi, Martinsried, Germany) and grown to 50% confluence. Pooled siRNA targeting zonula occludens-1 sequences GCAAAGACAUUGAUAGAAA, GAGAAGAAGUGACCAUAUU, CAAAAGAUCUGCAUCCUUA, and CCUGAACCAUGGUCGGAAA (siGENOME SMARTpool human TJP1 siRNA; Thermo Fisher Scientific, Lafayette, USA) were transfected using Lipofectamine® RNAiMAX transfection reagent (Life Technologies, Carlsbad, USA) according to the manufacturer’s instructions. Experiments were performed 72 h after incubation with siRNA.

For control measurements non-targeting siRNA sequences UAAGGCUAUGAAGAGAUAC, AUGUAUUGGCCUGUAUUAG, AUGAACGUGAAUUGCUCAA, UGGUUUACAUGUCGACUAA (siGENOME non-targeting siRNA Pool #2, GE Healthcare, Lafayette, USA) were applied.

### Cell labelling

Cells were grown on Petri dishes (μ-Dish 35 mm, low Grid-500; ibidi, Martinsried, Germany), treated as desired, and fixed with 4% paraformaldehyde in phosphate buffered saline without calcium and magnesium (PBS) for 20 min. To block unspecific binding sites and to permeabilise the plasma membrane, cells were treated with blocking buffer (5% (w/v) bovine serum albumin (BSA), 0.3% (v/v) Triton X-100 in PBS) for 30 min. For ZO-1 labelling the fluorescently labelled primary antibody (ZO-1 Monoclonal Antibody (ZO1-1A12), Alexa Fluor 488; Thermo Fisher Scientific, Schwerte, Germany) was diluted with dilution buffer (1% (w/v) BSA, 0.3% (v/v) Triton X-100 in PBS) to a concentration of 5 μg/mL, and cells were incubated for 1 h at room temperature. E-cadherin was labelled by incubation of the permeabilised cells with a fluorescently labelled antibody (Alexa Fluor 488 or Alexa Fluor 555 Mouse anti-E-Cadherin (Clone 36), c = 5 µg/mL; BD Biosciences, Heidelberg, Germany). β-catenin staining was performed incubating permeabilised cells with mouse anti-β-catenin antibody (c = 5 µg/mL in dilution buffer; BD Biosciences, Heidelberg, Germany) for 1 h. The secondary antibody (Alexa Fluor 546-conjugated goat anti-mouse IgG; Life Technologies, Carlsbad, USA) was diluted with dilution buffer down to a concentration of 5 μg/mL and incubated with the cells for 45 min. F-actin was labelled by incubating cells with 165 nm Alexa Fluor 546-phalloidin or Alexa Fluor 647-phalloidin (Life Technologies, Carlsbad, USA) for 45 min. Between every labelling step, cells were rinsed three times with PBS for 5 min each on a vibratory plate (75 rpm).

Fluorescence imaging was carried out using a confocal laser scanning microscope (CLSM; FluoView1200; Olympus, Tokyo, Japan) using an 100x oil immersion objective (UPLFLN100xO2PH, NA = 1.3; Olympus, Tokyo, Japan). Fluorescence images are shown in false colours.

### AFM imaging

Cells were prepared as desired and fixed by incubation with 2.5% (v/v) glutaraldehyde in PBS for 20 min. Cell imaging was carried out using a Nanowizard® II, III or 4 atomic force microscope (JPK Instruments, Berlin, Germany) equipped with silicon nitride cantilevers (MLCT, k = 10 mN/m, Bruker AFM Probes, Camarillo, USA). The AFM was mounted on an inverse optical microscope (IX 81 or IX83, Olympus, Tokyo, Japan) to enable phase contrast microscopy during imaging. Cells were imaged in contact mode in PBS at room temperature with a scan rate of 0.5 Hz. Image processing was performed using software provided by the AFM manufacturer.

### AFM force indentation experiments

Force indentation cycles were recorded on the above-mentioned setup. Cantilevers (MLCT, k = 10 mN/m, Bruker AFM Probes, Camarillo, USA) were plasma cleaned (Argon, 30 s) and incubated with FITC-conjugated Concanavalin A (c = 2.5 mg/mL in PBS; Sigma-Aldrich, Steinheim, Germany) for 1.5 h. This enables strong interaction between the cantilever and the plasma membrane during indentation and results in pulling out membrane tethers upon retraction of the indenter. The exact spring constant of each cantilever was determined using the thermal noise method46. Petri dishes containing samples of living cells were mounted on an inverted microscope and kept at 37 °C during experimentation. Cells were indented up to a force of 1 nN. After a dwell time of 0.5 s the indenter was retracted from the plasma membrane and membrane nanotubes were frequently pulled out of the plasma membrane. The indentation and retraction speed was set to 2 µm/s. These force indentation experiments were performed while scanning laterally across the sample, referred to as force mapping25.

### Data analysis of force indentation and retraction experiments

Force indentation curves were analysed as described in our previous studies34. Briefly, Force (F)-indentation (δ) curves were fitted by a polynomial starting at the contact point, where the indenter touches the cell membrane:

$$F(\delta )=a\cdot \delta +b\cdot {\delta }^{3}$$

Prefactors a and b were used to calculate the overall tension T0 and apparent area compressibility modulus $${\hat{K}}_{{\rm{A}}}$$ taking the geometrical properties of the indenter with the half-opening angle $$\varphi$$ into account:

$${T}_{0}=\frac{a}{2\pi \,\cos ({\rm{\Theta }})},$$

and

$${\hat{K}}_{{\rm{A}}}=\frac{b}{2{\pi }^{2}(1-\,\cos ({\rm{\Theta }}))}\,\tan ({\rm{\Theta }}),$$

where $${\rm{\Theta }}=\frac{\pi }{2}-\varphi$$.

The advantage of this simplified model over more sophisticated ones, where the geometry of the cell is considered47, is its analytical tractability and that it only considers the area increase due to the presence of the indenter. It has been shown by Discher and co-workers that a polynomial fit (linear and cubic term) serves well to describe the experimental data48.

Membrane tension Tt was calculated from force steps (Ftether) according to:

$${T}_{{\rm{t}}}=\frac{{F}_{{\rm{tether}}}^{2}}{8{\pi }^{2}\kappa }$$

$$\kappa =2.7\cdot {10}^{-27}{\rm{J}}$$ was chosen25,49,50.

An example of a force indentation and retraction cycle and the fitting procedure is shown in the supplementary information (suppl. Fig. S2).

## Data Availability

The datasets generated and analysed during the current study are available from the corresponding author on reasonable request.

## References

1. 1.

Balda, M. S. & Matter, K. Tight junctions at a glance. J. Cell Sci. 121, 3677–3682 (2008).

2. 2.

Harris, T. J. C. & Tepass, U. Adherens junctions: from molecules to morphogenesis. Nat. Rev. Mol. Cell Biol. 11, 502–514 (2010).

3. 3.

Hatzfeld, M., Keil, R. & Magin, T. M. Desmosomes and Intermediate Filaments: Their Consequences for Tissue Mechanics. Cold Spring Harb. Perspect. Biol. 9, a029157 (2017).

4. 4.

Kumar, N. M. & Gilula, N. B. The Gap Junction Communication Channel. Cell 84, 381–388 (1996).

5. 5.

Rodriguez-Boulan, E. & Nelson, W. J. Morphogenesis of the polarized epithelial cell phenotype. Science 245, 718–725 (1989).

6. 6.

Leoni, G., Neumann, P. A., Sumagin, R., Denning, T. L. & Nusrat, A. Wound repair: role of immune–epithelial interactions. Mucosal Immunol. 8, 959–968 (2015).

7. 7.

Clark, A. G. & Vignjevic, D. M. Modes of cancer cell invasion and the role of the microenvironment. Curr. Opin. Cell Biol. 36, 13–22 (2015).

8. 8.

Shen, L., Weber, C. R., Raleigh, D. R., Yu, D. & Turner, J. R. Tight Junction Pore and Leak Pathways: A Dynamic Duo. Annu. Rev. Physiol. 73, 283–309 (2011).

9. 9.

Van Itallie, C. M., Fanning, A. S., Bridges, A. & Anderson, J. M. ZO-1 stabilizes the tight junction solute barrier through coupling to the perijunctional cytoskeleton. Mol. Biol. Cell 20, 3930–3940 (2009).

10. 10.

Zihni, C., Mills, C., Matter, K. & Balda, M. S. Tight junctions: from simple barriers to multifunctional molecular gates. Nat. Rev. Mol. Cell Biol. 17, 564–580 (2016).

11. 11.

Fanning, A. S., Van Itallie, C. M. & Anderson, J. M. Zonula occludens-1 and -2 regulate apical cell structure and the zonula adherens cytoskeleton in polarized epithelia. Mol. Biol. Cell 23, 577–590 (2012).

12. 12.

Cartagena-Rivera, A. X., Van Itallie, C. M., Anderson, J. M. & Chadwick, R. S. Apical surface supracellular mechanical properties in polarized epithelium using noninvasive acoustic force spectroscopy. Nature Commun. 8, 1030 (2017).

13. 13.

Brasch, J., Harrison, O. J., Honig, B. & Shapiro, L. Thinking outside the cell: how cadherins drive adhesion. Trends Cell Biol. 22, 299–310 (2012).

14. 14.

Rimm, D. L., Koslov, E. R., Kebriaei, P., Cianci, C. D. & Morrow, J. S. Alpha 1(E)-catenin is an actin-binding and -bundling protein mediating the attachment of F-actin to the membrane adhesion complex. Proc. Natl. Acad. Sci. USA 92, 8813–8817 (1995).

15. 15.

Takeichi, M. Dynamic contacts: rearranging adherens junctions to drive epithelial remodelling. Nat. Rev. Mol. Cell Biol. 15, 397–410 (2014).

16. 16.

Watabe-Uchida, M. et al. α-Catenin-Vinculin Interaction Functions to Organize the Apical Junctional Complex in Epithelial Cells. J. Cell Biol. 142, 847–857 (1998).

17. 17.

Bazellières, E. et al. Control of cell–cell forces and collective cell dynamics by the intercellular adhesome. Nat. Cell Biol. 17, 409–420 (2015).

18. 18.

Broussard, J. A. et al. The desmoplakin–intermediate filament linkage regulates cell mechanics. Mol. Biol. Cell 28, 3156–3164 (2017).

19. 19.

Müller, D. J. & Dufrêne, Y. F. Atomic force microscopy: a nanoscopic window on the cell surface. Trends Cell Biol. 21, 461–469 (2011).

20. 20.

Staunton, J. R., Doss, B. L., Lindsay, S. & Ros, R. Correlating confocal microscopy and atomic force indentation reveals metastatic cancer cells stiffen during invasion into collagen I matrices. Sci. Rep. 6, 19686 (2016).

21. 21.

Tee, S.-Y., Bausch, A. R. & Janmey, P. A. The mechanical cell. Curr. Biol. 19, R745–R748 (2009).

22. 22.

Pelham, R. J. Jr. & Wang, Y. Cell locomotion and focal adhesions are regulated by substrate flexibility. Proc. Natl. Acad. Sci. USA 94, 13661–13665 (1997).

23. 23.

Yeung, T. et al. Effects of substrate stiffness on cell morphology, cytoskeletal structure, and adhesion. Cell Motil. Cytoskeleton 60, 24–34 (2005).

24. 24.

Darling, E. M., Zauscher, S., Block, J. A. & Guilak, F. A Thin-Layer Model for Viscoelastic, Stress-Relaxation Testing of Cells Using Atomic Force Microscopy: Do Cell Properties Reflect Metastatic Potential? Biophys. J. 92, 1784–1791 (2007).

25. 25.

Pietuch, A., Brückner, B. R., Fine, T., Mey, I. & Janshoff, A. Elastic properties of cells in the context of confluent cell monolayers: impact of tension and surface area regulation. Soft Matter 9, 11490–11502 (2013).

26. 26.

Brückner, B. R., Pietuch, A., Nehls, S., Rother, J. & Janshoff, A. Ezrin is a Major Regulator of Membrane Tension in EpithelialCells. Sci. Rep. 5, 14700 (2015).

27. 27.

Ozawa, M., Hoschützky, H., Herrenknecht, K. & Kemler, R. A possible new adhesive site in the cell-adhesion molecule uvomorulin. Mech. Dev. 33, 49–56 (1990).

28. 28.

Ni, H., Li, A., Simonsen, N. & Wilkins, J. A. Integrin Activation by Dithiothreitol or Mn2+ Induces a Ligand-occupied Conformation and Exposure of a Novel NH2-terminal Regulatory Site on the β1Integrin Chain. J. Biol. Chem. 273, 7981–7987 (1998).

29. 29.

Buckley, C. D. et al. The minimal cadherin-catenin complex binds to actin filaments under force. Science 346, 1254211 (2014).

30. 30.

Chu, Y.-S. et al. Force measurements in E-cadherin–mediated cell doublets reveal rapid adhesion strengthened by actin cytoskeleton remodeling through Rac and Cdc42. J. Cell Biol. 167, 1183–1194 (2004).

31. 31.

Campbell, H. K., Maiers, J. L. & DeMali, K. A. Interplay between tight junctions & adherens junctions. Exp. Cell Res. 358, 39–44 (2017).

32. 32.

Hatte, G., Prigent, C. & Tassan, J.-P. Tight junctions negatively regulate mechanical forces applied to adherens junctions in vertebrate epithelial tissue. J. Cell Sci. 131, jcs208736 (2018).

33. 33.

Dumbauld, D. W. et al. How vinculin regulates force transmission. Proc. Natl. Acad. Sci. USA 110, 9788–9793 (2013).

34. 34.

Karsch, S., Kong, D., Großhans, J. & Janshoff, A. Single-Cell Defects Cause a Long-Range Mechanical Response in a Confluent Epithelial Cell Layer. Biophys. J. 113, 2601–2608 (2017).

35. 35.

Yonemura, S. Cadherin–actin interactions at adherens junctions. Curr. Opin. Cell Biol. 23, 515–522 (2011).

36. 36.

Yonemura, S., Wada, Y., Watanabe, T., Nagafuchi, A. & Shibata, M. α-Catenin as a tension transducer that induces adherens junction development. Nat. Cell Biol. 12, 533–542 (2010).

37. 37.

Ikenouchi, J., Umeda, K., Tsukita, S., Furuse, M. & Tsukita, S. Requirement of ZO-1 for the formation of belt-like adherens junctions during epithelial cell polarization. J. Cell Biol. 176, 779–786 (2007).

38. 38.

Capaldo, C. T. & Macara, I. G. Depletion of E-cadherin disrupts establishment but not maintenance of cell junctions in Madin-Darby canine kidney epithelial cells. Mol. Biol. Cell 18, 189–200 (2007).

39. 39.

Tao, Q., Nandadasa, S., McCrea, P. D., Heasman, J. & Wylie, C. G-protein-coupled signals control cortical actin assembly by controlling cadherin expression in the early Xenopus embryo. Development 134, 2651–2661 (2007).

40. 40.

Gates, J. & Peifer, M. Can 1000 Reviews Be Wrong? Actin, α-Catenin, and Adherens Junctions. Cell 123, 769–772 (2005).

41. 41.

42. 42.

Liu, Z. et al. Mechanical tugging force regulates the size of cell–cell junctions. Proc. Natl. Acad. Sci. USA 107, 9944–9949 (2010).

43. 43.

Papusheva, E. & Heisenberg, C. P. Spatial organization of adhesion: force‐dependent regulation and function in tissue morphogenesis. EMBO J. 29, 2753–2768 (2010).

44. 44.

Diz-Munoz, A., Fletcher, D. A. & Weiner, O. D. Use the force: membrane tension as an organizer of cell shape and motility. Trends Cell Biol. 23, 47–53 (2013).

45. 45.

van Furden, D., Johnson, K., Segbert, C. & Bossinger, O. The C. elegans ezrin-radixin-moesin protein ERM-1 is necessary for apical junction remodelling and tubulogenesis in the intestine. Dev. Biol. 272, 262–276 (2004).

46. 46.

Hutter, J. L. & Bechhoefer, J. Calibration of atomic‐force microscope tips. Rev. Sci. Instrum. 64, 1868–1873 (1993).

47. 47.

Sen, S., Subramanian, S. & Discher, D. E. Indentation and adhesive probing of a cell membrane with AFM: theoretical model and experiments. Biophys. J. 89, 3203–3213 (2005).

48. 48.

Hategan, A., Law, R., Kahn, S. & Discher, D. E. Adhesively-Tensed Cell Membranes: Lysis Kinetics and Atomic Force Microscopy Probing. Biophys. J. 85, 2746–2759 (2003).

49. 49.

Brochard-Wyart, F., Borghi, N., Cuvelier, D. & Nassoy, P. Hydrodynamic narrowing of tubes extruded from cells. Proc. Natl. Acad. Sci. USA 103, 7660–7663 (2006).

50. 50.

Dai, J. & Sheetz, M. P. Membrane tether formation from blebbing cells. Biophys. J. 77, 3363–3370 (1999).

## Acknowledgements

The authors thank Angela Ruebeling for cell service and technical assistance. Financial support by the DFG through SPP 1782 and CRC 937 (A14) is acknowledged. BRB acknowledges financial support by a scholarship of the Konrad Adenauer Foundation.

## Author information

Authors

### Contributions

B.R.B. carried out the experiments, analysed the data and prepared the figures. A.J. wrote the computer programs for data analysis. Both authors wrote the manuscript.

### Corresponding author

Correspondence to Andreas Janshoff.

## Ethics declarations

### Competing Interests

The authors declare no competing interests.

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Reprints and Permissions

Brückner, B.R., Janshoff, A. Importance of integrity of cell-cell junctions for the mechanics of confluent MDCK II cells. Sci Rep 8, 14117 (2018). https://doi.org/10.1038/s41598-018-32421-2

• Accepted:

• Published:

### Keywords

• Cell-cell Junctions
• Cortical Tension
• Individual Epithelial Cells
• Tight Junctions (TJs)

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Frontiers in Cell and Developmental Biology (2020)


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These are chat archives for Codewars/codewars.com/kata-solving-help

7th
Dec 2016
Ze-Marcos
@MarkSchemmer
Dec 07 2016 02:33
I refactored my code again .
using System;
using System.Collections.Generic;
using System.Text.RegularExpressions;
using System.Linq;

public static class Kata
{

public static long NextSmaller(long n)
{
Console.WriteLine("Max Number is : " + n);
string order = n.ToString().ToCharArray().OrderBy(x => x).Aggregate("", (cur, nex) => cur + nex);
long L = Convert.ToInt64(order);
Console.WriteLine( "i must be bigger than : " + L);

for(var i = n - 2; i >= L; i--)
{
string z = i.ToString().ToCharArray().OrderBy(x => x).Aggregate("", (cur, nex) => cur + nex);

if ( order == z )
{
Console.WriteLine(\$"The next lowest number from {n} is {i} ");
return i;
}
}

Console.WriteLine("There is no next lower number so we return : -1");
return -1;
}
}
but it is failing now on time out error
Does anybody know how to speed this algorithm up ?
Ze-Marcos
@MarkSchemmer
Dec 07 2016 03:01
Please if there is someone I could really use some help
I have refactored my code a little bit more
using System;
using System.Collections.Generic;
using System.Text.RegularExpressions;
using System.Linq;

public static class Kata
{

public static long NextSmaller(long n)
{
string order = n.ToString().ToCharArray().AsParallel().AsOrdered().OrderBy(x => x).Aggregate("", (cur, nex) => cur + nex);
long L = Convert.ToInt64(order);

for(var i = n - 2; i >= L; i--)
{
string z = i.ToString().ToCharArray().AsParallel().AsOrdered().OrderBy(x => x).Aggregate("", (cur, nex) => cur + nex);

if(order == z)
return i;

}

return -1;
}
}
Benjamin Kästner
@bkaestner
Dec 07 2016 05:40
Your algorithm isn't efficient. Every of your ToString() lines needs to create a string, an array, sort them, and the recreate a string.
Ze-Marcos
@MarkSchemmer
Dec 07 2016 05:41
Yes
It is my algorithm
and it is about 630 ms slow I guess
Benjamin Kästner
@bkaestner
Dec 07 2016 05:41
Also, you check a lot of numbers.
For example, if the input is 987654321, you would start with 123456789.
Ze-Marcos
@MarkSchemmer
Dec 07 2016 05:42
Yes but how would I shorten that ?
that is what I am trying to understand right
now
Benjamin Kästner
@bkaestner
Dec 07 2016 05:43
You don't check all numbers. Instead, you check how you could modify the number to make it smaller.
How would you make 321 smaller?
Ze-Marcos
@MarkSchemmer
Dec 07 2016 05:44
My first thought is this var query = "321".ToCharArray().OrderByDecending(x => x);
Benjamin Kästner
@bkaestner
Dec 07 2016 05:44
How would you make 12321 smaller?
Ze-Marcos
@MarkSchemmer
Dec 07 2016 05:44
then I would convert it to a long integer
Benjamin Kästner
@bkaestner
Dec 07 2016 05:45
No. As a human. Not as a machine.
Ze-Marcos
@MarkSchemmer
Dec 07 2016 05:46
That is a harder question
I guess
Benjamin Kästner
@bkaestner
Dec 07 2016 05:46
Which one? The human one? Or 12321?
Ze-Marcos
@MarkSchemmer
Dec 07 2016 05:47
No. As a human ? and how to make 12321
smaller
Benjamin Kästner
@bkaestner
Dec 07 2016 05:47
=> 12312
Ze-Marcos
@MarkSchemmer
Dec 07 2016 05:48
I would say that 11223
Benjamin Kästner
@bkaestner
Dec 07 2016 05:48
But that's not the next smaller.
Ze-Marcos
@MarkSchemmer
Dec 07 2016 05:48
bu the next smallest is
Benjamin Kästner
@bkaestner
Dec 07 2016 05:48
Right.
Ze-Marcos
@MarkSchemmer
Dec 07 2016 05:49
12231
?
I think that is correct
Benjamin Kästner
@bkaestner
Dec 07 2016 05:49
12321 => 12312
Ze-Marcos
@MarkSchemmer
Dec 07 2016 05:50
it is the same
Benjamin Kästner
@bkaestner
Dec 07 2016 05:51
$12312 \neq 12231$
Ze-Marcos
@MarkSchemmer
Dec 07 2016 05:51
Yes
Benjamin Kästner
@bkaestner
Dec 07 2016 05:51
What's the "next smaller" number of 9111111111111111111?
Ze-Marcos
@MarkSchemmer
Dec 07 2016 05:52
I do not know really
well can I guess ?
Natacha Beck
@natacha-beck
Dec 07 2016 05:57
you really have a smallest int ? with the same digits ?
Ze-Marcos
@MarkSchemmer
Dec 07 2016 05:58
I pass all the tests
The only problem is that my algorithm is slow
Benjamin Kästner
@bkaestner
Dec 07 2016 05:59
Yeah, the algorithm is "correct", but it's runtime is linear in the size of the number, not linear in the number of digits (aka logarithmic in the size of the number).
Ze-Marcos
@MarkSchemmer
Dec 07 2016 06:03
Yes you are totally right
I am trying to understand
and I get that I need to change my for loop to make the count shorter
and to make my Linq query statements faster as well
Benjamin Kästner
@bkaestner
Dec 07 2016 06:05
I'm not sure whether Linq is appropriate here.
Ze-Marcos
@MarkSchemmer
Dec 07 2016 06:07
you are probably right because linq is slow when iterating through big lists compared to for loops I guess
but I would like to find a way
though to make it possible
do you have a article I could read
Ze-Marcos
@MarkSchemmer
Dec 07 2016 06:12
or maybe a idea of what I could do
?
Giacomo Sorbi
@GiacomoSorbi
Dec 07 2016 07:59
oh, working on the spin off of my most appreciated kata
and wondering if I'm doing correctly:
1. Write the function
2. Write the test case
Is that correct? And am I testing the right way
const howManyLightsabersDoYouOwn = (name) => {
return name === 'Zach' ? 18 : 0;
}

describe('howManyLightsabersDoYouOwn', function(name) {
it ('should return 18 if name === "Zach"', function() {
var name = 'Zach';
Test.expect(howManyLightsabersDoYouOwn(name) === 18);
})
it ('should return 0 for somene else', function() {
var name = 'Me';
Test.expect(howManyLightsabersDoYouOwn(name) === 0);
})
})
thanks first time writing a test case
Luke Murray
@lukemurray77
Dec 07 2016 12:13

Hey guys, im am attempting this kata, https://www.codewars.com/kata/coding-meetup-number-15-higher-order-functions-series-find-the-odd-names/train/javascript

I think I have solved it- my solution passes the basic tests but fails on the attempt. Can someone offer some advice? here is my code:

function findOddNames(list) {
var count = 0;
list.forEach(function(element, index){
var name = element.firstName.split('');
count = name.reduce(function(result, element, index){
return result += element.charCodeAt(0);
}, 0)

if (count % 2 === 0){
element.firstName = 1;
}
})
return list.filter(function(elem, index){
return elem.firstName !== 1;
})
}
asterixorobelix
@asterixorobelix
Dec 07 2016 12:45
Is anyone else having problems with the Fighters Kata in Javascript? I'm passing all the tests, except for one. However, when I run the code on my own machine, it passes all the tests. https://www.codewars.com/kata/two-fighters-one-winner/train/javascript
My code passes all the asserts, except for the third one, were it says it expected a result "Harald" and it didnt get that returned. However, when I run that test in the terminal on my machine, it does return "Harald"
Giacomo Sorbi
@GiacomoSorbi
Dec 07 2016 18:18
@dantesolis, test cases have been written for you, just submit


text:
# zbMATH — the first resource for mathematics

A classification of generalised state space reduction methods for linear multivariable systems. (English) Zbl 0859.93024
The aim of this paper is to establish two algorithms which reduce a linear multivariable system $$\Sigma$$, described by a polynomial matrix model of the form: $A(\rho)\beta(t)= B(\rho)u(t), \qquad y(t)= C(\rho)\beta(t)+ D(\rho)u(t), \tag{$$\Sigma$$}$ to an equivalent model in generalized state space form. Here $$\rho=d/dt$$, $$A(\rho)\in\mathbb{R} [\rho]^{r\times r}$$ with $$\text{rank}_\mathbb{R} A(\rho)=r$$, $$B(\rho)\in \mathbb{R}[\rho]^{r\times m}$$, $$C(\rho)\in \mathbb{R}[\rho]^{p\times r}$$, $$D(\rho)\in \mathbb{R}[\rho]^{p\times m}$$, $$\beta(t)$$ is the pseudostate of $$\Sigma$$, $$u(t)$$ the input vector and $$y(t)$$ the output vector.
More precisely, the authors solve the problem of determining a system: $E\dot x(t)= Ax(t)+ Bu(t), \qquad y(t)= Cx(t)+ Du(t) \tag{$${\Sigma_R}$$}$ equivalent to $$\Sigma$$. This equivalence means that they exhibit identical system properties. The first algorithm is based on the realization of $${\mathcal T}(s)$$ defined by: ${\mathcal T}(\rho)=\begin{pmatrix} A(\rho) &B(\rho) &0\\ -C(\rho) &D(\rho) &I_p\\ 0 &-I_m &0\end{pmatrix}\in \mathbb{R}[\rho]^{\bar r\times\bar r}, \qquad \bar r=r+p+m,$ while the second algorithm is based on a realization of $${\mathcal T}(s)^{-1}$$. In fact, all the known reduction algorithms can be classified by these two different theoretical reduction algorithms which are mentioned above.

##### MSC:
 93C05 Linear systems in control theory 93C35 Multivariable systems, multidimensional control systems 93B11 System structure simplification
Full Text:
##### References:
 [1] B. D. O. Anderson W. A. Coppel, D. J. Cullen: Strong system equivalence (I). J. Austral. Math. Soc. Ser. B 27 (1985), 194-222. · Zbl 0594.93015 [2] O. H. Bosgra, A. J. J. Van Der Weiden: Realisations in generalised state-space form for polynomial system matrices, and the defìnition of poles, zeros and decoupling zeros at infinity. Internat. J. Control 33 (1981), 393-411. · Zbl 0464.93021 [3] G. E. Hayton A. B. Walker, A. C. Pugh: Infinite frequency structure preserving transformations for general polynomial system matrices. Internat. J. Control 52 (1990), 1-14. · Zbl 0702.93021 [4] N. P. Karampetakis, A. I. G. Vardulakis: Matrix fractions and full system equivalence. IMA J. Math. Control Inform. 9 (1992), 147-160. · Zbl 0777.93052 [5] N. P. Karampetakis, A. I. G. Vardulakis: Generalized state-space system matrix equivalents of a Rosenbrock system matrix. IMA J. Math. Control Inform. 10 (1993), 323-344. · Zbl 0807.93009 [6] T. Shaohua, J. Vandewalle: A singular system realisation for arbitrary matrix fraction descriptions. ISCAS’88, pp. 615-618. [7] A. I. G. Vardulakis: Linear Multivariable Control, Algebraic Analysis and Synthesis Methods. Nelson-Wiley, London 1991. · Zbl 0751.93002 [8] A. I. G. Vardulakis: On the transformation of a polynomial matrix model of a linear multivariable system to generalised state space form. Proceedings of the 30th IEEE Conference on Decision and Control, Brighton 1991, U.K., pp. 11-13. [9] G. C. Verghese: Infinite Frequency Behavior in Generalized Dynamical Systems. P\?.D. Dissertation, Stanford Univ., Stanford, CA 1978. [10] W. A. Wolovich: Linear Multivariable Systems. Springer-Verlag, New York 1974. · Zbl 0291.93002
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
Volume 36, Issue 3
Estimates of Dirichlet Eigenvalues for One-Dimensional Fractal Drums

Anal. Theory Appl., 36 (2020), pp. 243-261.

Published online: 2020-09

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• Abstract

Let $\Omega$, with finite Lebesgue measure $|\Omega|$, be a non-empty open subset of $\mathbb{R}$, and $\Omega=\bigcup_{j=1}^\infty\Omega_j$, where the open sets $\Omega_j$ are pairwise disjoint and the boundary $\Gamma=\partial\Omega$ has Minkowski dimension $D\in (0,1)$.  In this paper we study the Dirichlet eigenvalues problem on the domain $\Omega$ and give the exact second asymptotic term for the eigenvalues, which is related to the Minkowski dimension $D$. Meanwhile, we give sharp lower bound estimates for Dirichlet eigenvalues for such one-dimensional fractal domains.

• Keywords

One-dimensional fractal drum, Dirichlet eigenvalues, Pόlya conjecture, Minkowski dimension.

• AMS Subject Headings

52B10, 65D18, 68U05, 68U07

• Copyright

COPYRIGHT: © Global Science Press

• Email address
• BibTex
• RIS
• TXT
@Article{ATA-36-243, author = {Hua Chen , and Jinning Li , }, title = {Estimates of Dirichlet Eigenvalues for One-Dimensional Fractal Drums}, journal = {Analysis in Theory and Applications}, year = {2020}, volume = {36}, number = {3}, pages = {243--261}, abstract = {

Let $\Omega$, with finite Lebesgue measure $|\Omega|$, be a non-empty open subset of $\mathbb{R}$, and $\Omega=\bigcup_{j=1}^\infty\Omega_j$, where the open sets $\Omega_j$ are pairwise disjoint and the boundary $\Gamma=\partial\Omega$ has Minkowski dimension $D\in (0,1)$.  In this paper we study the Dirichlet eigenvalues problem on the domain $\Omega$ and give the exact second asymptotic term for the eigenvalues, which is related to the Minkowski dimension $D$. Meanwhile, we give sharp lower bound estimates for Dirichlet eigenvalues for such one-dimensional fractal domains.

}, issn = {1573-8175}, doi = {https://doi.org/10.4208/ata.OA-SU7}, url = {http://global-sci.org/intro/article_detail/ata/18285.html} }
TY - JOUR T1 - Estimates of Dirichlet Eigenvalues for One-Dimensional Fractal Drums AU - Hua Chen , AU - Jinning Li , JO - Analysis in Theory and Applications VL - 3 SP - 243 EP - 261 PY - 2020 DA - 2020/09 SN - 36 DO - http://doi.org/10.4208/ata.OA-SU7 UR - https://global-sci.org/intro/article_detail/ata/18285.html KW - One-dimensional fractal drum, Dirichlet eigenvalues, Pόlya conjecture, Minkowski dimension. AB -

Let $\Omega$, with finite Lebesgue measure $|\Omega|$, be a non-empty open subset of $\mathbb{R}$, and $\Omega=\bigcup_{j=1}^\infty\Omega_j$, where the open sets $\Omega_j$ are pairwise disjoint and the boundary $\Gamma=\partial\Omega$ has Minkowski dimension $D\in (0,1)$.  In this paper we study the Dirichlet eigenvalues problem on the domain $\Omega$ and give the exact second asymptotic term for the eigenvalues, which is related to the Minkowski dimension $D$. Meanwhile, we give sharp lower bound estimates for Dirichlet eigenvalues for such one-dimensional fractal domains.

Hua Chen & Jinning Li. (2020). Estimates of Dirichlet Eigenvalues for One-Dimensional Fractal Drums. Analysis in Theory and Applications. 36 (3). 243-261. doi:10.4208/ata.OA-SU7
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# $3\times 3$ Orthogonal Matrices with an Analysis of Eigenvalues

It is said that one can prove that all 3x3 orthogonal matrices correspond to linear operators on $R^3$ of the following types:

1. Rotations about lines through he origin
2. Reflections about planes through the origin
3. A rotation about a line through the origin followed by a reflection about the plane through the origin that is perpendicular to the line

Obviously, we can tell whether a 2x2 matrix $A$ represents a rotation or reflection by looking at its determinant: that is, a rotation if $det(A)=1$ and a reflection if $det(A)=-1$. But for a 3x3 matrix, this is different. My textbook says that if $det(A)=1$, it is a rotation. If $det(A)=-1$, it is either of type 2 or 3.

My question is: How to tell whether a 3x3 orthogonal matrix with determinant $-1$ represents a type 2 or a type 3 ? I heard that this has to do with an analysis of eigenvectors and eigenvalus, but could anyone shed some light on this please?

For example, how can one tell whether the following operator is of type 2 or type 3? (Actually I was quite sure, as I "invented" it, if no mistake, that the following linear operator is neither a rotation, nor a reflection$-$it is of type 3.) \begin{bmatrix} \frac{1}{3} & 0 & \frac{4\sqrt{2}}{6} \\ \frac{2}{3} & \frac{1}{\sqrt{2}} & \frac{-\sqrt{2}}{6} \\ \frac{2}{3} & \frac{-1}{\sqrt{2}} & \frac{-\sqrt{2}}{6} \\ \end{bmatrix}

Eigenvalues (or strictly speaking, eigendecomposition) can be used to establish the fact that the three listed cases are the only possible ones. They are not necessary in differentiating type 2 and type 3 matrices.

If $A$ is a reflection about a plane through the origin, then by applying $A$ again, the reflected image will be reflected back to its original. Therefore $A^2=I$.

If $A$ is a rotation followed by a reflection through the rotation plane, pick any vector $u$ that has a nonzero component on the plane of rotation (i.e. pick any $u$ whose orthogonal projection on the rotation plane is nonzero). Decompose $u$ into the sum $v+w$, where $v$ lies on the plane of rotation and $w$ is parallel to the rotation axis (and hence orthogonal to $v$. Then $A^2u=A^2v+w$. When the angle of rotation is not an odd multiple of $\pi$, we have $A^2v\ne v$ and hence $A^2u\ne u$ and hence $A^2\ne I$. When the angle of rotation is an odd multiple of $\pi$, we get $A=-I$.

Therefore, when $A$ is real orthogonal and $\det A=-1$, it is of type 2 if and only if $A^2=I$ and $A\ne-I$, and it is of type 3 if and only if $A^2\ne I$ or $A=-I$.

• I have read your answer and amd's answer already. As you pointed out, we can determine the type of the matrix operator by checking if $A^2 = I$ and $A \ne I$ or not; if this is the case, it is of type2, if this is not the case, it is of another type. But I was just wondering how can I relate this to an analysis of eigenvectors and eigenvalues as amd's pointed out in his post. – IgNite Apr 16 '16 at 8:02
• Or more precisely, why the matrix operator is of type 3 if one of its eigenvalues is $1$ and the other two aren't $\pm1$? – IgNite Apr 16 '16 at 8:03
• @Ignite What you say cannot happen. As $\det A=-1$ and one of the eigenvalues must be -1 (because a reflection occurs), if one of the eigenvalues is 1, the other two must be -1 and 1. – user1551 Apr 16 '16 at 8:40
• @Ignite Anyway, take a unit vector $u$ parallel to the axis of rotation and take two mutually orthogonal unit vectors $v$ and $w$ from the plane normal to $u$. Then $\{u,v,w\}$ is an orthonormal basis of $\mathbb R^3$. With respect to this basis, the matrix becomes $\pmatrix{-1\\ &R}$, where $R$ is a 2x2 rotation matrix. $R=I$ when it is of type 2, and $R\ne I$ when it is of type 3. The eigenvalues of a 2x2 rotation matrix are $\cos\theta\pm i\sin\theta$, where $\theta$ is the angle of rotation. If $R^2=I$ but $R\ne I$, $\theta$ must be an odd multiple of $\pi$, i.e. $R=-I$. – user1551 Apr 16 '16 at 8:44
• Oh I'm very sorry, there's a typo. It should be why the matrix operator is of type 3 if one of its eigenvalues is $-1$ and the other two aren't $\pm1$? – IgNite Apr 16 '16 at 8:55

If there’s a rotation involved, two of the eigenvalues will be a complex conjugate pair. If it’s a pure reflection, all of the eigenvalues will be real. There’s one exception to this—a rotation through an angle of $\pi$—but I’ll cover that below.

In more detail, a pure 3-D rotation matrix will have eigenvalues $1$ and $\cos\theta\pm i\sin\theta$, where $\theta$ is the rotation angle. If $\theta=n\pi$, then this complex conjugate pair will show up as $\pm1$. If positive, you just have the identity transformation; if negative, you have a 180-degree rotation (which is equivalent to a reflection relative the axis of rotation). The eigenspace of $1$ in the non-identity case is one-dimensional and represents the rotation axis. If you follow this rotation with a reflection in the plane normal to this axis, the eigenvalue $1$ will become $-1$.

On the other hand, the eigenvalues of a reflection will all be $\pm1$. The multiplicity of $-1$ tells you what sort of a reflection this is: a single $-1$ is a reflection in a plane, with the corresponding eigenspace normal to this plane; a multiplicity of 2 is a reflection in a line, which is equivalent to a rotation through an angle of $\pi$; a multiplicity of 3 represents releflection through the origin, which we see can be decomposed into a 180-degree rotation followed by a reflection relative to a plane.

To take your example, the matrix has $-1$ as an eigenvalue with multiplicity one, while the other eigenvalues aren’t $\pm1$, so it’s type 3. The axis of rotation is found by computing the kernel of $A+I$, giving $(2+\sqrt2,-2,-2-2\sqrt2)^T$. The angle of rotation can be found by finding the other eigenvalues and converting to polar form. (I did some quick calculations in a CAS program and it looks like the angle might be $\frac\pi4$.)

• Working on this. I'm figuring out, in the third paragraph, why the number of the multiplicity of eigenvalues $-1$ can tell what sort of a reflection is... – IgNite Apr 16 '16 at 7:31
• Could you explain a little bit more why the matrix operator is of type 3 if one of its eigenvalues is $-1$ and the other two aren't $\pm1$? – IgNite Apr 16 '16 at 8:56
• @Ignite Looks like user1551 covered it in the comments to his answer. Another way to understand this is to recall the motivation for the concept of eigenvectors: what lines are mapped to themselves? The corresponding eigenvalues are the “scale factors” for those fixed lines. For an orthogonal matrix, each $-1$ corresponds to an axis that is flipped by the transformation and each $1$ leaves the corresponding axis unchanged. If there’s some other eigenvalue, then something else—a rotation in this case—is going on. – amd Apr 16 '16 at 16:51