2016 AMC 10A Problems/Problem 19

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 $3+5+12=\boxed{\textbf{(E) }20}$

Solution 5 (Cheap Solution)

Use your ruler (it is recommended you bring ruler and protractor to AMC10 tests) and accurately draw the diagram as one in solution 1, then measure the length of the segments, you should get a ratio of $r:s:t$ being $\frac{5}{3}:1:\frac{12}{3}$ , multiplying each side by three the result is $r+s+t = 5+3+12 = E(20)$

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

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