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

(Solution)
(Solution)
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Solution 2: (choices+faster)
 
Solution 2: (choices+faster)
 
As in the above solution, let's find the area. By heron's, it is indeed <math>12\sqrt{5}</math>. Because <math>\overline {AH}</math> is an altitude and we are given a base, let's find <math>\overline {AH}</math>. We set up the area is equal to <math>\frac {1}{2}
 
As in the above solution, let's find the area. By heron's, it is indeed <math>12\sqrt{5}</math>. Because <math>\overline {AH}</math> is an altitude and we are given a base, let's find <math>\overline {AH}</math>. We set up the area is equal to <math>\frac {1}{2}
cdot b \cdot h \implies </math>12/sqrt5=\overline {AH}\cdot \frac {1}{2}\cdot 8 \implies \overline {AH}=3\sqrt5<math> and since the answer should have </math>\sqrt 5<math> in it so our answer is </math>D$.
+
cdot b \cdot h \implies 12/sqrt5=\overline {AH}\cdot \frac {1}{2}\cdot 8 \implies \overline {AH}=3\sqrt5</math> and since the answer should have <math>\sqrt 5</math> in it so our answer is <math>D</math>.
  
 
==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}}

Revision as of 18:01, 8 August 2016

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

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 angle bisector theorem on triangle $ACH$ and triangle $ABH$, we get $AP:PH = 9:6$ and $AQ:QH = 7:2$, respectively. From now, you can simply use the answer choices because only choice $\textbf{D}$ has $\sqrt{5}$ in it and we know that $AH = 3\sqrt{5}$ the segments on it all have integral lengths so that $\sqrt{5}$ will remain there. However, by scaling up the length ratio: $AH:AP:PH = 45:27:18$ and $AQ:QH =45:35:10$. we get $AH:PQ = 45:(18 - 10) = 45 : 8$. \[PQ = 3\sqrt{5} * \frac{8}{45} = \boxed{\textbf{(D)}\frac{8}{15}\sqrt{5}}\]

Solution 2: (choices+faster) As in the above solution, let's find the area. By heron's, it is indeed $12\sqrt{5}$. Because $\overline {AH}$ is an altitude and we are given a base, let's find $\overline {AH}$. We set up the area is equal to $\frac {1}{2} cdot b \cdot h \implies 12/sqrt5=\overline {AH}\cdot \frac {1}{2}\cdot 8 \implies \overline {AH}=3\sqrt5$ and since the answer should have $\sqrt 5$ in it so our answer is $D$.

See Also

2016 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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 12 Problems and Solutions

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