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Difference between revisions of "1983 AHSME Problems/Problem 19"

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==Solution==
 
==Solution==
  
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Let <math>AD = y</math>. Since <math>AD</math> bisects <math>\angle{BAC}</math>, the Angle Bisector Theorem gives <math>\frac{DB}{CD} = \frac{AB}{AC} = 2</math>, so write <math>CD = x</math> and <math>DB = 2x</math>. Applying the Law of Cosines to <math>\triangle CAD</math> gives <math>x^2 = 3^2 + y^2 - 3y</math>, and to <math>\triangle DAB</math> gives <math>(2x)^2 = 6^2 + y^2 - 6y</math>. Subtracting <math>4</math> times the first equation from the second equation therefore yields <math>0 = 6y - 3y^2 \Rightarrow y(y-2) = 0</math>, so <math>y</math> is <math>0</math> or <math>2</math>. But since <math>y \neq 0</math> (<math>y</math> is the length of a side of a triangle), <math>y</math> must be <math>2</math>, so the answer is <math>\boxed{\textbf{A}}</math>.
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Let <math>AD = y</math>. Since <math>AD</math> bisects <math>\angle{BAC}</math>, the Angle Bisector Theorem gives <math>\frac{DB}{CD} = \frac{AB}{AC} = 2</math>, so let <math>CD = x</math> and <math>DB = 2x</math>. Applying the Law of Cosines to <math>\triangle CAD</math> gives <math>x^2 = 3^2 + y^2 - 3y</math>, and to <math>\triangle DAB</math> gives <math>(2x)^2 = 6^2 + y^2 - 6y</math>. Subtracting <math>4</math> times the first equation from the second equation therefore yields <math>0 = 6y - 3y^2 \Rightarrow y(y-2) = 0</math>, so <math>y</math> is <math>0</math> or <math>2</math>. But since <math>y \neq 0</math> (<math>y</math> is the length of a side of a triangle), <math>y</math> must be <math>2</math>, so the answer is <math>\boxed{\textbf{(A)} \ 2}</math>.
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==See Also==
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{{AHSME box|year=1983|num-b=18|num-a=20}}
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{{MAA Notice}}

Latest revision as of 23:57, 19 February 2019

Problem

Point $D$ is on side $CB$ of triangle $ABC$. If $\angle{CAD} = \angle{DAB} = 60^\circ\mbox{, }AC = 3\mbox{ and }AB = 6$, then the length of $AD$ is

$\textbf{(A)} \ 2 \qquad \textbf{(B)} \ 2.5 \qquad \textbf{(C)} \ 3 \qquad \textbf{(D)} \ 3.5 \qquad \textbf{(E)} \ 4$

Solution

Let $AD = y$. Since $AD$ bisects $\angle{BAC}$, the Angle Bisector Theorem gives $\frac{DB}{CD} = \frac{AB}{AC} = 2$, so let $CD = x$ and $DB = 2x$. Applying the Law of Cosines to $\triangle CAD$ gives $x^2 = 3^2 + y^2 - 3y$, and to $\triangle DAB$ gives $(2x)^2 = 6^2 + y^2 - 6y$. Subtracting $4$ times the first equation from the second equation therefore yields $0 = 6y - 3y^2 \Rightarrow y(y-2) = 0$, so $y$ is $0$ or $2$. But since $y \neq 0$ ($y$ is the length of a side of a triangle), $y$ must be $2$, so the answer is $\boxed{\textbf{(A)} \ 2}$.

See Also

1983 AHSME (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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions


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