AMC 10 · 2025 · #23
Grade 8 geometry-2dPick an answer.
AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The whole problem is unlocked by one careful picture, so Tool #1 (Draw a Diagram) is the spine: drawing the altitude CD and the bisector shows that P lies on CD, and that the small right triangle BDP (right angle at D) has BP as its hypotenuse. That reframing means we no longer fight the whole triangle; Tool #7 (Identify Subproblems) breaks the job into three clean pieces: find the foot distance BD, find the altitude CD, then find DP. Tool #4 (Introduce a Variable) names BD=x so the shared altitude, written two ways by the Pythagorean theorem, collapses into a single linear equation. With both legs BD and DP in hand, one last Pythagorean step gives BP.
Frame the small right triangle
Let D be the foot of the altitude from C; P sits on CD, so BP is the hypotenuse of right triangle BDP — just find the legs BD and DP.
Drawing the altitude turns the messy full triangle into one small right triangle whose hypotenuse is exactly the length we want.
8.G.A.5Draw A DiagramFind BD with the shared altitude
With BD=x and AD=80-x, writing CD² both ways gives 45²-x²=75²-(80-x)²; the x² cancels, so 160x=2800 and BD=.
Setting the one shared altitude equal to itself makes the squared unknown cancel, leaving a single easy linear equation.
8.EE.C.7Introduce A VariableFind the altitude CD
In right triangle CDB with hypotenuse BC=45, CD²=2025-=, so CD=.
Once one leg BD is known, the Pythagorean theorem hands you the other leg of the very same right triangle.
8.G.B.7Identify SubproblemsLocate P with the angle bisector theorem
Inside right triangle BDC the bisector splits DC as DP:PC=BD:BC=7:18, so DP is of CD: DP=.
A bisector cutting the opposite side always splits it in the same ratio as the two sides that meet at that corner.
A bisector cutting the opposite side splits it in the same ratio as the two sides meeting at that corner.
▸ Why?
The two pieces sit in triangles of the same shape, so their sides keep one fixed ratio.
▸ Why?
A ratio fixes only relative sizes, so one common factor scales both pieces at once.
Finish with the Pythagorean theorem
Triangle BDP is right-angled at D, so BP²=+=441 and BP=21, choice (D).
With both legs known, one last Pythagorean step gives the hypotenuse BP directly.
8.G.B.7Identify SubproblemsDrop the altitude to spot the tiny right triangle BDP; find its two legs BD and DP, then let the Pythagorean theorem give BP=21.
- Frame the small right triangle
- Find BD with the shared altitude
- Find the altitude CD
- Locate P with the angle bisector theorem
- Finish with the Pythagorean theorem
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