AMC 10 · 2025 · #23
Grade 8 geometry-2dTriangle △ABC has side lengths AB=80, BC=45, and AC=75. The bisector of ∠B and the altitude to side AB intersect at point P. What is BP?
Pick an answer.
AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Triangle $ABC$ has side lengths $AB=80$, $BC=45$, and $AC=75$. Drop the altitude from $C$ to side $AB$, and also draw the bisector of $\angle B$. These two lines cross at a point $P$. Find the length $BP$.
Givens: The side lengths are $AB=80$, $BC=45$, and $AC=75$; One line is the altitude from $C$ to side $AB$; call its foot $D$, so $CD\perp AB$ and $D$ lies on segment $AB$; The other line is the bisector of $\angle B$, which splits $\angle ABC$ into two equal angles; $P$ is the point where the altitude and the bisector meet; Answer choices: (A) $18$, (B) $19$, (C) $20$, (D) $21$, (E) $22$
Unknowns: The length $BP$
Understand
Restated: Triangle $ABC$ has side lengths $AB=80$, $BC=45$, and $AC=75$. Drop the altitude from $C$ to side $AB$, and also draw the bisector of $\angle B$. These two lines cross at a point $P$. Find the length $BP$.
Givens: The side lengths are $AB=80$, $BC=45$, and $AC=75$; One line is the altitude from $C$ to side $AB$; call its foot $D$, so $CD\perp AB$ and $D$ lies on segment $AB$; The other line is the bisector of $\angle B$, which splits $\angle ABC$ into two equal angles; $P$ is the point where the altitude and the bisector meet; Answer choices: (A) $18$, (B) $19$, (C) $20$, (D) $21$, (E) $22$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #4 Introduce a Variable
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 $BP$.
Execute — Answer: D
8.G.A.5 Step 1 Frame the small right triangle
- Drop the altitude from $C$ to $AB$ and call the foot $D$, so $CD\perp AB$ and $D$ lies on $AB$.
- The bisector of $\angle B$ meets this altitude at $P$, so $P$ lies on segment $CD$.
- Because $D$ is on ray $BA$, the bisector of $\angle ABC$ is exactly the bisector of $\angle DBC$ inside right triangle $BDC$.
- Look at triangle $BDP$: it has a right angle at $D$, so $BP$ is its hypotenuse and $BP=\sqrt{BD^2+DP^2}$.
- The plan is now just to find the two 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.EE.C.7 Step 2 Find BD with the shared altitude
- The altitude $CD$ is a common leg of the two right triangles $CDB$ and $CDA$.
- Let $BD=x$, so $AD=80-x$.
- Writing $CD^2$ by the Pythagorean theorem in each triangle gives $45^2-x^2=75^2-(80-x)^2$.
- The $x^2$ terms cancel, leaving a linear equation: expanding the right side gives $2025=-775+160x$, so $160x=2800$ and $x=\tfrac{35}{2}$.
- Thus $BD=\tfrac{35}{2}$.
💡 Setting the one shared altitude equal to itself makes the squared unknown cancel, leaving a single easy linear equation.
8.G.B.7 Step 3 Find the altitude CD
- Now use the Pythagorean theorem in right triangle $CDB$, where $BC=45$ is the hypotenuse and $BD=\tfrac{35}{2}$ is a leg: $CD^2=45^2-\left(\tfrac{35}{2}\right)^2=2025-\tfrac{1225}{4}=\tfrac{6875}{4}$.
- Taking the square root, $CD=\tfrac{25\sqrt{11}}{2}$.
💡 Once one leg $BD$ is known, the Pythagorean theorem hands you the other leg of the very same right triangle.
8.G.A.4 Step 4 Locate P with the angle bisector theorem
- Work inside right triangle $BDC$.
- The bisector from $B$ meets the opposite side $DC$ at $P$, so the angle bisector theorem splits $DC$ in the ratio of the two adjacent sides: $\dfrac{DP}{PC}=\dfrac{BD}{BC}=\dfrac{35/2}{45}=\dfrac{7}{18}$.
- Since $DP+PC=DC$, the piece $DP$ is the fraction $\dfrac{7}{7+18}=\dfrac{7}{25}$ of $DC$, so $DP=\dfrac{7}{25}\cdot\dfrac{25\sqrt{11}}{2}=\dfrac{7\sqrt{11}}{2}$.
💡 A bisector cutting the opposite side always splits it in the same ratio as the two sides that meet at that corner.
8.G.B.7 Step 5 Finish with the Pythagorean theorem
- Triangle $BDP$ is right-angled at $D$, so $BP^2=BD^2+DP^2=\left(\tfrac{35}{2}\right)^2+\left(\tfrac{7\sqrt{11}}{2}\right)^2=\tfrac{1225}{4}+\tfrac{539}{4}=\tfrac{1764}{4}=441$.
- Therefore $BP=\sqrt{441}=21$, and the answer is (D).
💡 With both legs known, one last Pythagorean step gives the hypotenuse $BP$ directly.
8.G.A.5 Drop the altitude from $C$ to $AB$ and call the foot $D$, so $CD\perp AB$ and $D 8.EE.C.7 The altitude $CD$ is a common leg of the two right triangles $CDB$ and $CDA$. Le 8.G.B.7 Now use the Pythagorean theorem in right triangle $CDB$, where $BC=45$ is the hy 8.G.A.4 Work inside right triangle $BDC$. The bisector from $B$ meets the opposite side 8.G.B.7 Triangle $BDP$ is right-angled at $D$, so $BP^2=BD^2+DP^2=\left(\tfrac{35}{2}\ri Review
Reasonableness: The final $BP^2=441=21^2$ is a clean perfect square, a reassuring sign on a competition. It must also hold that $BP$ exceeds the leg $BD=17.5$, since $BP$ is the hypotenuse of right triangle $BDP$; indeed $21>17.5$. Numerically $DP=\tfrac{7\sqrt{11}}{2}\approx11.6$, so $BP\approx\sqrt{306.25+134.75}=\sqrt{441}=21$, which lands squarely on choice (D) among the options $18$ through $22$.
Alternative: Use coordinates. Put $B=(0,0)$ and $A=(80,0)$ on the $x$-axis, so the foot is $D=(\tfrac{35}{2},0)$ and $C=(\tfrac{35}{2},\tfrac{25\sqrt{11}}{2})$. The direction of $BA$ is $(1,0)$ and the unit vector along $BC$ is $\left(\tfrac{7}{18},\tfrac{5\sqrt{11}}{18}\right)$; adding them gives a bisector direction proportional to $(5,\sqrt{11})$. The bisector $t(5,\sqrt{11})$ meets the altitude line $x=\tfrac{35}{2}$ when $5t=\tfrac{35}{2}$, i.e. $t=\tfrac{7}{2}$, so $P=\left(\tfrac{35}{2},\tfrac{7\sqrt{11}}{2}\right)$ and $BP=\sqrt{\left(\tfrac{35}{2}\right)^2+\left(\tfrac{7\sqrt{11}}{2}\right)^2}=\sqrt{441}=21$, the same answer.
CCSS standards used (min grade 8)
8.G.A.5Use informal arguments to establish facts about angle sum and exterior angles (Recognizing that the altitude makes a right angle at $D$ and that the bisector of $\angle ABC$ is the bisector of $\angle DBC$ inside right triangle $BDC$, so $BP$ is the hypotenuse of right triangle $BDP$.)8.EE.C.7Solve linear equations in one variable (Writing the shared altitude two ways and solving the resulting linear equation $160x=2800$ for $BD=\tfrac{35}{2}$ after the $x^2$ terms cancel.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the altitude $CD=\tfrac{25\sqrt{11}}{2}$ in right triangle $CDB$ and then computing $BP=\sqrt{BD^2+DP^2}=21$ in right triangle $BDP$.)8.G.A.4Understand that a two-dimensional figure is similar to another using transformations (Applying the angle bisector theorem (which follows from similar triangles) to split $DC$ in the ratio $\tfrac{DP}{PC}=\tfrac{BD}{BC}=\tfrac{7}{18}$ and get $DP=\tfrac{7\sqrt{11}}{2}$.)
⭐ Drop the altitude to spot the tiny right triangle $BDP$; find its two legs $BD$ and $DP$, then let the Pythagorean theorem give $BP=21$.
⭐ Drop the altitude to spot the tiny right triangle $BDP$; find its two legs $BD$ and $DP$, then let the Pythagorean theorem give $BP=21$.
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