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

Revision as of 23:24, 18 January 2024

Problem

In the figure, $ABCD$ is an orthogonal trapezium with $\angle A= \angle D=90^\circ$ 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 2: / Line 2: @@
 [[Image:2007 CyMO-22.PNG|200px]]
-In the figure, <math>ABCD</math> is an orthogonal trapezium with <math>\ang A= \ang D=90^\circ</math> and bases <math>AB = a</math> , <math>DC = 2a</math> . If <math>AD = 3a</math> and <math>M</math> is the midpoint of the side <math>BC</math>, then <math>AM</math> equals to
+In the figure, <math>ABCD</math> is an orthogonal trapezium with <math>\angle A= \angle D=90^\circ</math> and bases <math>AB = a</math> , <math>DC = 2a</math> . If <math>AD = 3a</math> and <math>M</math> is the midpoint of the side <math>BC</math>, then <math>AM</math> equals to
 <math> \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</math>

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 23:24, 18 January 2024

Problem

Solution

See also