2007 Cyprus MO/Lyceum/Problem 15

Problem

2007 CyMO-15.PNG

The reflex angles of the concave octagon $ABCDEFGH$ measure $240^\circ$ each. Diagonals $AE$ and $GC$ are perpendicular, bisect each other, and are both equal to $2$.

The area of the octagon is

$\mathrm{(A) \ } \frac{6-2\sqrt{3}}{3}\qquad \mathrm{(B) \ } 8\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } \frac{6+2\sqrt{3}}{3}\qquad \mathrm{(E) \ } \mathrm{None\;of\;these}$

Solution

The problem statement apparently misses one crucial piece of information: the fact that all the sides of the octagon are equal. Without this fact the octagon area is not uniquely determined. For example, we could move point $B$ along a suitable arc (the locus of all points $X$ such that $AXC$ is $120^\circ$), and as this would change the height from $B$ to $AC$, it would change the area of the triangle $ABC$, and hence the area of the octagon.

With this additional assumption we can compute the area of $ABCDEFGH$ as the area of the square $ACEG$ (which is obviously $2$), minus four times the area of $ABC$.

[asy] unitsize(4cm); defaultpen(0.8); pair a=(0,1), c=(1,0), bb=(a+c)/2, b=bb+dir(225)/sqrt(6); draw (a -- b -- c -- cycle); draw (a -- (0,0) -- c); draw (b -- bb); label ("$A$", a, N ); label ("$B$", b, SW ); label ("$C$", c, S ); label ("$B'$", bb, NE ); label ("$1$", a/2, W ); [/asy]

In the triangle $ABC$, we have $AC=\sqrt 2$. Let $B'$ be the foot of the height from $B$ onto $AC$. As $AB=BC$, $B'$ bisects $AC$. As the angle $ABC$ is $120^\circ$, the angle $ABB'$ is $60^\circ$.

We now have $\frac{AB'}{BB'} = \tan 60^\circ = \sqrt 3$. Hence $BB'=\frac{AB'}{\sqrt 3} = \frac{1}{\sqrt 6}$.

Then the area of triangle $ABC$ is $\frac{BB' \cdot AC}2 = \frac{\sqrt 2}{2\sqrt 6} = \frac 1{2\sqrt 3} = \frac{\sqrt 3}6$.

Hence the area of the given octogon is $2 - 4\cdot \frac{\sqrt 3}6 = \boxed{\frac{6 - 2\sqrt 3}3}$.

See also

2007 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 14
Followed by
Problem 16
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