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

 
 
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== Problem ==
 
== Problem ==
−
Let <math> ABC </math> be a triangle with sides 3, 4, and 5, and <math> DEFG </math> be a 6-by-7 rectangle. A segment is drawn to divide triangle <math> ABC </math> into a triangle <math> U_1 </math> and a trapezoid <math> V_1 </math> and another segment is drawn to divide rectangle <math> DEFG </math> into a triangle <math> U_2 </math> and a trapezoid <math> V_2 </math> such that <math> U_1 </math> is similar to <math> U_2 </math> and <math> V_1 </math> is similar to <math> V_2. </math> The minimum value of the area of <math> U_1 </math> can be written in the form <math> m/n, </math> where <math> m </math> and <math> n </math>are relatively prime positive integers. Find <math> m+n. </math>
+
Let <math> ABC </math> be a [[triangle]] with sides 3, 4, and 5, and <math> DEFG </math> be a 6-by-7 [[rectangle]]. A segment is drawn to divide triangle <math> ABC </math> into a triangle <math> U_1 </math> and a trapezoid <math> V_1 </math> and another segment is drawn to divide rectangle <math> DEFG </math> into a triangle <math> U_2 </math> and a trapezoid <math> V_2 </math> such that <math> U_1 </math> is similar to <math> U_2 </math> and <math> V_1 </math> is similar to <math> V_2. </math> The minimum value of the area of <math> U_1 </math> can be written in the form <math> m/n, </math> where <math> m </math> and <math> n</math> are [[relatively prime]] positive integers. Find <math> m+n. </math>
  
 
== Solution ==
 
== Solution ==
 +
We let <math>AB=3, AC=4, DE=6, DG=7</math> for the purpose of labeling. Clearly, the dividing segment in <math>DEFG</math> must go through one of its vertices, [[without loss of generality]] <math>D</math>. The other endpoint (<math>D'</math>) of the segment can either lie on <math>\overline{EF}</math> or <math>\overline{FG}</math>. <math>V_2</math> is a trapezoid with a right angle then, from which it follows that <math>V_1</math> contains one of the right angles of <math>\triangle ABC</math>, and so <math>U_1</math> is similar to <math>ABC</math>. Thus <math>U_1</math>, and hence <math>U_2</math>, are <math>3-4-5\,\triangle</math>s. 
 +
 +
Suppose we find the ratio <math>r</math> of the smaller base to the larger base for <math>V_2</math>, which consequently is the same ratio for <math>V_1</math>. By similar triangles, it follows that <math>U_1 \sim \triangle ABC</math> by the same ratio <math>r</math>, 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 <math>[U_1] = r^2 \cdot [ABC] = 6r^2</math>.
 +
<center><table><tr><td>
 +
 +
<asy> defaultpen(linewidth(0.7));
 +
pair A=(0,0),B=(0,3),C=(4,0);
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draw(MP("A",A)--MP("B",B,N)--MP("C",C)--cycle);
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draw((9*4/14,0)--(9*4/14,5*3/14),dashed);
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label("\(V_1\)",(1,1.2)); label("\(U_1\)",(3,0.3));
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</asy>
 +
<asy> defaultpen(linewidth(0.7)); pointpen = black;
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pair D=(0,0),E=(0,6),F=(7,6),G=(7,0),H=(4.5,6);
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draw(MP("D",D)--MP("E",E,N)--MP("F",F,N)--MP("G",G)--cycle);
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draw(D--D(MP("D'",H,N)),dashed);
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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);
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</asy>
 +
 +
</td></tr><tr><td>
 +
 +
<asy> defaultpen(linewidth(0.7));
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pair A=(0,0),B=(0,3),C=(4,0);
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draw(MP("A",A)--MP("B",B,N)--MP("C",C)--cycle);
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draw((3.5,0)--(3.5,3/8),dashed);
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label("\(V_1\)",(1.5,1)); label("\(U_1\)",(3.8,0.4));
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</asy>
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<asy> defaultpen(linewidth(0.7));  pointpen = black;
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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);
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draw(D--D(MP("D'",H,NW)),dashed);
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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));
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</asy>
 +
 +
</td></tr></table></center>
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 +
*If <math>D'</math> lies on <math>\overline{EF}</math>, then <math>ED' = \frac{9}{2},\, 8</math>; the latter can be discarded as extraneous. Therefore, <math>D'F = \frac{5}{2}</math>, and the ratio <math>r = \frac{D'F}{DG} = \frac{5}{14}</math>. The area of <math>[U_1] = 6\left(\frac{5}{14}\right)^2 </math> in this case.
 +
 +
*If <math>D'</math> lies on <math>\overline{FG}</math>, then <math>GD' = \frac{21}{4},\, \frac{28}{3}</math>; the latter can be discarded as extraneous. Therefore, <math>D'F = \frac{3}{4}</math>, and the ratio <math>r = \frac{D'F}{DE} = \frac{1}{8}</math>. The area of <math>[U_1] = 6\left(\frac{1}{8}\right)^2</math> in this case.
 +
 +
Of the two cases, the second is smaller; the answer is <math>\frac{3}{32}</math>, and <math>m+n = \boxed{035}</math>.
  
 
== See also ==
 
== See also ==
−
* [[2004 AIME I Problems]]
+
{{AIME box|year=2004|n=I|num-b=8|num-a=10}}
 +
{{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)
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Problem 8 		Followed by
Problem 10
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