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 "2016 AMC 10A Problems/Problem 19"

m
Line 36: Line 36:
  
 
Extend <math>AF</math> to meet <math>CD</math> at point <math>T</math>. Since <math>FC=1</math> and <math>BF=2</math>, <math>TC=3</math> by similar triangles <math>\triangle TFC</math> and <math>\triangle AFB</math>. It follows that <math>\frac{BQ}{QD}=\frac{BP+PQ}{QD}=\frac{2}{3}</math>. Now, using similar triangles <math>\triangle BEP</math> and <math>\triangle DAP</math>, <math>\frac{BP}{PD}=\frac{BP}{PQ+QD}=\frac{1}{3}</math>. WLOG let <math>BP=1</math>. Solving for <math>PQ, QD</math> gives <math>PQ=\frac{3}{5}</math> and <math>QD=\frac{12}{5}</math>. So our desired ratio is <math>5:3:12</math> and <math>5+3+12=\boxed{\textbf{(E) } 20}</math>.
 
Extend <math>AF</math> to meet <math>CD</math> at point <math>T</math>. Since <math>FC=1</math> and <math>BF=2</math>, <math>TC=3</math> by similar triangles <math>\triangle TFC</math> and <math>\triangle AFB</math>. It follows that <math>\frac{BQ}{QD}=\frac{BP+PQ}{QD}=\frac{2}{3}</math>. Now, using similar triangles <math>\triangle BEP</math> and <math>\triangle DAP</math>, <math>\frac{BP}{PD}=\frac{BP}{PQ+QD}=\frac{1}{3}</math>. WLOG let <math>BP=1</math>. Solving for <math>PQ, QD</math> gives <math>PQ=\frac{3}{5}</math> and <math>QD=\frac{12}{5}</math>. So our desired ratio is <math>5:3:12</math> and <math>5+3+12=\boxed{\textbf{(E) } 20}</math>.
 +
 +
==Solution 4==
 +
 +
Mass Points:
 +
Draw line segment <math>AC</math>, and call the intersection between <math>AC</math> and <math>BD</math> point <math>K</math>. In <math>\delta ABC</math>, observe that <math>BE:EC=1:2</math> and <math>AK:KC=1:1</math>. Using mass points, find that <math>BP:PK=1:1</math>. Again utilizing <math>\delta ABC</math>, observe that <math>BF:FC=2:1</math> and <math>AK:KC=1:1</math>. Use mass points to find that <math>BQ:QK=4:1</math>. Now, draw a line segment with points <math>B</math>,<math>P</math>,<math>Q</math>, and <math>K</math> ordered from left to right. Set the values <math>BP=x</math>,<math>PK=x</math>,<math>BQ=4y</math> and <math>QK=y</math>. Setting both sides segment <math>BK</math> equal, we get <math>y= \frac{2}{5}x</math>. Plugging in and solving gives <math>QK= \frac{2}{5}x</math>, <math>PQ=\frac{3}{5}x</math>,<math>BP=x</math>. The question asks for <math>BP:PQ:QD</math>, so we add <math>2x</math> to <math>QK</math> and multiply the ratio by <math>5</math> to create integers. This creates <math>5(1:\frac{3}{5}:\frac{12}{5})= 5:3:12</math>. This sums up to <math>3+5+12=\boxed{\textbf{(E) }20}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=A|num-b=18|num-a=20}}
 
{{AMC10 box|year=2016|ab=A|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:16, 18 December 2017

Problem

In rectangle $ABCD,$ $AB=6$ and $BC=3$. Point $E$ between $B$ and $C$, and point $F$ between $E$ and $C$ are such that $BE=EF=FC$. Segments $\overline{AE}$ and $\overline{AF}$ intersect $\overline{BD}$ at $P$ and $Q$, respectively. The ratio $BP:PQ:QD$ can be written as $r:s:t$ where the greatest common factor of $r,s,$ and $t$ is $1.$ What is $r+s+t$?

$\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20$

Solution 1

[asy] size(6cm); pair D=(0,0), C=(6,0), B=(6,3), A=(0,3); draw(A--B--C--D--cycle); draw(B--D); draw(A--(6,2)); draw(A--(6,1)); label("$A$", A, dir(135)); label("$B$", B, dir(45)); label("$C$", C, dir(-45)); label("$D$", D, dir(-135)); label("$Q$", extension(A,(6,1),B,D),dir(-90)); label("$P$", extension(A,(6,2),B,D), dir(90)); label("$F$", (6,1), dir(0)); label("$E$", (6,2), dir(0)); [/asy]

Use similar triangles. Since $\triangle APD \sim \triangle EPB,$ $\frac{DP}{PB}=\frac{AD}{BE}=3.$ Similarly, $\frac{DQ}{QB}=\frac{3}{2}$. This means that ${DQ}=\frac{3\cdot BD}{5}$. As $\triangle ADP$ and $\triangle BEP$ are similar, we see that $\frac{PD}{PB}=\frac{3}{1}$. Thus $PB=\frac{BD}{4}$. Therefore, $r:s:t=\frac{1}{4}:\frac{2}{5}-\frac{1}{4}:\frac{3}{5}=5:3:12,$ so $r+s+t=\boxed{\textbf{(E) }20.}$

Solution 2

Coordinate Bash: We can set coordinates for the points. $D=(0,0), C=(6,0), B=(6,3),$ and $A=(0,3)$. The line $BD$'s equation is $y = \frac{1}{2}x$, line $AE$'s equation is $y = -\frac{1}{6}x + 3$, and line $AF$'s equation is $y = -\frac{1}{3}x + 3$. Adding the equations of lines $BD$ and $AE$, we find that the coordinates of $P$ are $(\frac{9}{2},\frac{9}{4})$. Furthermore we find that the coordinates of $Q$ are $(\frac{18}{5}, \frac{9}{5})$. Using the Pythagorean Theorem, we get that the length of $QD$ is $\sqrt{(\frac{18}{5})^2+(\frac{9}{5})^2} = \sqrt{\frac{405}{25}} = \frac{\sqrt{405}}{5} = \frac{9\sqrt{5}}{5}$, and the length of $DP$ is $\sqrt{(\frac{9}{2})^2+(\frac{9}{4})^2} = \sqrt{\frac{81}{4} + \frac{81}{16}} = \sqrt{\frac{405}{16}} = \frac{\sqrt{405}}{4} = \frac{9\sqrt{5}}{4}.$ $PQ = DP - DQ = \frac{9\sqrt{5}}{5} - \frac{9\sqrt{5}}{4} = \frac{9\sqrt{5}}{20}.$ The length of $DB = \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt{5}$. Then $BP= 3\sqrt{5} - \frac{9\sqrt{5}}{4} = \frac{3\sqrt{5}}{4}.$ The ratio $BP : PQ : QD = \frac{3\sqrt{5}}{4} : \frac{9\sqrt{5}}{20} : \frac{9\sqrt{5}}{5} = 15\sqrt{5} : 9\sqrt{5} : 36\sqrt{5} = 15 : 9 : 36 = 5 : 3 : 12.$ Then $r, s,$ and $t$ is $5, 3,$ and $12$, respectively. The problem tells us to find $r + s + t$, so $5 + 3 + 12 = \boxed{\textbf{(E) }20}$

An alternate solution is to perform the same operations, but only solve for the x-coordinates. By similar triangles, the ratios will be the same.

Solution 3

Extend $AF$ to meet $CD$ at point $T$. Since $FC=1$ and $BF=2$, $TC=3$ by similar triangles $\triangle TFC$ and $\triangle AFB$. It follows that $\frac{BQ}{QD}=\frac{BP+PQ}{QD}=\frac{2}{3}$. Now, using similar triangles $\triangle BEP$ and $\triangle DAP$, $\frac{BP}{PD}=\frac{BP}{PQ+QD}=\frac{1}{3}$. WLOG let $BP=1$. Solving for $PQ, QD$ gives $PQ=\frac{3}{5}$ and $QD=\frac{12}{5}$. So our desired ratio is $5:3:12$ and $5+3+12=\boxed{\textbf{(E) } 20}$.

Solution 4

Mass Points: Draw line segment $AC$, and call the intersection between $AC$ and $BD$ point $K$. In $\delta ABC$, observe that $BE:EC=1:2$ and $AK:KC=1:1$. Using mass points, find that $BP:PK=1:1$. Again utilizing $\delta ABC$, observe that $BF:FC=2:1$ and $AK:KC=1:1$. Use mass points to find that $BQ:QK=4:1$. Now, draw a line segment with points $B$,$P$,$Q$, and $K$ ordered from left to right. Set the values $BP=x$,$PK=x$,$BQ=4y$ and $QK=y$. Setting both sides segment $BK$ equal, we get $y= \frac{2}{5}x$. Plugging in and solving gives $QK= \frac{2}{5}x$, $PQ=\frac{3}{5}x$,$BP=x$. The question asks for $BP:PQ:QD$, so we add $2x$ to $QK$ and multiply the ratio by $5$ to create integers. This creates $5(1:\frac{3}{5}:\frac{12}{5})= 5:3:12$. This sums up to $3+5+12=\boxed{\textbf{(E) }20}$

See Also

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 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=2016_AMC_10A_Problems/Problem_19&oldid=88965"