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Difference between revisions of "2023 WSMO Speed Round Problems/Problem 8"

(Created page with "==Problem== In regular octagon <math>ABCDEFGH</math> of sidelength <math>4,</math> quadrilaterals <math>ACEG</math> and <math>BDFH</math> are drawn. Find the square of the ar...")
 
 
(One intermediate revision by the same user not shown)
Line 4: Line 4:
  
 
==Solution==
 
==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 <math>X = BH\cap AC</math> and <math>Y = BH\cap AG</math>. Suppose that <math>BX = s</math>. From symmetry, we have <math>AX=AY=HY=s</math>. From the Pythagorean Theorem on <math>AXY,</math> we have <math>XY = s\sqrt{2}</math>. So, <cmath>BH = HY+YX+XB = s+s\sqrt{2}+s = s(2+\sqrt{2}),</cmath> meaning <cmath>XY = s\sqrt{2} = \frac{BH\sqrt{2}}{2+\sqrt{2}} = \frac{BH}{1+\sqrt{2}}</cmath>
 +
From the Law of Cosines on <math>ABH</math>, we have
 +
<cmath>\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*}</cmath>
 +
Now, from the formula of the area of an octagon, we have
 +
<cmath>\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*}</cmath>
 +
meaning our answer is <cmath>(32\sqrt{2})^2 = \boxed{2048}.</cmath>

Latest revision as of 12:46, 12 September 2025

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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