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Difference between revisions of "2008 AIME I Problems/Problem 2"

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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==
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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>.
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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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label("\(10\)",(M+E)/2,S);
 
label("\(10\)",(M+E)/2,S);
 
</asy></center>
 
</asy></center>
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Let <math>GE</math> meet <math>AI</math> at <math>X</math> and let <math>GM</math> meet <math>AI</math> at <math>Y</math>. Clearly, <math>XY=6</math> since the area of [[trapezoid]] <math>XYME</math> is <math>80</math>. Also, <math>\triangle GXY \sim \triangle GEM</math>. Let the height of <math>GXY</math> be <math>h</math>.
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Let <math>GE</math> meet <math>AI</math> at <math>X</math> and let <math>GM</math> meet <math>AI</math> at <math>Y</math>. Clearly, <math>XY=6</math> since the area of [[trapezoid]] <math>XYME</math> is <math>80</math>. Also, <math>\triangle GXY \sim \triangle GEM</math>.
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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>.
 
  
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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>.
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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>.
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}
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this could have been a #1-#5 on the amc 10 lol

Latest revision as of 18:37, 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 $h + 10 = \boxed{025}$.

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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