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2025 AMC 10A Problems/Problem 23: Difference between revisions

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To find <imath>DP</imath>, we use the Angle Bisector Theorem. The ratio between <imath>BD</imath> and <imath>BC</imath> is <imath>\frac{7}{18}</imath>, so <imath>DP = \frac{7}{25} \cdot CD = \frac{7}{2}\sqrt{11}</imath>. Finally, we use the Pythagorean Theorem to get <cmath>BD^2+DP^2=(\frac{35}{2})^2+(\frac{7}{2}\sqrt{11})^2=\frac{1764}{4}=441=BP^2</cmath>
To find <imath>DP</imath>, we use the Angle Bisector Theorem. The ratio between <imath>BD</imath> and <imath>BC</imath> is <imath>\frac{7}{18}</imath>, so <imath>DP = \frac{7}{25} \cdot CD = \frac{7}{2}\sqrt{11}</imath>. Finally, we use the Pythagorean Theorem to get <cmath>BD^2+DP^2=(\frac{35}{2})^2+(\frac{7}{2}\sqrt{11})^2=\frac{1764}{4}=441=BP^2</cmath>
so <imath>BP=\boxed{21}</imath>.
so <imath>BP=\boxed{21}</imath>.
~ChickensEatGrass
~ChickensEatGrass

Revision as of 14:49, 6 November 2025

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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