2025 AMC 12A Problem 16

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.