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 $ACEG$ (which is obviously $2$ ), minus four times the area of $ABC$ .

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 $60^\circ$ .

(to be finished in a few hours)

This problem needs a solution. If you have a solution for it, please help us out by adding it.

2007 Cyprus MO, Lyceum (Problems)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30

Page

Toolbox

Search

2007 Cyprus MO/Lyceum/Problem 15

Problem

Solution

See also