2007 Cyprus MO/Lyceum/Problem 15

Problem

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 $2 - 4\cdot \frac{\sqrt 3}6 = \boxed{\frac{6 - 2\sqrt 3}3}$ .

2007 Cyprus MO, Lyceum (Problems)
Preceded by Problem 14	Followed by Problem 16
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2007 Cyprus MO/Lyceum/Problem 15

Problem

Solution

See also