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$ABCDEFG$ is a regular heptagon inscribed in a unit circle centered at $O$. $\ell$ is the line tangent to the circumcircle of $ABCDEFG$ at $A$, and $P$ is a point on $\ell$ such that triangle $AOP$ is isosceles. Let $p$ denote the value of $AP\cdot BP \cdot CP \cdot DP \cdot EP \cdot FP \cdot GP$. How do we determine the value of the value of $p^2$?

I have tried this problem and couldn't come up with a solution, but I did find a solution using complex numbers on the internet for this problem. If anyone can solve this without complex numbers I would appreciate it!

Link to complex number solution: http://www.artofproblemsolving.com/Wiki/index.php/Mock_AIME_1_Pre_2005_Problems/Problem_10

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Complex numbers are really the natural setting for the solution to this problem. Without them, you'll need to deploy a lot of trigonometric identities. It's surely possible, but I think it won't be pretty. –  Rahul Mar 10 '12 at 23:51
    
Maybe you could work it out from $\prod_{k=0}^6\left((1-\cos\frac{2\pi k}{7})^2+(1-\sin\frac{2\pi k}{7})^2\right)$. But I agree with rahul. It's not just the complex numbers, but their relation to roots of polynomials that makes it so natural for this problem. –  bgins Mar 10 '12 at 23:57

1 Answer 1

One can use the Law of Cosines. Let $\theta = \frac{2\pi}{7}$. Assume that $OA$ is on the positive $x$-axis and $P$ is in the first quadrant.

First note that $OP = \sqrt{2}$. Measuring from $OP$, the angles to the vertices are $n\theta - \pi/4$ starting with $n=1$ up to $7$. Fix a vertex $V$. We want to solve triangle VOP that has sides $1$ and $\sqrt{2}$, which determine an angle of $n\theta - \pi/4$. Let $x$ denote the distance from the vertex to the point $P$. $$ x^2 = 1 + 2 - 2(1)\sqrt{2} \cos(n\theta - \pi/4) = 3 + 2\sqrt{2}\cos(n\theta - \pi/4). $$ Therefore the square of the products of the distances is $$ \Pi_{n=1}^7 \;\;3 + 2 \sqrt{2}\cos(n\theta - \pi/4). $$

This is easy to approximate with a calculator.

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