Problem

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 $\triangle CDP$ .

Solution 1

$[asy] pair A,B,C,D,M,n,O,P; A=(0,42);B=(84,42);C=(84,0);D=(0,0);M=(0,21);n=(28,42);O=(12,18);P=(32,13); fill(C--D--P--cycle,lightgray); draw(A--B--C--D--cycle); draw(C--M); draw(D--n); draw(B--P); draw(D--P); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$M$",M,W); label("$N$",n,N); label("$O$",O,(-0.5,1)); label("$P$",P,N); dot(A); dot(B); dot(C); dot(D); dot(M); dot(n); dot(O); dot(P); label("28",(0,42)--(28,42),N); label("56",(28,42)--(84,42),N); [/asy]$ Impose a coordinate system on the diagram where point $D$ is the origin. Therefore $A=(0,42)$ , $B=(84,42)$ , $C=(84,0)$ , and $D=(0,0)$ . Because $M$ is a midpoint and $N$ is a trisection point, $M=(0,21)$ and $N=(28,42)$ . The equation for line $DN$ is $y=\frac{3}{2}x$ and the equation for line $CM$ is $\frac{1}{84}x+\frac{1}{21}y=1$ , so their intersection, point $O$ , is $(12,18)$ . Using the shoelace formula on quadrilateral $BCON$ , or drawing diagonal $\overline{BO}$ and using $\frac12bh$ , we find that its area is $2184$ . Therefore the area of triangle $BCP$ is $\frac{2184}{2} = 1092$ . Using $A = \frac 12 bh$ , we get $2184 = 42h$ . Simplifying, we get $h = 52$ . This means that the x-coordinate of $P = 84 - 52 = 32$ . Since P lies on $\frac{1}{84}x+\frac{1}{21}y=1$ , you can solve and get that the y-coordinate of $P$ is $13$ . Therefore the area of $CDP$ is $\frac{1}{2} \cdot 13 \cdot 84=\boxed{546}$ .

Solution Altered By conantwiz2023 and IronicNinja~

Solution 2 (No Coordinates)

Since the problem tells us that segment $\overline{BP}$ bisects the area of quadrilateral $BCON$ , let us compute the area of $BCON$ by subtracting the areas of $\triangle{AND}$ and $\triangle{DOC}$ from rectangle $ABCD$ . To do this, drop altitude $\overline{OE}$ onto side $\overline{DC}$ and draw a horizontal segment $\overline{MQ}$ from side $\overline{AD}$ to $\overline{ND}$ . Since $M$ is the midpoint of side $\overline{AD}$ , $\overline{MQ}=14$ . Denote $\overline{OE}$ as $a$ . Noting that $\triangle{MOQ}$ and $\triangle{COD}$ are similar, we can write the statement $\frac{\overline{DC}}{a}=\frac{\overline{MQ}}{21-a}\implies \frac{84}{a}=\frac{14}{21-a}\implies a=18$ . Using this information, the area of $\triangle{DOC}$ and $\triangle{AND}$ are $\frac{18\cdot 84}{2}=756$ and $\frac{28\cdot 42}{2}=588$ , respectively. Thus, the area of quadrilaterial $BCON$ is $84\cdot 42-588-756=2184$ . Now, it is clear that point $P$ lies on side $\overline{MC}$ , so the area of $\triangle{BPC}$ is $\frac{2184}{2}=1092$ . Given this, drop altitude $\overline{PF}$ (let's call it $b$ ) onto $\overline{BC}$ . Therefore, $\frac{42b}{2}=1092\implies b=52$ . From here, drop an altitude $\overline{PG}$ onto $\overline{DC}$ . Recognizing that $\overline{PF}=\overline{GC}$ and that $\triangle{MDC}$ and $\triangle{PGC}$ are similar, we write $\frac{\overline{PG}}{\overline{GC}}=\frac{\overline{MD}}{\overline{DC}}\implies \frac{\overline{PG}}{52}=\frac{21}{84}\implies \overline{PG}=13$ . The area of $\triangle{CDP}$ is given by $\frac{\overline{DC}\cdot \overline{PG}}{2}=\frac{84\cdot 13}{2}=\boxed{546}$ ~blitzkrieg21

2017 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2017 AIME II Problems/Problem 10

Contents

Problem

Solution 1

Solution 2 (No Coordinates)

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