2007 Cyprus MO/Lyceum/Problem 22

Problem

2007 CyMO-22.PNG

In the figure, $ABCD$ is an orthogonal trapezium with $\ang A= \ang D=90^\circ$ (Error compiling LaTeX. ! Undefined control sequence.) 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 $\mathrm{B}$.

See also

2007 Cyprus MO, Lyceum (Problems)
Preceded by
Problem 21
Followed by
Problem 23
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Invalid username
Login to AoPS