Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2017 AIME II Problems/Problem 10"

(→‎Solution 1)
m (→‎Solution 2 (No Coordinates))
Line 36: Line 36:
 
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>.
 
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 horizontal segment <math>\overline{MQ}</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
+
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} 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}$ is given by <cmath>\frac{\overline{DC}\cdot \overline{PG}}{2}=\frac{84\cdot 13}{2}=\boxed{546}</cmath> ~blitzkrieg21 and jdong2006
  
 
=See Also=
 
=See Also=
 
{{AIME box|year=2017|n=II|num-b=9|num-a=11}}
 
{{AIME box|year=2017|n=II|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

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}$, <cmath>\overline{MQ}=14</cmath> Denote$\overline{OE}$as$a$. Noting that$\triangle{MOQ}~\triangle{COD}$, we can write the statement <cmath>\frac{\overline{DC}}{a}=\frac{\overline{MQ}}{21-a}</cmath> <cmath>\implies \frac{84}{a}=\frac{14}{21-a}</cmath> <cmath>\implies a=18</cmath> Using this information, the area of$\triangle{DOC}$and$\triangle{AND}$are <cmath>\frac{18\cdot 84}{2}=756</cmath> and <cmath>\frac{28\cdot 42}{2}=588</cmath> respectively. Thus, the area of quadrilateral$BCON$is <cmath>84\cdot 42-588-756=2184</cmath> Now, it is clear that point$P$lies on side$\overline{MC}$, so the area of$\triangle{BPC}$is <cmath>\frac{2184}{2}=1092</cmath> Given this, drop altitude$\overline{PF}$(let's call it$b$) onto$\overline{BC}$. Therefore, <cmath>\frac{42b}{2}=1092\implies b=52</cmath> From here, drop an altitude$\overline{PG}$onto$\overline{DC}$. Recognizing that$\overline{PF}=\overline{GC}$and that$\triangle{MDC}$and$\triangle{PGC}$are similar, we write <cmath>\frac{\overline{PG}}{\overline{GC}}=\frac{\overline{MD}}{\overline{DC}}</cmath> <cmath>\implies \frac{\overline{PG}}{52}=\frac{21}{84}</cmath> <cmath>\implies \overline{PG}=13</cmath> The area of$\triangle{CDP}$ is given by \[\frac{\overline{DC}\cdot \overline{PG}}{2}=\frac{84\cdot 13}{2}=\boxed{546}\] ~blitzkrieg21 and jdong2006

See Also

2017 AIME II (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_10&oldid=188146"