2006 Cyprus MO/Lyceum/Problem 22

Problem

$ABCD$ is rectangular and the points $K,L,M,N$ lie on the sides $AB, BC, CD, DA$ respectively so that $\frac{AK}{KB}=\frac{BL}{LC}=\frac{CM}{MD}=\frac{DN}{NA}=2$ . If $E_1$ is the area of $KLMN$ and $E_2$ is the area of the rectangle $ABCD$ , the ratio $\frac{E_1}{E_2}$ equals

A. $\frac{5}{9}$

B. $\frac{1}{3}$

C. $\frac{9}{5}$

D. $\frac{3}{5}$

E. None of these

Solution

Let $AB = CD = x$ , $BC = AD = y$ . Using the Pythagorean Theorem, $KM = \sqrt{\frac{x^2}{9} + y^2}$ , $LN = \sqrt{x^2 + \frac{y^2}{9}}$ . Using the formula $A = \frac{1}{2}d_1d_2$ for a rhombus, we get $\frac{1}{2}\sqrt{\left(x^2 + \frac{y^2}{9}\right)\left(x^2 + \frac{y^2}{9}\right)} = \frac{1}{2}\sqrt{\frac{x^4}{9} + \frac{y^4}{9} + \frac{82}{81}x^2y^2} = \frac{\sqrt{9x^4 + 9y^4 + 82x^2y^2}}{18}$ . Thus the ratio is $\frac{\sqrt{9x^4 + 9y^4 + 82x^2y^2}}{18xy}$ . There is no way we can simplify this further, and in fact we can plug in different values of $x,y$ to see that the answer is $\mathrm{E}$ .

Be careful not to just try a couple of simple examples like $ABCD$ being a square, where we will get the answer $5/9$ , which is incorrect in general.

2006 Cyprus MO, Lyceum (Problems)
Preceded by Problem 21	Followed by Problem 23
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2006 Cyprus MO/Lyceum/Problem 22

Problem

Solution

See also