2025 AMC 10A Problem 23

The altitude to side $\overline{AB}$ is the perpendicular from the opposite vertex $C$; call its foot $H$, so $H$ lies on $\overline{AB}$ and $CH \perp AB$. The point $P$ is where the bisector from $B$ meets the line $CH$.

Two things are then needed: the distance $BH$, and the angle at $B$ that the bisector makes with $\overline{AB}$.

The cosine of $B$. By the law of cosines on side $AC$,
$$\cos B = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC} = \frac{6400 + 2025 - 5625}{2 \cdot 80 \cdot 45} = \frac{2800}{7200} = \frac{7}{18}.$$

The foot of the altitude. In right triangle $BHC$, $BH = BC \cos B = 45 \cdot \frac{7}{18} = \frac{35}{2}$. (Since $0 < \tfrac{35}{2} < 80$, $H$ lies strictly between $B$ and $A$, as expected.)

The half angle. The half-angle formula gives
$$\cos^2\frac{B}{2} = \frac{1 + \cos B}{2} = \frac{1 + \frac{7}{18}}{2} = \frac{25}{36}, \qquad \text{so } \cos\frac{B}{2} = \frac56,$$
taking the positive root because $\tfrac B2 < 90^\circ$.

Finish. Triangle $BHP$ has its right angle at $H$ and the angle $\angle PBH = \tfrac B2$ at $B$, because $P$ lies on the bisector of $\angle B$ and $H$ lies on $\overline{BA}$. Hence $BH = BP\cos\frac{B}{2}$, so
$$BP = \frac{BH}{\cos\frac{B}{2}} = \frac{35/2}{5/6} = \frac{35}{2}\cdot\frac65 = 21.$$

The answer is $\boxed{\textbf{(D)}\ 21}$.

An altitude to a side and a cevian from a vertex on that side always meet in a right triangle whose hypotenuse is the piece of the cevian being asked for, so the whole problem collapses to one leg and one angle. The leg $BH$ is the projection of $BC$ onto $BA$, which is what $\cos B$ means; the angle is half of $\angle B$, which is what the half-angle formula supplies. The numbers here are chosen so that $\cos B = \tfrac{7}{18}$ makes $\tfrac{1+\cos B}{2} = \tfrac{25}{36}$ a perfect square — a signal that the intended route is the half-angle identity rather than decimals. The same three-step pattern (law of cosines, projection, half angle) handles any "bisector meets altitude" question.

Coordinates are just as fast. Put $B = (0,0)$ and $A = (80, 0)$. Then $C = (x, y)$ with $x^2 + y^2 = 2025$ and $(x-80)^2 + y^2 = 5625$; subtracting gives $x = \tfrac{35}{2}$ and $y = \tfrac{25\sqrt{11}}{2}$. The altitude to $AB$ is the vertical line $x = \tfrac{35}{2}$.

The bisector from $B$ has direction equal to the sum of the unit vectors along $BA$ and $BC$, namely $(1,0) + \tfrac{1}{45}\left(\tfrac{35}{2}, \tfrac{25\sqrt{11}}{2}\right) = \tfrac{5}{18}\left(5, \sqrt{11}\right)$, so it is the line $y = \tfrac{\sqrt{11}}{5}x$. At $x = \tfrac{35}{2}$ this gives $y = \tfrac{7\sqrt{11}}{2}$, and
$$BP = \frac12\sqrt{35^2 + 49 \cdot 11} = \frac12\sqrt{1764} = 21.$$

• Reading 'the altitude to side AB' as a segment drawn from B, which would put P at B; the altitude meant is the one from the opposite vertex C, and its foot H is what fixes BP.
• Dropping the square root in the half-angle formula and using $\cos\frac B2 = \frac{25}{36}$, which gives $BP = 25.2$ and matches no choice.
• Computing the angle bisector as far as side $\overline{AC}$ instead of stopping at the altitude: that cevian has length $48$, since $AD = 48$, $DC = 27$ and $BD^2 = 80 \cdot 45 - 48 \cdot 27$.