Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 22"
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==Solution== | ==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>\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>. |
==See also== | ==See also== | ||
{{CYMO box|year=2007|l=Lyceum|num-b=21|num-a=23}} | {{CYMO box|year=2007|l=Lyceum|num-b=21|num-a=23}} |
Revision as of 02:30, 7 June 2007
Problem
In the figure, is an orthogonal trapezium with $\ang A= \ang D=90^\circ$ (Error compiling LaTeX. Unknown error_msg) and bases
,
. If
and
is the midpoint of the side
, then
equals to
Solution
Let the midpoint of be
. The length of
is the average of the bases, or
. The length of
is also
. Since
is a 45-45-90 triangle, the length of
is
.
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 |