2006 Cyprus MO/Lyceum/Problem 22

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Problem

2006 CyMO-22.PNG

$AB\Gamma \Delta$ is rectangular and the points $K,\Lambda ,M,N$ lie on the sides $AB, B\Gamma , \Gamma \Delta, \Delta A$ respectively so that $\frac{AK}{KB}=\frac{BL}{L\Gamma}=\frac{\Gamma M}{M\Delta}=\frac{\Delta N}{NA}=2$. If $E_1$ is the area of $K\Lambda MN$ and $E_2$ is the area of the rectangle $AB\Gamma \Delta$, the ratio $\frac{E_1}{E_2}$ equals

$\mathrm{(A)}\ \frac{5}{9}\qquad\mathrm{(B)}\ \frac{1}{3}\qquad\mathrm{(C)}\ \frac{9}{5}\qquad\mathrm{(D)}\ \frac{3}{5}\qquad\mathrm{(E)}\ \text{None of these}$

Solution

Let $AB = \Gamma \Delta = x$, $B\Gamma = A\Delta = y$. Using the Pythagorean Theorem, $KM = \sqrt{\frac{x^2}{9} + y^2}$, $\Lambda N = \sqrt{x^2 + \frac{y^2}{9}}$. 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 $AB\Gamma \Delta$ being a square, where we will get the answer $5/9$, which is incorrect in general.

See also

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