1967 AMC 12 Problem 21

Question

In right triangle

ABC

the hypotenuse

AB=5

and leg

AC=3.

The bisector of angle

A

meets the opposite side in

A_1.

A second right triangle

PQR

is then constructed with hypotenuse

PQ=A_1B

and leg

PR=A_1C.

If the bisector of angle

P

meets the opposite side in

P_1,

the length of

PP_1

is:

\dfrac{3\sqrt6}{4}

\dfrac{3\sqrt5}{4}

\dfrac{3\sqrt3}{4}

\dfrac{3\sqrt2}{2}

\dfrac{15\sqrt2}{16}

Use the angle-bisector theorem in the

3

-

4

-

5

triangle

The second triangle is a half-scale copy of the first

Since

BC=4,

the angle-bisector theorem gives

A_1B:A_1C=5:3.

Hence

A_1B=\frac{5}{2}

and

A_1C=\frac{3}{2},

so triangle

PQR

is a half-scale copy of

ABC.

For the angle bisector from

A

in the original triangle,

\begin{aligned} AA_1^2 &=5\cdot3\left(1-\frac{4^2}{8^2}\right)\\ &=\frac{45}{4}, \end{aligned}

so

AA_1=\frac{3\sqrt5}{2}.

Therefore

PP_1

is half of this, or

\frac{3\sqrt5}{4}.

Thus, the correct answer is B.

Problem 21 in other years