2025 AMC 10 A

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Problem 23 Easy

Triangle \Delta A B C has side lengths A B = 8 0, B C = 4 5, and A C = 7 5. The bisector of \angle B and the altitude to side \overline{AB} intersect at point P. What is BP?

  • A.

    18

  • B.

    19

  • C.

    20

  • D.

    21

  • E.

    22

Answer:D

 

Set BD = x, making AD = 80 - x. BC^2 - BD^2 = CD^2 = AC^2 - AD^2

This gives 45^2 - x^2 = 75^2 - (80-x)^2.

Solving: x = \frac{35}{2}, and CD^2 = 45^2 - \left(\frac{35}{2}\right)^2 = \frac{25 \times 275}{4}, so CD=\dfrac{25\sqrt{11}}{2}.

By the angle bisector theorem, \dfrac{DP}{PC}=\dfrac{BD}{BC}=\dfrac{7}{18}

So DP=\dfrac{7}{18+7}\times CD=\dfrac{7\sqrt{11}}{2}

Therefore: BP^2 = BD^2 + DP^2 = 441

Thus BP = 21.

Related Problems
Problem 21 2025 AMC 10 A
Problem 22 2025 AMC 10 A
Problem 24 2025 AMC 10 A
Problem 25 2025 AMC 10 A
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