2023 WSMO Speed Round Problems/Problem 8

Problem

In regular octagon $ABCDEFGH$ of sidelength $4,$ quadrilaterals $ACEG$ and $BDFH$ are drawn. Find the square of the area of the overlap of the two quadrilaterals.

Solution

$[asy] pair a = dir(112.5); pair b = dir(67.5); pair c = dir(22.5); pair d = dir(337.5); pair e = dir(292.5); pair f = dir(247.5); pair g = dir(202.5); pair h = dir(157.5); label("$A$", a, NW); label("$B$", b, NE); label("$C$", c, E); label("$D$", d, E); label("$E$", e, SE); label("$F$", f, SW); label("$G$", g, W); label("$H$", h, W); draw(a--b--c--d--e--f--g--h--cycle,black+linewidth(2)); draw(a--c--e--g--cycle,blue+linewidth(1)); draw(b--d--f--h--cycle,blue+linewidth(1)); pair X = intersectionpoint(b--h,a--c); dot(X); label("$X$", X, S); pair Y = intersectionpoint(b--h,a--g); dot(Y); label("$Y$", Y, SE); [/asy]$

Let $X = BH\cap AC$ and $Y = BH\cap AG$ . Suppose that $BX = s$ . From symmetry, we have $AX=AY=HY=s$ . From the Pythagorean Theorem on $AXY,$ we have $XY = s\sqrt{2}$ . So, $BH = HY+YX+XB = s+s\sqrt{2}+s = s(2+\sqrt{2}),$ meaning $XY = s\sqrt{2} = \frac{BH\sqrt{2}}{2+\sqrt{2}} = \frac{BH}{1+\sqrt{2}}$ From the Law of Cosines on $ABH$ , we have $\begin{align*} BH &= \sqrt{AH^2+AB^2-2(AB)(AH)\cos(135^{\circ})}\\ &= \sqrt{4^2+4^2-2(4)(4)\left(-\frac{\sqrt{2}}{2}\right)}\\ &= \sqrt{32+16\sqrt{2}}. \end{align*}$ Now, from the formula of the area of an octagon, we have $\begin{align*} \text{area}&=(XY)^2(2+2\sqrt{2})\\ &=\frac{BH^2}{(1+\sqrt{2})^2}\cdot(2+2\sqrt{2})\\ &=\frac{32+16\sqrt{2}}{(1+\sqrt{2})^2}\cdot(2+2\sqrt{2})\\ &=16\sqrt{2}\cdot2 = 32\sqrt{2}, \end{align*}$ meaning our answer is $(32\sqrt{2})^2 = \boxed{2048}.$

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2023 WSMO Speed Round Problems/Problem 8

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