AMC 10 · 2025 · #16

Grade 10 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
pythagorean-theoremangle-bisector-theoremdifference-of-squaressystems-of-equations identify-subproblems ↑ Prerequisites: pythagorean-theoremangle-bisector-theorem
📏 Long solution 💡 3 insights
Problem
In triangle ABC the sides are AB = 80, BC = 45, and AC = 75. From vertex B draw the bisector of angle B. From vertex C drop the altitude to side AB (the perpendicular that lands on line AB). These two lines cross at a point P. Find the length BP.

Pick an answer.

(A)
18
(B)
19
(C)
20
(D)
21
(E)
22
How to solve
Strategy Draw a Diagram

The problem mixes two different lines from the B-side of the triangle, so the first move is to draw the altitude and name its foot D on AB. The picture then shows P sitting on segment CD, which turns one hard length into a chain of small right-triangle questions (Identify Subproblems): first where D lands on AB, then how tall the altitude CD is, then where P sits on CD, and finally BP itself. Each of the first two pieces is a Pythagorean setup that I turn into equations with named unknowns (Introduce a Variable), and the third piece is the Angle Bisector Theorem hiding inside the small right triangle BDC.

1STEP 1

Draw the altitude and name its foot

Let D be the foot of the altitude from C to AB, so CD ⊥ AB; since P lies on CD, pinning down D and CD is the whole game.

CD ⊥ AB, AD + BD = AB = 80
2STEP 2

Subtract the two Pythagorean equations

Subtracting the two Pythagorean relations cancels the shared CD², leaving AC² − BC² = (AD+BD)(AD−BD) = 80(AD−BD).

AC² - BC² = AD² - BD² = (AD+BD)(AD-BD) = 80 (AD - BD)
3STEP 3

Solve for BD

Plugging in, AC² − BC² = 3600, so AD − BD = 45; combined with AD + BD = 80 this gives AD = 1252\frac{125}{2} and BD = 352\frac{35}{2}, both positive.

AD - BD = 3600/80 = 45 → AD = 125/2, BD = 35/2
4STEP 4

Find the altitude CD

Back in right triangle CDB, CD² = BC² − BD² = 68754\frac{6875}{4}, and since 6875 = 625·11 the root is clean: CD = 25112\frac{25\sqrt{11}}{2}.

CD² = BC² - BD² = 2025 - 1225/4 = 6875/4, CD = 25√(11)/2
5STEP 5

Split CD with the Angle Bisector Theorem

Since D lies between A and B, angle DBC equals angle B, so the bisector splits DC by DPPC\frac{DP}{PC} = BDBC\frac{BD}{BC} = 718\frac{7}{18}; with DC known, DP = 7112\frac{7\sqrt{11}}{2}.

DP/PC = BD/BC = 35/2/45 = 7/18, 25k = 25√(11)/2 → DP = 7k = 7√(11)/2
6STEP 6

Use Pythagoras for BP

Triangle BDP is right-angled at D, so BP² = BD² + DP² = 12254\frac{1225}{4} + 5394\frac{539}{4} = 441, giving BP = 21 — choice (D).

BP² = BD² + DP² = 1225/4 + 539/4 = 1764/4 = 441, BP = 21
Answer
21
BP = 21 is shorter than BC = 45, which must be true because P lies strictly inside the small triangle BDC, closer to B than C is. Numerically BD = 17.5 and CD is about 41.5, so DP is about 11.6; then sqrt(17.5² + 11.6²) is about sqrt(306 + 135) = sqrt(441) = 21, matching. The answer lands cleanly on choice (D) with no rounding drift.
💡Key takeaway

Drop the altitude to turn the triangle into two right triangles, find where its foot lands, then let the angle bisector split that vertical leg in the ratio of the two sides around angle B.

  • Draw the altitude and name its foot
  • Subtract the two Pythagorean equations
  • Solve for BD
  • Find the altitude CD
  • Split CD with the Angle Bisector Theorem
  • Use Pythagoras for BP

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