Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 22"

Revision as of 02:19, 24 April 2008

Problem

In the figure, $ABCD$ is an orthogonal trapezium with $\ang A= \ang D=90^\circ$ (Error compiling LaTeX. Unknown error_msg) and bases $AB = a$ , $DC = 2a$ . If $AD = 3a$ and $M$ is the midpoint of the side $BC$ , then $AM$ equals to

$\mathrm{(A) \ } \frac{3a}{2}\qquad \mathrm{(B) \ } \frac{3a}{\sqrt{2}}\qquad \mathrm{(C) \ } \frac{5a}{2}\qquad \mathrm{(D) \ } \frac{3a}{\sqrt{3}}\qquad \mathrm{(E) \ } 2a$

Solution

Let the midpoint of $AD$ be $N$ . The length of $MN$ is the average of the bases, or $\frac{3a}{2}$ . The length of $AN$ is also $\frac{3a}{2}$ .

Since $AMN$ is a $45-45-90$ triangle, the length of $AM$ is $\frac{3a}{\sqrt{2}}$ , and the answer is $\mathrm{B}$ .

@@ Line 7: / Line 7: @@
 ==Solution==
-Let the midpoint of <math>AD</math> be <math>N</math>. The length of <math>MN</math> is the average of the bases, or <math>\frac{3a}{2}</math>. The length of <math>AN</math> is also <math>\frac{3a}{2}</math>. Since <math>\triangle AMN</math> is a 45-45-90 triangle, the length of <math>AM</math> is <math>\frac{3a}{\sqrt{2}}\Longrightarrow\mathrm{B}</math>.
+Let the midpoint of <math>AD</math> be <math>N</math>. The length of <math>MN</math> is the average of the bases, or <math>\frac{3a}{2}</math>. The length of <math>AN</math> is also <math>\frac{3a}{2}</math>.
+Since <math>AMN</math> is a <math>45-45-90</math> triangle, the length of <math>AM</math> is <math>\frac{3a}{\sqrt{2}}</math>, and the answer is <math>\mathrm{B}</math>.
 ==See also==
 {{CYMO box|year=2007|l=Lyceum|num-b=21|num-a=23}}

2007 Cyprus MO, Lyceum (Problems)
Preceded by Problem 21	Followed by Problem 23
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Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 22"

Revision as of 02:19, 24 April 2008

Problem

Solution

See also