2025 AMC 10A Problem 23

P21-25~6 minPrint

Triangles: Area & Pythagorean Trigonometry Apply an identity: SFFT, sum of squares, difference of cubes, Vieta Set up the equation/formula and compute; no special trick needed

Triangle $\triangle ABC$ has side lengths $AB = 80$ , $BC = 45$ , and $AC = 75$ . The bisector of $\angle B$ and the altitude to side $\overline{AB}$ intersect at point $P.$ What is $BP$ ?

Show solution without answering

Key insight

Use Pythagoras twice to find where the altitude from C meets AB. The bisector from B then cuts that altitude in the ratio of the sides beside angle B.

Solution

The point $P$ lies on two lines: the altitude from $C$ down to side $\overline{AB}$ , and the bisector of the angle at $B$ . So first pin down the altitude, then find where the bisector cuts it.

Name the foot of the altitude. The altitude to side $\overline{AB}$ comes from the opposite corner $C$ and meets $\overline{AB}$ at a right angle. Call that meeting point $H$ . Let $BH = x$ , so $AH = 80 - x$ .

2. Use Pythagoras twice. The altitude $CH$ is a side of two right triangles, $BHC$ and $AHC$ . Each one gives a value for $CH^2$ :
$CH^2 = 45^2 - x^2 \qquad \text{and} \qquad CH^2 = 75^2 - (80 - x)^2.$

3. Set them equal and solve.
$2025 - x^2 = 5625 - (6400 - 160x + x^2) = -775 + 160x - x^2.$
The $x^2$ terms cancel: $2025 = -775 + 160x$ , so $160x = 2800$ and
$BH = x = \frac{35}{2}.$
This is between $0$ and $80$ , so $H$ really is on side $\overline{AB}$ .

4. Find $CH$ .
$CH^2 = 2025 - \left(\frac{35}{2}\right)^2 = \frac{8100 - 1225}{4} = \frac{6875}{4}.$
Since $6875 = 625 \times 11$ , we get $CH = \dfrac{25\sqrt{11}}{2}$ .

Bring in the bisector. Look at the right triangle $BHC$ . Its angle at $B$ is the same as angle $B$ of the big triangle, because $H$ lies on $\overline{BA}$ . So the bisector from $B$ also cuts this small triangle's angle in half, and it hits the opposite side $\overline{HC}$ at $P$ .

6. Angle Bisector Theorem. A bisector splits the opposite side in the same ratio as the two sides next to the angle. In triangle $BHC$ the sides next to angle $B$ are $BH = \frac{35}{2}$ and $BC = 45$ , so
$\frac{HP}{PC} = \frac{BH}{BC} = \frac{35/2}{45} = \frac{35}{90} = \frac{7}{18}.$

7. Find $HP$ . The altitude is split into $7 + 18 = 25$ equal parts, and $HP$ gets $7$ of them:
$HP = \frac{7}{25} \cdot \frac{25\sqrt{11}}{2} = \frac{7\sqrt{11}}{2}.$

8. Finish with Pythagoras. Triangle $BHP$ has a right angle at $H$ , so
$BP^2 = \left(\frac{35}{2}\right)^2 + \left(\frac{7\sqrt{11}}{2}\right)^2 = \frac{1225 + 539}{4} = \frac{1764}{4} = 441.$
So $BP = 21$ .

The answer is $\boxed{\textbf{(D)}\ 21}$ .

Why this works

An altitude creates right triangles, and right triangles are where Pythagoras does the work. The Angle Bisector Theorem turns "cuts the angle in half" into a ratio of lengths, which is exactly what you need to locate a point on a segment. Next time a bisector crosses an altitude, look for the right triangle made by the vertex, the foot of the altitude and the crossing point.

Alternative approach

If you know the Law of Cosines, there is a shorter route.

$\cos B = \dfrac{80^2 + 45^2 - 75^2}{2 \cdot 80 \cdot 45} = \dfrac{2800}{7200} = \dfrac{7}{18}$ .
In right triangle $BHC$ , $BH = 45 \cos B = 45 \cdot \frac{7}{18} = \frac{35}{2}$ , the same as before.
The half-angle formula gives $\cos^2 \frac{B}{2} = \dfrac{1 + \cos B}{2} = \dfrac{25}{36}$ , so $\cos \frac{B}{2} = \frac{5}{6}$ .
In right triangle $BHP$ the angle at $B$ is $\frac{B}{2}$ , so $BH = BP \cos \frac{B}{2}$ , and $BP = \frac{35}{2} \div \frac{5}{6} = 21$ .

The fact that $\frac{25}{36}$ is a perfect square is a hint that the problem was built to work out neatly this way.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The altitude to side $\overline{AB}$ is the perpendicular from the opposite vertex $C$ ; call its foot $H$ , so $H$ lies on $\overline{AB}$ and $CH \perp AB$ . The point $P$ is where the bisector from $B$ meets the line $CH$ .

Two things are then needed: the distance $BH$ , and the angle at $B$ that the bisector makes with $\overline{AB}$ .

The cosine of $B$ . By the law of cosines on side $AC$ ,
$\cos B = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC} = \frac{6400 + 2025 - 5625}{2 \cdot 80 \cdot 45} = \frac{2800}{7200} = \frac{7}{18}.$

The foot of the altitude. In right triangle $BHC$ , $BH = BC \cos B = 45 \cdot \frac{7}{18} = \frac{35}{2}$ . (Since $0 < \tfrac{35}{2} < 80$ , $H$ lies strictly between $B$ and $A$ , as expected.)

The half angle. The half-angle formula gives
$\cos^2\frac{B}{2} = \frac{1 + \cos B}{2} = \frac{1 + \frac{7}{18}}{2} = \frac{25}{36}, \qquad \text{so } \cos\frac{B}{2} = \frac56,$
taking the positive root because $\tfrac B2 < 90^\circ$ .

Finish. Triangle $BHP$ has its right angle at $H$ and the angle $\angle PBH = \tfrac B2$ at $B$ , because $P$ lies on the bisector of $\angle B$ and $H$ lies on $\overline{BA}$ . Hence $BH = BP\cos\frac{B}{2}$ , so
$BP = \frac{BH}{\cos\frac{B}{2}} = \frac{35/2}{5/6} = \frac{35}{2}\cdot\frac65 = 21.$

The answer is $\boxed{\textbf{(D)}\ 21}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 18: Introduction to Trigonometry
Introduction to Algebra, Chapter 11: Special Factorizations
Introduction to Geometry, Chapter 6: Right Triangles (§6.1, §6.2, §6.3)