1967 AHSME Problems/Problem 21

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Problem

In right triangle $ABC$ the hypotenuse $\overline{AB}=5$ and leg $\overline{AC}=3$. The bisector of angle $A$ meets the opposite side in $A_1$. A second right triangle $PQR$ is then constructed with hypotenuse $\overline{PQ}=A_1B$ and leg $\overline{PR}=A_1C$. If the bisector of angle $P$ meets the opposite side in $P_1$, the length of $PP_1$ is:

$\textbf{(A)}\ \frac{3\sqrt{6}}{4}\qquad \textbf{(B)}\ \frac{3\sqrt{5}}{4}\qquad \textbf{(C)}\ \frac{3\sqrt{3}}{4}\qquad \textbf{(D)}\ \frac{3\sqrt{2}}{2}\qquad \textbf{(E)}\ \frac{15\sqrt{2}}{16}$

Solution

$\fbox{B}$

See also

1967 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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