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Difference between revisions of "2004 AIME I Problems/Problem 9"

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draw(MP("A",A)--MP("B",B,N)--MP("C",C)--cycle);
 
draw(MP("A",A)--MP("B",B,N)--MP("C",C)--cycle);
 
draw((3.5,0)--(3.5,3/8),dashed);
 
draw((3.5,0)--(3.5,3/8),dashed);
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label("\(U_1\)",(1.5,1)); label("\(V_1\)",(3.8,0.4));
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label("\(V_1\)",(1.5,1)); label("\(U_1\)",(3.8,0.4));
 
</asy>
 
</asy>
 
<asy> defaultpen(linewidth(0.7));  pointpen = black;
 
<asy> defaultpen(linewidth(0.7));  pointpen = black;
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draw(MP("D",D)--MP("E",E,N)--MP("F",F,N)--MP("G",G)--cycle);
 
draw(MP("D",D)--MP("E",E,N)--MP("F",F,N)--MP("G",G)--cycle);
 
draw(D--D(MP("D'",H,NW)),dashed);
 
draw(D--D(MP("D'",H,NW)),dashed);
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label("\(U_2\)",(2,3)); label("\(V_2\)",(5,1)); MP("7",(D+G)/2,S); MP("6",(D+E)/2,W); MP("21/4",(G+H)/2,(-1,0));
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label("\(V_2\)",(2,3)); label("\(U_2\)",(5,1)); MP("7",(D+G)/2,S); MP("6",(D+E)/2,W); MP("21/4",(G+H)/2,(-1,0));
 
</asy>
 
</asy>
  
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== See also ==
 
== See also ==
 
{{AIME box|year=2004|n=I|num-b=8|num-a=10}}
 
{{AIME box|year=2004|n=I|num-b=8|num-a=10}}
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{{MAA Notice}}

Latest revision as of 19:01, 4 July 2013

Problem

Let $ABC$ be a triangle with sides 3, 4, and 5, and $DEFG$ be a 6-by-7 rectangle. A segment is drawn to divide triangle $ABC$ into a triangle $U_1$ and a trapezoid $V_1$ and another segment is drawn to divide rectangle $DEFG$ into a triangle $U_2$ and a trapezoid $V_2$ such that $U_1$ is similar to $U_2$ and $V_1$ is similar to $V_2.$ The minimum value of the area of $U_1$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution

We let $AB=3, AC=4, DE=6, DG=7$ for the purpose of labeling. Clearly, the dividing segment in $DEFG$ must go through one of its vertices, without loss of generality $D$. The other endpoint ($D'$) of the segment can either lie on $\overline{EF}$ or $\overline{FG}$. $V_2$ is a trapezoid with a right angle then, from which it follows that $V_1$ contains one of the right angles of $\triangle ABC$, and so $U_1$ is similar to $ABC$. Thus $U_1$, and hence $U_2$, are $3-4-5\,\triangle$s.

Suppose we find the ratio $r$ of the smaller base to the larger base for $V_2$, which consequently is the same ratio for $V_1$. By similar triangles, it follows that $U_1 \sim \triangle ABC$ by the same ratio $r$, and since the ratio of the areas of two similar figures is equal to the square of the ratio of their corresponding lengths, it follows that $[U_1] = r^2 \cdot [ABC] = 6r^2$.

[asy] defaultpen(linewidth(0.7)); pair A=(0,0),B=(0,3),C=(4,0); draw(MP("A",A)--MP("B",B,N)--MP("C",C)--cycle); draw((9*4/14,0)--(9*4/14,5*3/14),dashed); label("\(V_1\)",(1,1.2)); label("\(U_1\)",(3,0.3)); [/asy] [asy] defaultpen(linewidth(0.7)); pointpen = black; pair D=(0,0),E=(0,6),F=(7,6),G=(7,0),H=(4.5,6); draw(MP("D",D)--MP("E",E,N)--MP("F",F,N)--MP("G",G)--cycle); draw(D--D(MP("D'",H,N)),dashed); label("\(U_2\)",(1,3)); label("\(V_2\)",(5,3)); MP("7",(D+G)/2,S); MP("6",(D+E)/2,W); MP("9/2",(E+H)/2,N); [/asy]

[asy] defaultpen(linewidth(0.7)); pair A=(0,0),B=(0,3),C=(4,0); draw(MP("A",A)--MP("B",B,N)--MP("C",C)--cycle); draw((3.5,0)--(3.5,3/8),dashed); label("\(V_1\)",(1.5,1)); label("\(U_1\)",(3.8,0.4)); [/asy] [asy] defaultpen(linewidth(0.7)); pointpen = black; pair D=(0,0),E=(0,6),F=(7,6),G=(7,0),H=(7,21/4); draw(MP("D",D)--MP("E",E,N)--MP("F",F,N)--MP("G",G)--cycle); draw(D--D(MP("D'",H,NW)),dashed); label("\(V_2\)",(2,3)); label("\(U_2\)",(5,1)); MP("7",(D+G)/2,S); MP("6",(D+E)/2,W); MP("21/4",(G+H)/2,(-1,0)); [/asy]

  • If $D'$ lies on $\overline{EF}$, then $ED' = \frac{9}{2},\, 8$; the latter can be discarded as extraneous. Therefore, $D'F = \frac{5}{2}$, and the ratio $r = \frac{D'F}{DG} = \frac{5}{14}$. The area of $[U_1] = 6\left(\frac{5}{14}\right)^2$ in this case.
  • If $D'$ lies on $\overline{FG}$, then $GD' = \frac{21}{4},\, \frac{28}{3}$; the latter can be discarded as extraneous. Therefore, $D'F = \frac{3}{4}$, and the ratio $r = \frac{D'F}{DE} = \frac{1}{8}$. The area of $[U_1] = 6\left(\frac{1}{8}\right)^2$ in this case.

Of the two cases, the second is smaller; the answer is $\frac{3}{32}$, and $m+n = \boxed{035}$.

See also

2004 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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