Difference between revisions of "2008 AIME I Problems/Problem 2"
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By the similarity, <math>\dfrac{h}{6} = \dfrac{h + 10}{10}</math>, we get <math>h = 15</math>. Thus, the height of <math>GEM</math> is <math>h + 10 = \boxed{025}</math>. | By the similarity, <math>\dfrac{h}{6} = \dfrac{h + 10}{10}</math>, we get <math>h = 15</math>. Thus, the height of <math>GEM</math> is <math>h + 10 = \boxed{025}</math>. | ||
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+ | Note that we know that the altitude of the triangle is greater than ten because if it's not, the maximum intersection of the areas of the triangle and the square is <math>5\cdot 10=50</math>. | ||
== See also == | == See also == |
Revision as of 21:37, 24 March 2008
Problem
Square has sides of length
units. Isosceles triangle
has base
, and the area common to triangle
and square
is
square units. Find the length of the altitude to
in
.
Solution
![[asy] pair E=(0,0), M=(10,0), I=(10,10), A=(0,10); draw(A--I--M--E--cycle); pair G=(5,25); draw(G--E--M--cycle); label("\(G\)",G,N); label("\(A\)",A,NW); label("\(I\)",I,NE); label("\(M\)",M,NE); label("\(E\)",E,NW); label("\(10\)",(M+E)/2,S); [/asy]](http://latex.artofproblemsolving.com/6/2/1/621cda5459b0a04c3bb47fbec28d266740d0a08b.png)
Let meet
at
and let
meet
at
. Clearly,
since the area of trapezoid
is
. Also,
. Let the height of
be
.
By the similarity, , we get
. Thus, the height of
is
.
Note that we know that the altitude of the triangle is greater than ten because if it's not, the maximum intersection of the areas of the triangle and the square is .
See also
2008 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |