Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 7"
(→Problem) |
(sol) |
||
Line 10: | Line 10: | ||
B. <math>8</math> | B. <math>8</math> | ||
− | C. | + | C. <math>4</math> |
D. <math>3</math> | D. <math>3</math> | ||
Line 17: | Line 17: | ||
==Solution== | ==Solution== | ||
− | {{ | + | <math>\triangle EZ\Gamma</math> is a <math>30-60-90</math> [[right triangle]], so <math>Z\Gamma = \frac{\sqrt{3}}{\sqrt{3}} = 1</math>. Also <math>\angle ZE\Delta = 90 - 30 = 60^{\circ}</math>, so <math>\triangle ZE\Delta</math> also is a <math>30-60-90 \triangle</math>. Thus <math>\Delta Z = \sqrt{3} \cdot \sqrt{3} = 3</math>. Adding, <math>\Delta Z + Z\Gamma = 4</math>, and a side of <math>\triangle AB\Gamma</math> is <math>2 \Delta \Gamma = 8\ \mathrm{(B)}</math>. |
==See also== | ==See also== | ||
{{CYMO box|year=2006|l=Lyceum|num-b=6|num-a=8}} | {{CYMO box|year=2006|l=Lyceum|num-b=6|num-a=8}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] |
Revision as of 19:02, 19 October 2007
Problem
In the figure, is an equilateral triangle and
,
,
. If
, then the length of the side of the triangle
is
A.
B.
C.
D.
E.
Solution
is a
right triangle, so
. Also
, so
also is a
. Thus
. Adding,
, and a side of
is
.
See also
2006 Cyprus MO, Lyceum (Problems) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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 |