2001 AIME II Problems/Problem 4

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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

The coordinates of P can be written as $(a, 15a/8)$ and the coordinates of point Q can be written as $(b,3b/10)$. By the midpoint formula, we have $(a+b)/2=8$ and $15a/16+3b/20=6$. Solving for b gives $b=80/7$, so the point Q is $(80/7,24/7)$. The answer is twice the distance from Q to $(8,6)$, which by the distance formula is $60/7$. Thus, the answer is 67.

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