Difference between revisions of "2008 AIME I Problems/Problem 2"
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== Solution == | == Solution == | ||
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+ | Note that if the altitude of the triangle is at most <math>10</math>, then the maximum area of the intersection of the triangle and the square is <math>5\cdot10=50</math>. | ||
+ | This implies that vertex G must be located outside of square <math>AIME</math>. | ||
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<center><asy> | <center><asy> | ||
pair E=(0,0), M=(10,0), I=(10,10), A=(0,10); | pair E=(0,0), M=(10,0), I=(10,10), A=(0,10); | ||
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Let the height of <math>GXY</math> be <math>h</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>. | Let the height of <math>GXY</math> be <math>h</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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== See also == | == See also == |
Revision as of 22:34, 30 July 2018
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
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 .
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 .
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.