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| ==Problem 23==
| | #redirect [[2025 AMC 12A Problems/Problem 16]] |
| Triangle <imath>\triangle ABC</imath> has side lengths <imath>AB = 80</imath>, <imath>BC = 45</imath>, and <imath>AC = 75</imath>. The bisector <imath>\angle B</imath> and the altitude to side <imath>\overline{AB}</imath> intersect at point <imath>P</imath>. What is <imath>BP</imath>?
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| <imath>\textbf{(A)}~18\qquad\textbf{(B)}~19\qquad\textbf{(C)}~20\qquad\textbf{(D)}~21\qquad\textbf{(E)}~22</imath>
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| ==Solution 1==
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| Let <imath>D</imath> be the foot of the altitude from C to AB. We wish to find <imath>BD</imath> and <imath>DP</imath>.
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| First, notice that <imath>AD^2 + CD^2 = AC^2</imath> and <imath>BD^2+CD^2=BC^2</imath> by the Pythagorean Theorem. Subtracting the second equation from the first, we get <cmath>AD^2-BD^2=(AD-BD)(AD+BD)=AC^2-BC^2</cmath> Plugging in values, we see that <imath>AD-BD =45</imath>. So, <imath>AD = \frac{125}{2}</imath>, <imath>BD=\frac{35}{2}</imath>, and <imath>CD = \frac{25}{2}\sqrt{11}</imath>.
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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>.
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