Difference between revisions of "2001 AIME II Problems/Problem 4"

 
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== Problem ==
 
== Problem ==
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Let <math>R = (8,6)</math>. The lines whose equations are <math>8y = 15x</math> and <math>10y = 3x</math> contain points <math>P</math> and <math>Q</math>, respectively, such that <math>R</math> is the [[midpoint]] of <math>\overline{PQ}</math>. The length of <math>PQ</math> equals <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
  
 
== Solution ==
 
== Solution ==
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<center><asy>
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pointpen = black; pathpen = black+linewidth(0.7);
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pair R = (8,6), P = (32,60)/7, Q= (80,24)/7;
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D((0,0)--MP("x",(13,0),E),EndArrow(6)); D((0,0)--MP("y",(0,10),N),EndArrow(6));
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D((0,0)--(10/(15/8),10),EndArrow(6)); D((0,0)--(13,13 * 3/10),EndArrow(6));
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D(D(MP("P",P,NW))--D(MP("Q",Q),SE),linetype("4 4")); D(MP("R",R,NE));
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</asy></center>
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The coordinates of <math>P</math> can be written as <math>\left(8a, 15a\right)</math> and the coordinates of point <math>Q</math> can be written as <math>\left(10b, 3b\right)</math>. By the midpoint formula, we have <math>8a+10b=16</math> and <math>15a+3b=12</math>. Substituting <math>b=4-5a</math> we derive <math>a = \frac{4}{7}</math> and <math>b = \frac{8}{7}</math>. Thus <math>P</math> is <math>\left(\frac{32}7, \frac{60}7\right)</math>, Q is <math>\left(\frac{80}7, \frac{24}7\right)</math>, and the coordinates form a 3-4-5 triangle dilated by <math>\frac{12}7</math>. Finally the distance <math>PQ</math> must be <math>\frac{60}{7}</math> so the answer is <math>\boxed{067}</math>.
  
 
== See also ==
 
== See also ==
* [[2001 AIME II Problems]]
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{{AIME box|year=2001|n=II|num-b=3|num-a=5}}
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[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 04:01, 14 October 2025

Problem

Let $R = (8,6)$. The lines whose equations are $8y = 15x$ and $10y = 3x$ contain points $P$ and $Q$, respectively, such that $R$ is the midpoint of $\overline{PQ}$. The length of $PQ$ equals $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution

[asy] pointpen = black; pathpen = black+linewidth(0.7); pair R = (8,6), P = (32,60)/7, Q= (80,24)/7; D((0,0)--MP("x",(13,0),E),EndArrow(6)); D((0,0)--MP("y",(0,10),N),EndArrow(6)); D((0,0)--(10/(15/8),10),EndArrow(6)); D((0,0)--(13,13 * 3/10),EndArrow(6)); D(D(MP("P",P,NW))--D(MP("Q",Q),SE),linetype("4 4")); D(MP("R",R,NE)); [/asy]

The coordinates of $P$ can be written as $\left(8a, 15a\right)$ and the coordinates of point $Q$ can be written as $\left(10b, 3b\right)$. By the midpoint formula, we have $8a+10b=16$ and $15a+3b=12$. Substituting $b=4-5a$ we derive $a = \frac{4}{7}$ and $b = \frac{8}{7}$. Thus $P$ is $\left(\frac{32}7, \frac{60}7\right)$, Q is $\left(\frac{80}7, \frac{24}7\right)$, and the coordinates form a 3-4-5 triangle dilated by $\frac{12}7$. Finally the distance $PQ$ must be $\frac{60}{7}$ so the answer is $\boxed{067}$.

See also

2001 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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