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

(→‎Solution: add note)
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label("\(10\)",(M+E)/2,S);
 
label("\(10\)",(M+E)/2,S);
 
</asy></center>
 
</asy></center>
−
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>.
+
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>.
  
−
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>.
  
−
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>.
+
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>.
  
 
== See also ==
 
== See also ==

Revision as of 14:32, 19 April 2008

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

[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 $h + 10 = \boxed{025}$.

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 $5\cdot10=50$.

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
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