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 $BP?$

$18$

$19$

$20$

$21$

$22$

Solution:

By the Law of Cosines, $\cos B = \frac{80^2 + 45^2 - 75^2}{2 \cdot 80 \cdot 45} = \frac{2800}{7200} = \frac{7}{18}.$

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 $\cos\dfrac{B}{2} = \sqrt{\dfrac{1 + 7/18}{2}} = \sqrt{\dfrac{25}{36}} = \dfrac{5}{6},$ we get $BP = \dfrac{17.5}{5/6} = 21.$

Thus, the correct answer is D.

2025 AMC 12A Problem 16

Solution:

Problem 16 in Other Years