AIME 2017 II · 第 10 题

Solution 1

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 segment $\overline{MQ}$ parallel to $\overline{AN}$ 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 $\triangle{MOQ}~\triangle{COD}$, 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 quadrilateral $BCON$ 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 and jdong2006

Solution 3 (Coord bash)

Let $D = (0,0), C = (84,0), B = (84,42), A = (0,42)$. We can also find that $M = (0,21)$ by the midpoint formula. Now, lets find $O$. $O$ is the intersection of $ND$ and $CM$. So, we just need to find the equation of those lines, and solve. For $DN$, we first find the slope to be $\dfrac{42}{28} = \dfrac{3}{2}$. And obviously, the y-int is just $0$. So, our equation is:

$$y=\dfrac{3}{2}x$$ . Now, for $CM$, we have a slope of:

$$\dfrac{21}{-84} = -\dfrac{1}{4}$$ , and the y-int is just $21$, so our equation is:

$$y=-\dfrac{1}{4}x+21$$ , solving for $x$ and $y$, we get: $(12,18)$. Now, we use the fact that $\overline{BP}$ bisects the area of $BCON$. By shoelace, we find the area of $BCON$ to be $2184$. Now, we will use $\triangle{BPC}$. We have that its area is $\dfrac{2184}{2} = 1092$. So, by the area of a triangle formula the height $BCP$ is $52$. So, the $x$ coordinate of $P$ is $84-52 = 32$. Then, we use the fact that $P$ lies on $CM$, so using the equation of line $CM$ (which we found to be $y=-\dfrac{1}{4}x+21$, plugging $x=32$, we get $y = 13$, so we have the coordinates of $P$ to be $(32,13)$, so the height of $PCD$ is $13$, so the area is $\dfrac{13\cdot84}{2}=\boxed{546}$

~jb2015007

Solution 4 (Menelaus, Similarity)

From the givens, we know that $DM=MA=21$, $AN=28$, and $BN=56$. Extend $BA$ to meet $CM$ at point $X$. Since $M$ is the midpoint of $AD$, $\triangle{AXM}$ and $\triangle{DCM}$ are congruent, so $AX=84$. From Menelaus' Theorem on $\triangle{DAN}$, we have

$$\dfrac{DM}{MA} \cdot \dfrac{AX}{NX}\cdot \dfrac{NO}{OD}=1.$$ After plugging in the known values, we have that $\frac{NO}{OD}=\frac{4}{3}$. Let point $Q$ be the point on $DA$ such that $\triangle{DQO}\sim\triangle{DAN}$. Then, because of the ratio we just found, we know that $QO=12$. Thus, if we let point $Q^{\prime}$ be the point on $BC$ such that $OQ^{\prime} \parallel AB$, we have $OQ^{\prime}=72.$ This means that $[COB]=\frac{1}{2}(72)(42).$ We can do the same process vertically to find that $[NOB]=\frac{1}{2}(24)(56)$. Summing, we have that $[BCON]=12\cdot 182$. Then, $[CPB]=6\cdot 182$. Since $BC=42$, the altitude from $P$ to $BC$ is $52$. Let the altitude from $P$ to $CD$ be $h$. Then, from similar triangles,

$$\dfrac{h}{21}=\dfrac{52}{84}.$$ This gives that $h=13$, and thus the area is $\frac{1}{2}(84)(13)=\boxed{546}$.

- The_Other_Guy