2001 AIME II Problems/Problem 4

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 $\boxed{067}$ .

2001 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2001 AIME II Problems/Problem 4

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