AIME 1999 · 第 4 题

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,

$$\begin{aligned}x + y + \frac{43}{99} &= 1\\ x^2 + 2xy + y^2 &= \left(\frac{56}{99}\right)^2\end{aligned}$$ Substituting,

$$\begin{aligned}\left(\frac{43}{99}\right)^2 + 2xy &= \left(\frac{56}{99}\right)^2 \\ 2xy = \frac{(56 + 43)(56 - 43)}{99^2} &= \frac{13}{99} \end{aligned}$$ Thus, the area of the octagon is $1 - \frac{13}{99} = \frac{86}{99}$, so $m + n = \boxed{185}$.

Diagram