2025 AMC 10A Problems/Problem 23

Problem 23

Triangle $\triangle ABC$ has side lengths $AB = 80$ , $BC = 45$ , and $AC = 75$ . The bisector $\angle B$ and the altitude to side $\overline{AB}$ intersect at point $P$ . What is $BP$ ?

$\textbf{(A)}~18\qquad\textbf{(B)}~19\qquad\textbf{(C)}~20\qquad\textbf{(D)}~21\qquad\textbf{(E)}~22$

Solution 1

Let $D$ be the foot of the altitude from $C$ to $AB$ . We wish to find $BD$ and $DP$ . First, notice that $AD^2 + CD^2 = AC^2$ and $BD^2+CD^2=BC^2$ by the Pythagorean Theorem. Subtracting the second equation from the first, we get $AD^2-BD^2=(AD-BD)(AD+BD)=AC^2-BC^2$ Plugging in values, we see that $AD-BD =45$ . So, $AD = \frac{125}{2}$ , $BD=\frac{35}{2}$ , and $CD = \frac{25}{2}\sqrt{11}$ . To find $DP$ , we use the Angle Bisector Theorem. The ratio between $BD$ and $BC$ is $\frac{7}{18}$ , so $DP = \frac{7}{25} \cdot CD = \frac{7}{2}\sqrt{11}$ . Finally, we use the Pythagorean Theorem to get $BD^2+DP^2=(\frac{35}{2})^2+(\frac{7}{2}\sqrt{11})^2=\frac{1764}{4}=441=BP^2$ so $BP=\boxed{21}$ . ~ChickensEatGrass

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2025 AMC 10A Problems/Problem 23

Problem 23

Solution 1