# 1999 AIME Problems/Problem 4

## Problem

The two squares shown share the same center $O_{}$ and have sides of length 1. The length of $\overline{AB}$ is $43/99$ and the area of octagon $ABCDEFGH$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$ $[asy] //code taken from thread for problem real alpha = 25; pair W=dir(225), X=dir(315), Y=dir(45), Z=dir(135), O=origin; pair w=dir(alpha)*W, x=dir(alpha)*X, y=dir(alpha)*Y, z=dir(alpha)*Z; draw(W--X--Y--Z--cycle^^w--x--y--z--cycle); pair A=intersectionpoint(Y--Z, y--z), C=intersectionpoint(Y--X, y--x), E=intersectionpoint(W--X, w--x), G=intersectionpoint(W--Z, w--z), B=intersectionpoint(Y--Z, y--x), D=intersectionpoint(Y--X, w--x), F=intersectionpoint(W--X, w--z), H=intersectionpoint(W--Z, y--z); dot(O); label("O", O, SE); label("A", A, dir(O--A)); label("B", B, dir(O--B)); label("C", C, dir(O--C)); label("D", D, dir(O--D)); label("E", E, dir(O--E)); label("F", F, dir(O--F)); label("G", G, dir(O--G)); label("H", H, dir(O--H));[/asy]$

## Solution 1

Triangles $AOB$, $BOC$, $COD$, etc. are congruent by symmetry (you can prove it rigorously by using the power of a point to argue that exactly two chords of length $1$ in the circumcircle of the squares pass through $B$, etc.), and each area is $\frac{\frac{43}{99}\cdot\frac{1}{2}}{2}$. Since the area of a triangle is $bh/2$, the area of all $8$ of them is $\frac{86}{99}$ and the answer is $\boxed{185}$.

## Solution 2

Define the two possible distances from one of the labeled points and the corners of the square upon which the point lies as $x$ and $y$. The area of the octagon (by subtraction of areas) is $1 - 4\left(\frac{1}{2}xy\right) = 1 - 2xy$.

By the Pythagorean theorem, $$x^2 + y^2 = \left(\frac{43}{99}\right)^2$$

Also, \begin{align*}x + y + \frac{43}{99} &= 1\\ x^2 + 2xy + y^2 &= \left(\frac{56}{99}\right)^2\end{align*}

Substituting, \begin{align*}\left(\frac{43}{99}\right)^2 + 2xy &= \left(\frac{56}{99}\right)^2 \\ 2xy = \frac{(56 + 43)(56 - 43)}{99^2} &= \frac{13}{99} \end{align*}

Thus, the area of the octagon is $1 - \frac{13}{99} = \frac{86}{99}$, so $m + n = \boxed{185}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 