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Difference between revisions of "2016 AMC 10A Problems/Problem 19"

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<math>\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20</math>
 
<math>\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20</math>
  
−
== Solution 1==
+
== Solution 1 (Similar Triangles)==
  
 
<asy>
 
<asy>
Line 24: Line 24:
 
</asy>
 
</asy>
  
−
Use similar triangles. Our goal is to put the ratio in terms of <math>{BD}. Since </math>\triangle APD \sim \triangle EPB,<math> </math>\frac{DP}{PB}=\frac{AD}{BE}=3.<math> Similarly, </math>\frac{DQ}{QB}=\frac{3}{2}<math>. This means that </math>{DQ}=\frac{3\cdot BD}{5}<math>. As </math>\triangle ADP<math> and </math>\triangle BEP<math> are similar, we see that </math>\frac{PD}{PB}=\frac{3}{1}<math>. Thus </math>PB=\frac{BD}{4}<math>. Therefore, </math>r:s:t=\frac{1}{4}:\frac{2}{5}-\frac{1}{4}:\frac{3}{5}=5:3:12,<math> so </math>r+s+t=\boxed{\textbf{(E) }20.}$
+
Use similar triangles. Our goal is to put the ratio in terms of <math>{BD}</math>. Since <math>\triangle APD \sim \triangle EPB,</math> <math>\frac{DP}{PB}=\frac{AD}{BE}=3.</math> Therefore, <math>PB=\frac{BD}{4}</math>. Similarly, <math>\frac{DQ}{QB}=\frac{3}{2}</math>. This means that <math>{DQ}=\frac{3\cdot BD}{5}</math>. Therefore, <math>r:s:t=\frac{1}{4}:\frac{2}{5}-\frac{1}{4}:\frac{3}{5}=5:3:12,</math> so <math>r+s+t=\boxed{\textbf{(E) }20.}</math>
  
−
==Solution 2==
+
==Solution 2 (Mass points and Similar Triangles - Easy)==
 +
<asy>
 +
size(9cm);
 +
pair D=(0,0), C=(6,0), B=(6,3), A=(0,3), G=(2, 1), H=(9, 0);
 +
draw(A--B--C--D--cycle);
 +
draw(B--D);
 +
draw(A--(6,2));
 +
draw(A--(6,1));
 +
draw(A--H);
 +
draw((6,1)--G);
 +
draw(D--H);
 +
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(45));
 +
label("$E$", (6,2), dir(45));
 +
label("$H$", H, dir(0));
 +
label("$G$", G, dir(135));
 +
</asy>
  
−
Coordinate Bash:
+
This problem breaks down into finding <math>QP:PB</math> and <math>DQ:QB</math>. We can find the first using mass points, and the second using similar triangles.
−
We can set coordinates for the points. <math>D=(0,0), C=(6,0),  B=(6,3),</math> and <math>A=(0,3)</math>. The line <math>BD</math>'s equation is <math>y = \frac{1}{2}x</math>, line <math>AE</math>'s equation is <math>y = -\frac{1}{6}x + 3</math>, and line <math>AF</math>'s equation is <math>y = -\frac{1}{3}x + 3</math>. Adding the equations of lines <math>BD</math> and <math>AE</math>, we find that the coordinates of <math>P</math> are <math>(\frac{9}{2},\frac{9}{4})</math>. Furthermore we find that the coordinates of <math>Q</math> are <math>(\frac{18}{5}, \frac{9}{5})</math>. Using the [[Pythagorean Theorem]], we get that the length of <math>QD</math> is <math>\sqrt{(\frac{18}{5})^2+(\frac{9}{5})^2} = \sqrt{\frac{405}{25}} = \frac{\sqrt{405}}{5} = \frac{9\sqrt{5}}{5}</math>, and the length of <math>DP</math> is <math>\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}.</math> <math>PQ = DP - DQ = \frac{9\sqrt{5}}{5} - \frac{9\sqrt{5}}{4} = \frac{9\sqrt{5}}{20}.</math> The length of <math>DB = \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt{5}</math>. Then <math>BP= 3\sqrt{5} - \frac{9\sqrt{5}}{4} = \frac{3\sqrt{5}}{4}.</math> The ratio <math>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.</math> Then <math>r, s,</math> and <math>t</math> is <math>5, 3,</math> and <math>12</math>, respectively. The problem tells us to find <math>r + s + t</math>, so <math>5 + 3 + 12 = \boxed{\textbf{(E) }20}</math>
 
  
−
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.
+
Draw point <math>G</math> on <math>DB</math> such that <math>FG\parallel CD</math>. Then, by similar triangles <math>FG=BF\cdot 2=4</math>. Again, by similar triangles <math>AQB</math> and <math>FQG</math>, <math>AQ:FQ=AB:FG=6:4=3:2</math>. Now we begin Mass Points.
  
−
==Solution 3==
+
We will consider the triangle <math>ABF</math> with center <math>P</math>, so that <math>E</math> balances <math>B</math> and <math>F</math>, and <math>Q</math> balances <math>A</math> and <math>F</math>. Assign a mass of <math>1</math> to <math>B</math>. Then, <math>BE:FE=1:1</math> so <math>F=B=1</math>. By mass points addition, <math>E=B+F=2</math> since <math>E</math> balances <math>B</math> and <math>F</math>.
 +
Also, <math>AQ:QF=3:2</math> so <math>A=\frac{2}{3}F=\frac{2}{3}</math> so <math>Q=\frac{5}{3}</math>. Then, <math>QP:PB=1:\frac{5}{3}=3:5</math>.
 +
 
 +
To calculate <math>DQ:BQ</math>, extend <math>AF</math> past <math>F</math> to point <math>H</math> such that <math>H</math> lies on <math>BC</math>. Then <math>AFB</math> is similar to <math>HAD</math> so <math>DH=3\cdot AD=9</math>. Also, <math>AQB</math> is similar to <math>HQD</math> so <math>DQ:BQ=9:6=3:2</math>
 +
 
 +
Now, we wish to get <math>DQ:QP:PB</math>. Observe that <math>BQ=QP+PB</math>. So, <math>DQ:BQ=DQ:QP+PB=3:2</math> so (since <math>QP:PB=3:5</math> has sum <math>3+5=8</math>), <math>DQ:QP+PB=3:2=12:8</math>. now, we may combine the two and get <math>DQ:QP:PB=12:3:5</math> so <math>r+s+t=12+3+5=\boxed{\textbf{(E) }20}</math>.
 +
 
 +
~Firebolt360(minor edits by vadava_lx)
 +
 
 +
==Solution 3(Coordinate Bash)==
 +
 
 +
We can set coordinates for the points. <math>D=(0,0), C=(6,0),  B=(6,3),</math> and <math>A=(0,3)</math>. The line <math>BD</math>'s equation is <math>y = \frac{1}{2}x</math>, line <math>AE</math>'s equation is <math>y = -\frac{1}{6}x + 3</math>, and line <math>AF</math>'s equation is <math>y = -\frac{1}{3}x + 3</math>. Adding the equations of lines <math>BD</math> and <math>AE</math>, we find that the coordinates of <math>P</math> are <math>\left(\frac{9}{2},\frac{9}{4}\right)</math>. Furthermore we find that the coordinates of <math>Q</math> are <math>\left(\frac{18}{5}, \frac{9}{5}\right)</math>. Using the [[Pythagorean Theorem]], we get that the length of <math>QD</math> is <math>\sqrt{\left(\frac{18}{5}\right)^2+\left(\frac{9}{5}\right)^2} = \sqrt{\frac{405}{25}} = \frac{\sqrt{405}}{5} = \frac{9\sqrt{5}}{5}</math>, and the length of <math>DP</math> is <math>\sqrt{\left(\frac{9}{2}\right)^2+\left(\frac{9}{4}\right)^2} = \sqrt{\frac{81}{4} + \frac{81}{16}} = \sqrt{\frac{405}{16}} = \frac{\sqrt{405}}{4} = \frac{9\sqrt{5}}{4}.</math> <math>PQ = DP - DQ = \frac{9\sqrt{5}}{4} - \frac{9\sqrt{5}}{5} = \frac{9\sqrt{5}}{20}.</math> The length of <math>DB = \sqrt{6^2 + 3^2} = \sqrt{45} = 3\sqrt{5}</math>. Then <math>BP= 3\sqrt{5} - \frac{9\sqrt{5}}{4} = \frac{3\sqrt{5}}{4}.</math> The ratio <math>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.</math> Then <math>r, s,</math> and <math>t</math> is <math>5, 3,</math> and <math>12</math>, respectively. The problem tells us to find <math>r + s + t</math>, so <math>5 + 3 + 12 = \boxed{\textbf{(E) }20}</math> ~ minor <math>\LaTeX</math> edits by dolphin7
 +
 
 +
==Solution 4==
  
 
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==
+
==Solution 5 (Mass Points)==
  
−
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>
 
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>
  
−
==Solution 5 (Cheap Solution)==
+
==Solution 6 (Easy Coord Bash)==
−
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 <math>r:s:t</math> being <math>\frac{5}{3}:1:\frac{12}{3}</math>, multiplying each side by three the result is <math>r+s+t = 5+3+12 = \boxed{\textbf{(E) }20}</math>
+
We set coordinates for the points. Let <math>A=(0,3),  B=(6,3), C=(6,0)</math> and <math>D=(0,0)</math>. Then the equation of line <math>AE</math> is <math>y = -\frac{1}{6}x + 3,</math> the equation of line <math>AF</math> is <math>y = -\frac{1}{3}x + 3,</math> and the equation of line <math>BD</math> is <math>y = \frac{1}{2}x</math>. We find that the x-coordinate of point <math>P</math> is <math>\frac 9 2</math> by solving <math> -\frac{1}{6}x + 3=\frac{1}{2}x.</math> Similarly we find that the x-coordinate of point <math>Q</math> is <math>\frac {18} 5</math> by solving <math>-\frac{1}{3}x + 3=\frac{1}{2}x.</math> It follows that <math>BP:PQ:QD=6-\frac 9 2 : \frac 9 2 - \frac {18} 5 : \frac {18} 5= \frac 3 2 : \frac 9 {10} : \frac {18} 5 = 5:3:12.</math> Hence <math>r,s,t=5,3,12</math> and <math>r+s+t=5+3+12=\boxed{\textbf{(E) } 20}.</math> ~ Solution by dolphin7
 +
 
 +
==Video Solution==
 +
 
 +
https://www.youtube.com/watch?v=aG9JiBMd0ag
 +
 
 +
==Video Solution 2==
 +
https://youtu.be/us3e2jMjWnw
 +
 
 +
~IceMatrix
 +
 
 +
== Video Solution 3 by OmegaLearn ==
 +
https://youtu.be/4_x1sgcQCp4?t=3406
 +
 
 +
~ pi_is_3.14
  
 
==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}}

Latest revision as of 17:49, 3 October 2024

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 (Similar Triangles)

[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. Our goal is to put the ratio in terms of ${BD}$. Since $\triangle APD \sim \triangle EPB,$ $\frac{DP}{PB}=\frac{AD}{BE}=3.$ Therefore, $PB=\frac{BD}{4}$. Similarly, $\frac{DQ}{QB}=\frac{3}{2}$. This means that ${DQ}=\frac{3\cdot BD}{5}$. 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 (Mass points and Similar Triangles - Easy)

[asy] size(9cm); pair D=(0,0), C=(6,0), B=(6,3), A=(0,3), G=(2, 1), H=(9, 0); draw(A--B--C--D--cycle); draw(B--D); draw(A--(6,2)); draw(A--(6,1)); draw(A--H); draw((6,1)--G); draw(D--H); 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(45)); label("$E$", (6,2), dir(45)); label("$H$", H, dir(0)); label("$G$", G, dir(135)); [/asy]

This problem breaks down into finding $QP:PB$ and $DQ:QB$. We can find the first using mass points, and the second using similar triangles.

Draw point $G$ on $DB$ such that $FG\parallel CD$. Then, by similar triangles $FG=BF\cdot 2=4$. Again, by similar triangles $AQB$ and $FQG$, $AQ:FQ=AB:FG=6:4=3:2$. Now we begin Mass Points.

We will consider the triangle $ABF$ with center $P$, so that $E$ balances $B$ and $F$, and $Q$ balances $A$ and $F$. Assign a mass of $1$ to $B$. Then, $BE:FE=1:1$ so $F=B=1$. By mass points addition, $E=B+F=2$ since $E$ balances $B$ and $F$. Also, $AQ:QF=3:2$ so $A=\frac{2}{3}F=\frac{2}{3}$ so $Q=\frac{5}{3}$. Then, $QP:PB=1:\frac{5}{3}=3:5$.

To calculate $DQ:BQ$, extend $AF$ past $F$ to point $H$ such that $H$ lies on $BC$. Then $AFB$ is similar to $HAD$ so $DH=3\cdot AD=9$. Also, $AQB$ is similar to $HQD$ so $DQ:BQ=9:6=3:2$

Now, we wish to get $DQ:QP:PB$. Observe that $BQ=QP+PB$. So, $DQ:BQ=DQ:QP+PB=3:2$ so (since $QP:PB=3:5$ has sum $3+5=8$), $DQ:QP+PB=3:2=12:8$. now, we may combine the two and get $DQ:QP:PB=12:3:5$ so $r+s+t=12+3+5=\boxed{\textbf{(E) }20}$.

~Firebolt360(minor edits by vadava_lx)

Solution 3(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 $\left(\frac{9}{2},\frac{9}{4}\right)$. Furthermore we find that the coordinates of $Q$ are $\left(\frac{18}{5}, \frac{9}{5}\right)$. Using the Pythagorean Theorem, we get that the length of $QD$ is $\sqrt{\left(\frac{18}{5}\right)^2+\left(\frac{9}{5}\right)^2} = \sqrt{\frac{405}{25}} = \frac{\sqrt{405}}{5} = \frac{9\sqrt{5}}{5}$, and the length of $DP$ is $\sqrt{\left(\frac{9}{2}\right)^2+\left(\frac{9}{4}\right)^2} = \sqrt{\frac{81}{4} + \frac{81}{16}} = \sqrt{\frac{405}{16}} = \frac{\sqrt{405}}{4} = \frac{9\sqrt{5}}{4}.$ $PQ = DP - DQ = \frac{9\sqrt{5}}{4} - \frac{9\sqrt{5}}{5} = \frac{9\sqrt{5}}{20}.$ 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 $5 + 3 + 12 = \boxed{\textbf{(E) }20}$ ~ minor $\LaTeX$ edits by dolphin7

Solution 4

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 5 (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 6 (Easy Coord Bash)

We set coordinates for the points. Let $A=(0,3), B=(6,3), C=(6,0)$ and $D=(0,0)$. Then the equation of line $AE$ is $y = -\frac{1}{6}x + 3,$ the equation of line $AF$ is $y = -\frac{1}{3}x + 3,$ and the equation of line $BD$ is $y = \frac{1}{2}x$. We find that the x-coordinate of point $P$ is $\frac 9 2$ by solving $-\frac{1}{6}x + 3=\frac{1}{2}x.$ Similarly we find that the x-coordinate of point $Q$ is $\frac {18} 5$ by solving $-\frac{1}{3}x + 3=\frac{1}{2}x.$ It follows that $BP:PQ:QD=6-\frac 9 2 : \frac 9 2 - \frac {18} 5 : \frac {18} 5= \frac 3 2 : \frac 9 {10} : \frac {18} 5 = 5:3:12.$ Hence $r,s,t=5,3,12$ and $r+s+t=5+3+12=\boxed{\textbf{(E) } 20}.$ ~ Solution by dolphin7

Video Solution

https://www.youtube.com/watch?v=aG9JiBMd0ag

Video Solution 2

https://youtu.be/us3e2jMjWnw

~IceMatrix

Video Solution 3 by OmegaLearn

https://youtu.be/4_x1sgcQCp4?t=3406

~ pi_is_3.14

See Also

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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All AMC 10 Problems and Solutions

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