2017 AIME II Problems/Problem 10

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 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

See Also

2017 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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