2025 AMC 12A Problem 16

Question

Triangle

\triangle ABC

has side lengths

AB = 80,

BC = 45,

and

AC = 75.

The bisector of

\angle B

and the altitude to side

AB

intersect at point

P.

What is

BP?

18

19

20

21

22

\cos B = \dfrac{80^2 + 45^2 - 75^2}{2 \cdot 80 \cdot 45} = \dfrac{7}{18}

The altitude’s foot is

BC\cos B = 17.5

from

B

along

AB,

and

BP\cos\tfrac{B}{2} = 17.5

By the Law of Cosines,

\begin{aligned} \cos B &= \frac{80^2 + 45^2 - 75^2}{2 \cdot 80 \cdot 45} \\ &= \frac{2800}{7200} = \frac{7}{18}. \end{aligned}

The altitude to

AB

is drawn from

C,

and its foot is at distance

BC\cos B = 45 \cdot \dfrac{7}{18} = 17.5

from

B

along

AB.

Along the bisector from

B,

the component parallel to

AB

is

BP\cos\dfrac{B}{2},

which must reach the altitude’s foot:

BP\cos\dfrac{B}{2} = 17.5.

Since

\cos\dfrac{B}{2} = \sqrt{\dfrac{1 + \frac{7}{18}}{2}}

= \sqrt{\dfrac{25}{36}}

= \dfrac{5}{6},

we get

BP = \dfrac{17.5}{\frac{5}{6}} = 21.

Thus, the correct answer is D.

Problem 16 in other years