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 20:37, 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 | ||
