Latest revision as of 23:21, 26 January 2023

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

$[asy] pair A,B,C,D,M,n,O,P; A=(0,42);B=(84,42);C=(84,0);D=(0,0);M=(0,21);n=(28,42);O=(12,18);P=(32,13); fill(C--D--P--cycle,blue); draw(A--B--C--D--cycle); draw(C--M); draw(D--n); draw(B--P); draw(D--P); label("$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 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

@@ Line 2: / Line 2: @@
 Rectangle <math>ABCD</math> has side lengths <math>AB=84</math> and <math>AD=42</math>. Point <math>M</math> is the midpoint of <math>\overline{AD}</math>, point <math>N</math> is the trisection point of <math>\overline{AB}</math> closer to <math>A</math>, and point <math>O</math> is the intersection of <math>\overline{CM}</math> and <math>\overline{DN}</math>. Point <math>P</math> lies on the quadrilateral <math>BCON</math>, and <math>\overline{BP}</math> bisects the area of <math>BCON</math>. Find the area of <math>\triangle CDP</math>.
-==Solution==
+==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);
+fill(C--D--P--cycle,blue);
 draw(A--B--C--D--cycle);
 draw(C--M);
@@ Line 31: / Line 31: @@
 label("56",(28,42)--(84,42),N);
 </asy>
-Impose a coordinate system on the diagram where point <math>D</math> is the origin. Therefore <math>A=(0,42)</math>, <math>B=(84,42)</math>, <math>C=(84,0)</math>, and <math>D=(0,0)</math>. Because <math>M</math> is a midpoint and <math>M</math> is a trisection point, <math>M=(0,21)</math> and <math>N=(28,42)</math>. The equation for line <math>DN</math> is <math>y=\frac{3}{2}x</math> and the equation for line <math>CM</math> is <math>\frac{1}{84}x+\frac{1}{21}y=1</math>, so their intersection, point <math>O</math>, is <math>(12,18)</math>. Using the shoelace formula on quadrilateral <math>BCON</math>, its area is <math>2184</math>. Therefore the area of triangle <math>BCP</math> is <math>\frac{2184}{2}</math> and the distance from <math>P</math> to line <math>PC</math> is <math>52</math> and its <math>x</math>-coordinate is <math>32</math>. Because <math>P</math> lies on the equation <math>\frac{1}{84}x+\frac{1}{21}y=1</math>, <math>P</math>'s <math>y</math>-coordinate is <math>13</math>, which is also the height of <math>CDP</math>. Therefore the area of <math>CDP</math> is <math>\frac{1}{2}\times13\times84=\boxed{546}</math>.
+Impose a coordinate system on the diagram where point <math>D</math> is the origin. Therefore <math>A=(0,42)</math>, <math>B=(84,42)</math>, <math>C=(84,0)</math>, and <math>D=(0,0)</math>. Because <math>M</math> is a midpoint and <math>N</math> is a trisection point, <math>M=(0,21)</math> and <math>N=(28,42)</math>. The equation for line <math>DN</math> is <math>y=\frac{3}{2}x</math> and the equation for line <math>CM</math> is <math>\frac{1}{84}x+\frac{1}{21}y=1</math>, so their intersection, point <math>O</math>, is <math>(12,18)</math>. Using the shoelace formula on quadrilateral <math>BCON</math>, or drawing diagonal <math>\overline{BO}</math> and using <math>\frac12bh</math>, we find that its area is <math>2184</math>. Therefore the area of triangle <math>BCP</math> is <math>\frac{2184}{2} = 1092</math>. Using <math>A = \frac 12 bh</math>, we get <math>2184 = 42h</math>. Simplifying, we get <math>h = 52</math>. This means that the x-coordinate of <math>P = 84 - 52 = 32</math>. Since P lies on <math>\frac{1}{84}x+\frac{1}{21}y=1</math>, you can solve and get that the y-coordinate of <math>P</math> is <math>13</math>. Therefore the area of <math>CDP</math> is <math>\frac{1}{2} \cdot 13 \cdot 84=\boxed{546}</math>.
+==Solution 2 (No Coordinates)==
+Since the problem tells us that segment <math>\overline{BP}</math> bisects the area of quadrilateral <math>BCON</math>, let us compute the area of <math>BCON</math> by subtracting the areas of <math>\triangle{AND}</math> and <math>\triangle{DOC}</math> from rectangle <math>ABCD</math>.
+To do this, drop altitude <math>\overline{OE}</math> onto side <math>\overline{DC}</math> and draw a segment <math>\overline{MQ}</math> parallel to <math>\overline{AN}</math> from side <math>\overline{AD}</math> to <math>\overline{ND}</math>. Since <math>M</math> is the midpoint of side <math>\overline{AD}</math>, <cmath>\overline{MQ}=14</cmath> Denote <math>\overline{OE}</math> as <math>a</math>. Noting that <math>\triangle{MOQ}~\triangle{COD}</math>, we can write the statement <cmath>\frac{\overline{DC}}{a}=\frac{\overline{MQ}}{21-a}</cmath> <cmath>\implies \frac{84}{a}=\frac{14}{21-a}</cmath> <cmath>\implies a=18</cmath> Using this information, the area of <math>\triangle{DOC}</math> and <math>\triangle{AND}</math> are <cmath>\frac{18\cdot 84}{2}=756</cmath> and <cmath>\frac{28\cdot 42}{2}=588</cmath> respectively. Thus, the area of quadrilateral <math>BCON</math> is <cmath>84\cdot 42-588-756=2184</cmath> Now, it is clear that point <math>P</math> lies on side <math>\overline{MC}</math>, so the area of <math>\triangle{BPC}</math> is <cmath>\frac{2184}{2}=1092</cmath> Given this, drop altitude <math>\overline{PF}</math> (let's call it <math>b</math>) onto <math>\overline{BC}</math>. Therefore, <cmath>\frac{42b}{2}=1092\implies b=52</cmath> From here, drop an altitude <math>\overline{PG}</math> onto <math>\overline{DC}</math>. Recognizing that <math>\overline{PF}=\overline{GC}</math> and that <math>\triangle{MDC}</math> and <math>\triangle{PGC}</math> are similar, we write <cmath>\frac{\overline{PG}}{\overline{GC}}=\frac{\overline{MD}}{\overline{DC}}</cmath> <cmath>\implies \frac{\overline{PG}}{52}=\frac{21}{84}</cmath> <cmath>\implies \overline{PG}=13</cmath> The area of <math>\triangle{CDP}</math> is given by <cmath>\frac{\overline{DC}\cdot \overline{PG}}{2}=\frac{84\cdot 13}{2}=\boxed{546}</cmath> ~blitzkrieg21 and jdong2006
 =See Also=
 {{AIME box|year=2017|n=II|num-b=9|num-a=11}}
 {{MAA Notice}}

2017 AIME II (Problems • Answer Key • Resources)
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Difference between revisions of "2017 AIME II Problems/Problem 10"

Latest revision as of 23:21, 26 January 2023

Contents

Problem

Solution 1

Solution 2 (No Coordinates)

See Also