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Difference between revisions of "2001 AIME II Problems/Problem 4"
(fmt/asy)
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== Problem ==
 
== Problem ==
−
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>.
+
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 ==
−
The coordinates of P can be written as <math>(a, 15a/8)</math> and the coordinates of point Q can be written as <math>(b,3b/10)</math>. By the midpoint formula, we have <math>(a+b)/2=8</math> and <math>15a/16+3b/20=6</math>. Solving for b gives <math>b=80/7</math>, so the point Q is <math>(80/7,24/7)</math>. The answer is twice the distance from Q to <math>(8,6)</math>, which by the distance formula is <math>60/7</math>. Thus, the answer is 67.
+
<center><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></center>
 +
 
 +
The coordinates of <math>P</math> can be written as <math>\left(a, \frac{15a}8\right)</math> and the coordinates of point <math>Q</math> can be written as <math>\left(b,\frac{3b}{10}\right)</math>. By the midpoint formula, we have <math>\frac{a+b}2=8</math> and <math>\frac{15a}{16}+\frac{3b}{20}=6</math>. Solving for <math>b</math> gives <math>b= \frac{80}{7}</math>, so the point <math>Q</math> is <math>\left(\frac{80}7, \frac{24}7\right)</math>. The answer is twice the distance from <math>Q</math> to <math>(8,6)</math>, which by the distance formula is <math>\frac{60}{7}</math>. Thus, the answer is <math>\boxed{067}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2001|n=II|num-b=3|num-a=5}}
 
{{AIME box|year=2001|n=II|num-b=3|num-a=5}}
 +
 +
[[Category:Intermediate Geometry Problems]]

Revision as of 22:46, 25 July 2008

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 $\left(a, \frac{15a}8\right)$ and the coordinates of point $Q$ can be written as $\left(b,\frac{3b}{10}\right)$. By the midpoint formula, we have $\frac{a+b}2=8$ and $\frac{15a}{16}+\frac{3b}{20}=6$. Solving for $b$ gives $b= \frac{80}{7}$, so the point $Q$ is $\left(\frac{80}7, \frac{24}7\right)$. The answer is twice the distance from $Q$ to $(8,6)$, which by the distance formula is $\frac{60}{7}$. Thus, the answer is $\boxed{067}$.

See also

2001 AIME II (Problems • Answer Key • Resources)
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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