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Difference between revisions of "2016 AMC 12B Problems/Problem 17"

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Then <math>PQ  = \frac{8}{15} II' = </math> <cmath>\boxed{\textbf{(D)}\frac{8}{15}\sqrt{5}}</cmath>
 
Then <math>PQ  = \frac{8}{15} II' = </math> <cmath>\boxed{\textbf{(D)}\frac{8}{15}\sqrt{5}}</cmath>
  
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==Solution 3 (if you're running short on time)==
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==Solution 3 (FAST)==
 
<math>PQ</math> lies on altitude <math>AH</math>, which we find to have a length of <math>3\sqrt{5}</math> by Heron's Formula and dividing twice the area by <math>BC</math>. From H we can construct a segment <math>HX</math> with <math>X</math> on <math>CE</math> such that <math>HX</math> is parallel to <math>EB</math>. A similar construction gives <math>Y</math> on <math>BD</math> such that <math>HY</math> is parallel to <math>DC</math>. We can hence generate a system of ratios that will allow us to find <math>PQ/AH</math>. Note that such a system will generate a rational number for the ratio <math>PQ/AH</math>. Thus, we choose the only answer that has a <math>\sqrt{5}</math> term in it, giving us <math>\boxed{\textbf{(D)}} </math>.
 
<math>PQ</math> lies on altitude <math>AH</math>, which we find to have a length of <math>3\sqrt{5}</math> by Heron's Formula and dividing twice the area by <math>BC</math>. From H we can construct a segment <math>HX</math> with <math>X</math> on <math>CE</math> such that <math>HX</math> is parallel to <math>EB</math>. A similar construction gives <math>Y</math> on <math>BD</math> such that <math>HY</math> is parallel to <math>DC</math>. We can hence generate a system of ratios that will allow us to find <math>PQ/AH</math>. Note that such a system will generate a rational number for the ratio <math>PQ/AH</math>. Thus, we choose the only answer that has a <math>\sqrt{5}</math> term in it, giving us <math>\boxed{\textbf{(D)}} </math>.
  
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Let <math>h=AH</math> and <math>BH=x</math>.  Then, <math>CH=8-x</math>.  By the Pythagorean Theorem on right triangles <math>ABH</math> and <math>ACH</math>, we have <cmath>h^2+x^2=49</cmath> <cmath>x^2+(8-x)^2=81.</cmath>  Subtracting the prior from the latter yields <math>-16x+64=32\implies x=2</math>.  So, <math>BH=2</math>, <math>CH=6</math>, and <math>AH=3\sqrt{5}</math>.  Continue with Solution 1.
 
Let <math>h=AH</math> and <math>BH=x</math>.  Then, <math>CH=8-x</math>.  By the Pythagorean Theorem on right triangles <math>ABH</math> and <math>ACH</math>, we have <cmath>h^2+x^2=49</cmath> <cmath>x^2+(8-x)^2=81.</cmath>  Subtracting the prior from the latter yields <math>-16x+64=32\implies x=2</math>.  So, <math>BH=2</math>, <math>CH=6</math>, and <math>AH=3\sqrt{5}</math>.  Continue with Solution 1.
 +
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=ccB-z4_OHqw
 +
 +
==Video Solution by CanadaMath (Problem 11-20)==
 +
Fast Forward to 26:14 for problem 17
 +
https://www.youtube.com/watch?v=4osvFatUv1o
 +
 +
~THEMATHCANADIAN
 +
 +
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=B|num-b=16|num-a=18}}
 
{{AMC12 box|year=2016|ab=B|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:26, 10 November 2024

Problem

In $\triangle ABC$ shown in the figure, $AB=7$, $BC=8$, $CA=9$, and $\overline{AH}$ is an altitude. Points $D$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB}$, respectively, so that $\overline{BD}$ and $\overline{CE}$ are angle bisectors, intersecting $\overline{AH}$ at $Q$ and $P$, respectively. What is $PQ$?

[asy] import graph; size(9cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.381056062031275, xmax = 15.020004395092375, ymin = -4.051697595316909, ymax = 10.663513514111651; /* image dimensions */ draw((0.,0.)--(4.714285714285714,7.666518779999279)--(7.,0.)--cycle); /* draw figures */ draw((0.,0.)--(4.714285714285714,7.666518779999279)); draw((4.714285714285714,7.666518779999279)--(7.,0.)); draw((7.,0.)--(0.,0.)); label("7",(3.2916797119724284,-0.07831656949355523),SE*labelscalefactor); label("9",(2.0037562070503783,4.196493361737088),SE*labelscalefactor); label("8",(6.114150371695219,3.785453945272603),SE*labelscalefactor); draw((0.,0.)--(6.428571428571427,1.9166296949998194)); draw((7.,0.)--(2.2,3.5777087639996634)); draw((4.714285714285714,7.666518779999279)--(3.7058823529411766,0.)); /* dots and labels */ dot((0.,0.),dotstyle); label("$A$", (-0.2432592696221352,-0.5715638692509372), NE * labelscalefactor); dot((7.,0.),dotstyle); label("$B$", (7.0458397156813835,-0.48935598595804014), NE * labelscalefactor); dot((3.7058823529411766,0.),dotstyle); label("$E$", (3.8123296394941084,0.16830708038513573), NE * labelscalefactor); dot((4.714285714285714,7.666518779999279),dotstyle); label("$C$", (4.579603216894479,7.895848109917452), NE * labelscalefactor); dot((2.2,3.5777087639996634),linewidth(3.pt) + dotstyle); label("$D$", (2.1407693458718726,3.127790878929427), NE * labelscalefactor); dot((6.428571428571427,1.9166296949998194),linewidth(3.pt) + dotstyle); label("$H$", (6.004539860638023,1.9494778850645704), NE * labelscalefactor); dot((5.,1.49071198499986),linewidth(3.pt) + dotstyle); label("$Q$", (4.935837377830365,1.7302568629501784), NE * labelscalefactor); dot((3.857142857142857,1.1499778169998918),linewidth(3.pt) + dotstyle); label("$P$", (3.538303361851119,1.2370095631927964), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ \frac{5}{8}\sqrt{3} \qquad \textbf{(C)}\ \frac{4}{5}\sqrt{2} \qquad \textbf{(D)}\ \frac{8}{15}\sqrt{5} \qquad \textbf{(E)}\ \frac{6}{5}$

Solution 1

Get the area of the triangle by Heron's Formula: \[\sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(12)(3)(4)(5)} = 12\sqrt{5}\] Use the area to find the height $AH$ with known base $BC$: \[Area = 12\sqrt{5} = \frac{1}{2}bh = \frac{1}{2}(8)(AH)\] \[AH = 3\sqrt{5}\] \[BH = \sqrt{AB^2 - AH^2} = \sqrt{7^2 - (3\sqrt{5})^2} = 2\] \[CH = BC - BH = 8 - 2 = 6\] Apply the Angle Bisector Theorem on $\triangle ACH$ and $\triangle ABH$, we get $AP:PH = 9:6$ and $AQ:QH = 7:2$, respectively. To find $AP$, $PH$, $AQ$, and $QH$, apply variables, such that $AP:PH = 9:6$ is $\frac{3\sqrt{5} - x}{x} = \frac{9}{6}$ and $AQ:QH = 7:2$ is $\frac{3\sqrt{5} - y}{y} = \frac{7}{2}$. Solving them out, you will get $AP = \frac{9\sqrt{5}}{5}$, $PH = \frac{6\sqrt{5}}{5}$, $AQ = \frac{7\sqrt{5}}{3}$, and $QH = \frac{2\sqrt{5}}{3}$. Then, since $AP + PQ = AQ$ according to the Segment Addition Postulate, and thus manipulating, you get $PQ = AQ - AP = \frac{7\sqrt{5}}{3} - \frac{9\sqrt{5}}{5}$ = \[\boxed{\textbf{(D)}\frac{8}{15}\sqrt{5}}\]

Solution 2

Let the intersection of $BD$ and $CE$ be the point $I$. Then let the foot of the altitude from $I$ to $BC$ be $I'$. Note that $II'$ is an inradius and that $II' \cdot s = [ABC]$, where $s$ is the semiperimeter of the triangle.

Using Heron's Formula, we see that $II' \cdot 12 = \sqrt{12 \cdot 3 \cdot 4 \cdot 5} = 12\sqrt{5}$, so $II' = \sqrt{5}$.

Then since $II'$ and $AH$ are parallel, $\triangle CI'I \sim \triangle CHP$ and $\triangle BHQ \sim \triangle BI'I$.

Thus, $\frac{II'}{PQ + QH} = \frac{CI'}{CH}$ and $\frac{II'}{QH} = \frac{BI'}{BH}$, so $PQ = \frac{II' \cdot CH}{CI'} - \frac{II' \cdot BH}{BI'}$.

By the Dual Principle, $CI' = 5$ and $BI' = 3$. With the same method as Solution 1, $CH = 6$ and $BH = 2$. Then $PQ = \frac{8}{15} II' =$ \[\boxed{\textbf{(D)}\frac{8}{15}\sqrt{5}}\]

Solution 3 (FAST)

$PQ$ lies on altitude $AH$, which we find to have a length of $3\sqrt{5}$ by Heron's Formula and dividing twice the area by $BC$. From H we can construct a segment $HX$ with $X$ on $CE$ such that $HX$ is parallel to $EB$. A similar construction gives $Y$ on $BD$ such that $HY$ is parallel to $DC$. We can hence generate a system of ratios that will allow us to find $PQ/AH$. Note that such a system will generate a rational number for the ratio $PQ/AH$. Thus, we choose the only answer that has a $\sqrt{5}$ term in it, giving us $\boxed{\textbf{(D)}}$.

Solution 4

Let $h=AH$ and $BH=x$. Then, $CH=8-x$. By the Pythagorean Theorem on right triangles $ABH$ and $ACH$, we have \[h^2+x^2=49\] \[x^2+(8-x)^2=81.\] Subtracting the prior from the latter yields $-16x+64=32\implies x=2$. So, $BH=2$, $CH=6$, and $AH=3\sqrt{5}$. Continue with Solution 1.

Video Solution

https://www.youtube.com/watch?v=ccB-z4_OHqw

Video Solution by CanadaMath (Problem 11-20)

Fast Forward to 26:14 for problem 17 https://www.youtube.com/watch?v=4osvFatUv1o

~THEMATHCANADIAN


See Also

2016 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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All AMC 12 Problems and Solutions

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