1983 AHSME Problems/Problem 19

Revision as of 00:57, 20 February 2019 by Sevenoptimus (talk | contribs) (Improved clarity and added box at the bottom)

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 $\boxed{\textbf{A}}$.

See Also

1983 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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