2007 Cyprus MO/Lyceum/Problem 14

Problem

2007 CyMO-14.PNG

In square $ABCD$ the segment $KB$ equals a side of the square. The ratio of areas $\frac{S_1}{S_2}$ is

$\mathrm{(A) \ } \frac{1}{3}\qquad \mathrm{(B) \ } \frac{2}{3}\qquad \mathrm{(C) \ } \frac{1}{\sqrt{2}}\qquad \mathrm{(D) \ } \sqrt2-1\qquad \mathrm{(E) \ } \frac{\sqrt{2}}4$

Solution

$\triangle KBC$ and $\triangle KCD$ have the same heights ($\perp BD$), so the ratio of their areas is simply the ratio of $KD:KB$.

Let the length of $KB$ be $x$. Then $KD=x\sqrt{2}-x$, and the ratio of $S_1:S_2$ is $x\sqrt{2}-x:x$, or $\sqrt{2}-1:1\Longrightarrow\mathrm{ D}$.

See also

2007 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 13
Followed by
Problem 15
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