Question

Rectangle $ABCD$ has side lengths $AB = 84$ and $AD = 42.$ Point $M$ is the midpoint of $\overline{AD},$ point $N$ is the trisection point of $\overline{AB}$ closer to $A,$ and point $O$ is the intersection of $\overline{CM}$ and $\overline{DN}.$ Point $P$ lies on the quadrilateral $BCON,$ and $\overline{BP}$ bisects the area of $BCON.$ Find the area of $\triangle CDP.$

Answer

Solution:

Place $A = (0, 0),$ $B = (84, 0),$ $C = (84, 42),$ $D = (0, 42),$ so $M = (0, 21)$ and $N = (28, 0).$ Line $CM$ is $y = \frac{x}{4} + 21$ and line $DN$ is $y = 42 - \frac{3x}{2},$ which meet at $O = (12, 24).$

By the Shoelace Formula, quadrilateral $BCON$ has area $2184,$ so each half must have area $1092.$ Triangle $BCO$ alone has area $\frac{1}{2} \cdot 42 \cdot 72 = 1512 \gt 1092$ (base $\overline{BC},$ and $O$ is at horizontal distance $72$ from it), so the bisecting segment ends at a point $P$ on $\overline{CO}.$ For $[BPC] = 1092,$ the distance $d$ from $P$ to line $BC$ must satisfy $\frac{1}{2} \cdot 42 \cdot d = 1092,$ so $d = 52,$ giving $P$ the $x$ -coordinate $84 - 52 = 32.$ Since $P$ lies on line $CO,$ which is $y = \frac{x}{4} + 21,$ we get $P = (32, 29).$

Triangle $CDP$ has base $CD = 84$ on the line $y = 42$ and height $42 - 29 = 13,$ so its area is $\frac{1}{2} \cdot 84 \cdot 13 = 546.$

2017 AIME II Problem 10

Solution:

Problem 10 in Other Years