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

m (Heron's formula and angle bisector theorem)
m (→‎Solution)
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<cmath>BH = \sqrt{AB^2 - AH^2} = \sqrt{7^2 - (3\sqrt{5})^2} = 2</cmath>
 
<cmath>BH = \sqrt{AB^2 - AH^2} = \sqrt{7^2 - (3\sqrt{5})^2} = 2</cmath>
 
<cmath>CH = BC - BH = 8 - 2 = 6</cmath>
 
<cmath>CH = BC - BH = 8 - 2 = 6</cmath>
−
Apply angle bisector theorem on triangle ACH and triangle ABH, we get AP:PH = 9:6 and AQ:QH = 7:2, respectively.
+
<math>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 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.
+
From now, you can simply use the answer choices because only choice 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.</math>
 
However, by scaling up the length ratio:
 
However, by scaling up the length ratio:
 
<math>AH:AP:PH = 45:27:18 and AQ:QH =45:35:10</math>.
 
<math>AH:AP:PH = 45:27:18 and AQ:QH =45:35:10</math>.

Revision as of 21:26, 21 February 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: \[Area = 12\sqrt{5} = \frac{1}{2}bh = \frac{1}{2}(8)(AH)\] \[AH = 3\sqrt{5}\] \[BH = \sqrt{AB^2 - AH^2} = \sqrt{7^2 - (3\sqrt{5})^2} = 2\] \[CH = BC - BH = 8 - 2 = 6\] $Apply angle bisector theorem on triangle ACH and triangle ABH, we get AP:PH = 9:6 and AQ:QH = 7:2, respectively. From now, you can simply use the answer choices because only choice D has \sqrt{5} in it and we know that AH = 3\sqrt{5} the segments on it all have integral lengths, so that \sqrt{5} will remain there.$ However, by scaling up the length ratio: $AH:AP:PH = 45:27:18 and AQ:QH =45:35:10$. we get $AH:PQ = 45:(18 - 10) = 45 : 8$. \[PQ = 3\sqrt{5} * \frac{8}{45} = \boxed{\textbf{(D)}\frac{8}{15}\sqrt{5}}\]

See Also

2016 AMC 12B (Problems • Answer Key • Resources)
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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