Difference between revisions of "2008 AIME I Problems/Problem 2"
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==Solution 2== | ==Solution 2== | ||
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== See also == | == See also == | ||
Revision as of 18:35, 29 December 2020
Contents
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 1
Note that if the altitude of the triangle is at most , then the maximum area of the intersection of the triangle and the square is 
.
This implies that vertex G must be located outside of square .
![[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 
.
Solution 2
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
this could have been a #1-#5 on the amc 10 lol
