2025 AMC 10A Problems/Problem 23

Problem 23

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 $BP$ ?

$\textbf{(A)}~18\qquad\textbf{(B)}~19\qquad\textbf{(C)}~20\qquad\textbf{(D)}~21\qquad\textbf{(E)}~22$

Solution 1(Takes some time)

Let $CH$ be the altitude from vertex $C$ to the side $\overline{AB}$ , so $H$ is a point on $AB$ and $\angle CHB = 90^\circ$ . Let $BD$ be the angle bisector of $\angle B$ , where $D$ is on $AC$ .

Point $P$ is the intersection of the altitude $CH$ and the angle bisector $BD$ .

We want to find the length of $BP$ .

Consider the triangle $\triangle BPH$ . Since $P$ lies on the altitude $CH$ , the angle $\angle BHP$ is the same as $\angle CHB$ , which is $90^\circ$ . Therefore, $\triangle BPH$ is a right-angled triangle.

In the right $\triangle BPH$ , the angle $\angle PBH$ is the angle formed by the angle bisector $BD$ and the side $AB$ . By definition of the angle bisector, $\angle PBH = \angle ABC / 2 = B/2$ .

Using trigonometry in the right $\triangle BPH$ , we have:

$\cos(\angle PBH) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{BH}{BP}$

$\cos(B/2) = \frac{BH}{BP}$

Rearranging this gives:

$BP = \frac{BH}{\cos(B/2)}$

To solve the problem, we need to find the lengths of $BH$ and the value of $\cos(B/2)$ .

1. Find the length of BH} $BH$ is the projection of side $BC$ onto $AB$ . In the right-angled triangle $\triangle CHB$ , $BH = BC \cdot \cos(B) = a \cdot \cos(B)$ . We can find $\cos(B)$ using the Law of Cosines on $\triangle ABC$ :

$b^2 = a^2 + c^2 - 2ac \cos(B)$

$AC^2 = BC^2 + AB^2 - 2(BC)(AB) \cos(B)$

$75^2 = 45^2 + 80^2 - 2(45)(80) \cos(B)$

$5625 = 2025 + 6400 - 7200 \cos(B)$

$5625 = 8425 - 7200 \cos(B)$

$7200 \cos(B) = 8425 - 5625$

$7200 \cos(B) = 2800$

$\cos(B) = \frac{2800}{7200} = \frac{28}{72} = \frac{7}{18}$

Now, we can find $BH$ :

$BH = a \cdot \cos(B) = 45 \cdot \left(\frac{7}{18}\right) = \frac{45 \cdot 7}{18} = \frac{(5 \cdot 9) \cdot 7}{2 \cdot 9} = \frac{35}{2}$

2. Find the value of $\cos(B/2)$ } We use the half-angle identity for cosine: $\cos(B) = 2\cos^2(B/2) - 1$ . We know $\cos(B) = 7/18$ :

$\frac{7}{18} = 2\cos^2(B/2) - 1$

$\frac{7}{18} + 1 = 2\cos^2(B/2)$

$\frac{25}{18} = 2\cos^2(B/2)$

$\cos^2(B/2) = \frac{25}{36}$

$\cos(B/2) = \sqrt{\frac{25}{36}} = \frac{5}{6}$

(We take the positive root because $B/2$ must be an acute angle).

Now we substitute our values for $BH$ and $\cos(B/2)$ into the equation for $BP$ :

$BP = \frac{BH}{\cos(B/2)} = \frac{35/2}{5/6}$

$BP = \frac{35}{2} \cdot \frac{6}{5} = \frac{35 \cdot 6}{2 \cdot 5} = \frac{210}{10} = 21$

The length of $BP$ is 21. ~Jonathanmo

Solution 2

Let $D$ be the foot of the altitude from $C$ to $AB$ . We wish to find $BD$ and $DP$ .

First, notice that $AD^2 + CD^2 = AC^2$ and $BD^2+CD^2=BC^2$ by the Pythagorean Theorem. Subtracting the second equation from the first, we get $AD^2-BD^2=(AD-BD)(AD+BD)=AC^2-BC^2$ Plugging in values and simplifying, we see that $AD-BD =45$ . Knowing that $AD+BD=80$ , we can solve the system of equations to get $AD = \frac{125}{2}$ , $BD=\frac{35}{2}$ . Plug those values back into their original equations and we find that $CD = \frac{25}{2}\sqrt{11}$ . To find $DP$ , we use the Angle Bisector Theorem. The ratio between $BD$ and $BC$ is $\frac{7}{18}$ , so $DP = \frac{7}{25} \cdot CD = \frac{7}{2}\sqrt{11}$ . Finally, we use the Pythagorean Theorem to get $BD^2+DP^2=(\frac{35}{2})^2+(\frac{7}{2}\sqrt{11})^2=\frac{1764}{4}=441=BP^2$ so $BP=\boxed{\textbf{(D)}~21}$ .

~ChickensEatGrass

Solution 3 (Heron's Formula)

We are asked to find $BP$ in the given triangle configuration.

Let $D$ be the intersection of the altitude from $C$ to $AB$. To simplify calculations, we divide all side lengths by $5$, and multiply by $5$ again at the end.

First, we use Heron’s Formula, $\sqrt{s(s-a)(s-b)(s-c)}$, to find the area. Let $[ABC]$ denote the area of $\triangle ABC$. By Heron’s Formula, $[ABC] = \sqrt{20 \cdot 5 \cdot 4 \cdot 11} = 20\sqrt{11}.$

Next, we find the altitude $CD$ using the formula for the area of a triangle, $\tfrac{1}{2}bh = \text{area}$: $\frac{1}{2}(16)(CD) = 20\sqrt{11} \quad \Rightarrow \quad CD = \frac{5\sqrt{11}}{2}.$

We can use the Pythagorean Theorem in $\triangle CDB$ to find $DB$: $DB^2 + \frac{25 \cdot 11}{4} = 81.$ $4DB^2 + 275 = 324 \quad \Rightarrow \quad DB^2 = \frac{49}{4}.$ Thus, $DB = \frac{7}{2}.$

Next, we use the Angle Bisector Theorem to find $PD$. Let $x = PC$ and $y = PD$. Since $x + y = \frac{5\sqrt{11}}{2}$, we have $x = \frac{5\sqrt{11}}{2} - y.$

From the given ratio, $\frac{9}{x} = \frac{7}{2y} \quad \Rightarrow \quad 18y = 7x.$ Substituting $x = \frac{5\sqrt{11}}{2} - y$, $18y = 7\left(\frac{5\sqrt{11}}{2} - y\right) \quad \Rightarrow \quad 25y = \frac{35\sqrt{11}}{2}.$ Hence, $y = \frac{7\sqrt{11}}{10}.$

Now, using the Pythagorean Theorem again to find $BP$: $\frac{49 \cdot 11}{100} + \frac{49}{4} = BP^2.$ $100BP^2 = 49(11 + 25) = 49 \cdot 36 \quad \Rightarrow \quad BP = \frac{42}{10} = \frac{21}{5}.$

Finally, multiplying the side lengths by $5$ again gives $BP = 21.$, or $\boxed{\textbf{(B) 21}}$ .

~Voidling

Page

Toolbox

Search

2025 AMC 10A Problems/Problem 23

Contents

Problem 23

Solution 1(Takes some time)

Solution 2

Solution 3 (Heron's Formula)