I had a funny idea for proving an identity in Euclidean geometry. While it didn't end up being a very nice proof strategy in my case, I would still like to collect nice examples of where the proof strategy works out, because I think such a collection of examples would be nice for impressing students of high school algebra.


I was trying to prove the "[Power of a point](http://en.wikipedia.org/wiki/Power_of_a_point)" theorem (although I did not know it by that name at the time). My idea for a proof strategy was as follows:


1. Set up a coordinate system where the circle in centered at the origin, and the point P is on the x-axis.
2. Define a function $f(y)$ as follows: First connect $P$ and $(0,y)$ with a line. This line intersects the circle in two points, call them $A$ and $B$. Then let $f(y) = (AP \times AB)^2$. It is not too hard to see that $f$ is a polynomial of degree at most 8 in $y$.
3. Find 9 values of $y$ where $f(y)$ is easy to compute - they should all have a common value.
4. Since a degree at most 8 polynomial is determined by 9 points, we see that $f$ must be constant, which proves the theorem.


After having this idea, we found the proof by similar triangles. While 3 values of $y$ are pretty easy to compute with, it was hard to find 9 easy ones.


So even though the method of proof didn't pan out in this case, it still seemed pretty cute, and hinted that "baby algebraic geometry" ideas could be presented to students in geometry or algebra classes this way.


I am asking for a list of similar situations where the proof strategy does pan out, so that they could be pulled out when teaching students of algebra or geometry.



Unexpected applications of the fact that nth degree polynomials are determined by n+1 points
If you want research level mathematics, the joint theorem is an excellent example of the polynomial technique that can be presented to high-school students. The statement is


$n$ lines in the space can form at most $Cn^{3/2}$ joints (the points where at least three non-coplanar lines intersect).


The proof (for an expert) consists of 3 "elementary" steps:


1) Let $(m+1)^3$ be the least cube that is greater than the number of joints formed. Then we can find a not identically zero polynomial $P(x,y,z)$ of degree $3m$ or less vanishing on all joints (the number of free parameters (coefficients) exceeds the number of linear conditions).


2) Take such polynomial of least degree. If every line contains at least $3m+1$ joints, then the polynomial must vanish on every line. Hence, at each joint, three independent directional derivatives must vanish, whence the differential vanishes. Thus some partial derivative $P\_x$, $P\_y$, or $P\_z$ is a lower degree polynomial with the same properties, which is impossible. Therefore, at least one line contains $3m$ joints or fewer.


3) Remove that line and repeat the argument. We see that each time we remove $3m$ joints or fewer, so we will have to remove at least $m^2/3$ lines before we get rid of all joints. So, we had at least that number of lines in the configuration.


Presenting it to high school kids may easily take two full lectures but there is nothing there beyond their comprehension.


If you want an "olympiad flavor" purely geometric problem, my vote goes to the famous pentagonal area identity: 


If you have a convex pentagon $A\_1\dots A\_5$, $S$ is its area, and $S\_j$ is the area of the triangle $A\_{j-1}A\_jA\_{j+1}$, then 
$$
S^2-S(S\_1+S\_2+S\_3+S\_4+S\_5)+S\_1S\_2+S\_2S\_3+S\_3S\_4+S\_4S\_5+S\_5S\_1=0
$$ 
The slickest proof I know is to start with a pentagon and to move $A\_1$ along the line parallel to $A\_5A\_2$ until you get a degenerated pentagon, which is actually a convex quadrilateral with one extra vertex on a side. Note that the left hand side is a linear function in this case, so if we have the identity at the endpoints, we have it throughout. For the degenerate pentagon repeat the trick moving the point on the side along it. You'll see that it suffices to prove the identity for a pentagon that is a convex quadrilateral with one vertex doubled (say $A\_1=A\_2$). But then it becomes obvious because $S\_1=S\_2=0$, so we are left with 
$$
S^2-S(S\_3+S\_4+S\_5)+S\_3S\_4+S\_4S\_5=0
$$
but $S\_3+S\_5=S$ (those two tile the quadrilateral now).


Again, there is a traditional proof, but it is much harder to find :)


