Question

Triangle $ABC$ lies in the Cartesian plane and has area $70.$ The coordinates of $B$ and $C$ are $(12, 19)$ and $(23, 20),$ respectively, and the coordinates of $A$ are $(p, q).$ The line containing the median to side $\overline{BC}$ has slope $-5.$ Find the largest possible value of $p + q.$

答案

Solution:

The median to $\overline{BC}$ passes through the midpoint $M = \left(\frac{35}{2}, \frac{39}{2}\right)$ of $\overline{BC}.$ The line through $M$ with slope $-5$ is $y = -5x + 107,$ and $A$ lies on this line, so $A = (p, -5p + 107)$ and $q = -5p + 107.$

By the shoelace formula with $B = (12, 19)$ and $C = (23, 20),$ $[ABC] = \frac{1}{2}\left|{-p} + 12\bigl(20 - q\bigr) + 23\bigl(q - 19\bigr)\right| = \frac{1}{2}\left|980 - 56p\right| = 70,$ so $|56p - 980| = 140,$ giving $p = 15$ or $p = 20.$

Since $p + q = p + (-5p + 107) = 107 - 4p,$ the smaller value $p = 15$ gives the larger sum $107 - 60 = 47.$

2005 AIME I Problem 10

Solution:

Problem 10 in Other Years