Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 14"

Revision as of 23:02, 8 May 2007

Problem

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

@@ Line 8: / Line 8: @@
 ==Solution==
-{{solution}}
+<math>\triangle KBC</math> and <math>\triangle KCD</math> have the same heights (<math>\perp BD</math>), so the ratio of their areas is simply the ratio of <math>KD:KB</math>.
+Let <math>KB=x</math>. Then <math>KD=x\sqrt{2}-x</math>, and the ratio of <math>S_1:S_2</math> is <math>x\sqrt{2}-x:x</math>, or <math>\sqrt{2}-1:1\Longrightarrow\mathrm{ D}</math>.
 ==See also==
 {{CYMO box|year=2007|l=Lyceum|num-b=13|num-a=15}}

2007 Cyprus MO, Lyceum (Problems)
Preceded by Problem 13	Followed by Problem 15
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Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 14"

Revision as of 23:02, 8 May 2007

Problem

Solution

See also