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Difference between revisions of "2008 AIME I Problems/Problem 2"
(→‎Solution)
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Square <math>AIME</math> has sides of length <math>10</math> units.  Isosceles triangle <math>GEM</math> has base <math>EM</math>, and the area common to triangle <math>GEM</math> and square <math>AIME</math> is <math>80</math> square units.  Find the length of the altitude to <math>EM</math> in <math>\triangle GEM</math>.
 
Square <math>AIME</math> has sides of length <math>10</math> units.  Isosceles triangle <math>GEM</math> has base <math>EM</math>, and the area common to triangle <math>GEM</math> and square <math>AIME</math> is <math>80</math> square units.  Find the length of the altitude to <math>EM</math> in <math>\triangle GEM</math>.
  
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== Solution ==
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== Solution 1==
  
 
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>.
 
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>.
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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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 +
==Solution 2==
 +
<center><asy>
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pair E=(0,0), M=(10,0), I=(10,10), A=(0,10);
 +
draw(A--I--M--E--cycle);
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pair G=(5,25);
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draw(G--E--M--cycle);
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label("\(G\)",G,N);
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label("\(A\)",A,NW);
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label("\(I\)",I,NE);
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label("\(M\)",M,NE);
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label("\(E\)",E,NW);
 +
label("\(10\)",(M+E)/2,S);
 +
</asy></center>
  
 
== See also ==
 
== See also ==

Revision as of 18:34, 29 December 2020

Problem

Square $AIME$ has sides of length $10$ units. Isosceles triangle $GEM$ has base $EM$, and the area common to triangle $GEM$ and square $AIME$ is $80$ square units. Find the length of the altitude to $EM$ in $\triangle GEM$.

Solution 1

Note that if the altitude of the triangle is at most $10$, then the maximum area of the intersection of the triangle and the square is $5\cdot10=50$. This implies that vertex G must be located outside of square $AIME$.

[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]

Let $GE$ meet $AI$ at $X$ and let $GM$ meet $AI$ at $Y$. Clearly, $XY=6$ since the area of trapezoid $XYME$ is $80$. Also, $\triangle GXY \sim \triangle GEM$.

Let the height of $GXY$ be $h$. By the similarity, $\dfrac{h}{6} = \dfrac{h + 10}{10}$, we get $h = 15$. Thus, the height of $GEM$ is $h + 10 = \boxed{025}$.

Solution 2

[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]

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. AMC logo.png

this could have been a #1-#5 on the amc 10 lol

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