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 $BP$ ?

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

Diagram

$[asy] import olympiad; size(350); pair A = (0,0); pair B = (80,0); pair C = (62.5,41.4578098794425); pair H = (62.5,0); // Angle bisector point D on AC: AD : DC = 48 : 27 pair D = (40.0,26.5329983228432); path BD = B--D; pair P = intersectionpoint(BD,H--C); // main triangle draw(A--B--C--cycle, linewidth(1)); draw(C--H); draw(B--D); // right-angle mark at H draw((H+(1.8,0))--(H+(1.8,1.8))--(H+(0,1.8)), linewidth(0.8)); dot(A); label("$A$",A,SW); dot(B); label("$B$",B,SE); dot(C); label("$C$",C,NE); dot(H); label("$H$",H,S); dot(D); label("$D$",D,NNW); dot(P); label("$P$",P,ENE); // side labels (closer, larger) label("$80$", midpoint(A--B), S+1.5*dir(270), fontsize(11)); label("$75$", midpoint(A--C) + (-2,2), fontsize(11)); label("$45$", midpoint(B--C) + (2,2), fontsize(11)); xlimits(-5,85); ylimits(-5,50); currentpicture.fit(); [/asy]$ ~Avs2010

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 $BH$ 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 $\boxed{\textbf{(D) 21}}$ .

~Voidling

Solution 4

The semiperimeter is $\frac{75+45+80}{2}=100.$ Heron's formula for area gives $\sqrt{100\cdot20\cdot25\cdot55}=\left[ABC\right]=500\sqrt{11}\Rightarrow CH=\frac{25\sqrt{11}}{2}.$ Pythagorean theorem gives $BH^2=BC^2-CH^2=5^2\left(\frac{4\cdot 9^2-25\cdot 11}{4}\right)\Rightarrow BH=35/2.$

Angle bisector theorem on $\overline{CH}$ gives $\frac{PH}{35/2}=\frac{PC}{45},$ giving $PH=\frac7{18}\cdot PC\Rightarrow PH=\frac7{25}\cdot CH.$

Angle bisector theorem on $\overline{AC}$ gives $AD=48$ and $DC=27.$ Then, notice that $\frac{27}{45}=\frac{45}{75},$ as well as the fact that $\angle DCB=\angle ACB$ is shared, so $\triangle ABC\sim\triangle BDC$ by SAS. Therefore, $\angle DAB=\angle DBA.$

Now we look for another similar pair of triangles, $\triangle CHA\sim\triangle PHB$ by $AA.$ Hence, $\frac{75}{CH}=\frac{BP}{PH}=\frac{BP}{\frac{7}{25}\cdot CH}\Rightarrow \frac{25}{7}\cdot BP=75\Rightarrow BP=\boxed{21}.$

-NinJaPro

Solution 5

By angle bisector theorem, we find $AD=48$ and $DC=27.$

Let $BH=x,$ $CH=h,$ and $HA=16-x.$ We have

$x^2 + h^2 = 45^2 = 2025$ and $(16-x)^2 +h^2 = 256 - 32x +x^2 + h^2 = 75^2 = 5625$

by Pythagorean Theorem. After some algebra, we find $x = \frac{35}{2},$ meaning

$BH = \frac{35}{2}, HA = \frac{125}{2}.$

By Menelaus' Theorem, $\frac{BH}{HA} \cdot \frac{AC}{CD} \cdot \frac{DP}{PB} = 1.$

We have $\frac{\frac{35}{2}}{\frac{125}{2}} \cdot \frac{75}{27} \cdot \frac{DP}{PB} = 1,$ meaning $\frac{DP}{PB} = \frac{9}{7},$ or $BP = BD \cdot \frac{7}{16}.$

Stewart's Theorem on triangle $ABC$ with diagonal $BD$ gives the equation

$27 \cdot 75 \cdot 48 + d^2 \cdot 75 = 45^2 \cdot 48 + 80^2 \cdot 27.$

This gives $d = 48$ after some algebra.

Now, $BP = 48 \cdot \frac{7}{16} = \boxed{21}.$

-Jayjay2010

Solution 6

Use the compass and the ruler to construct the diagram accurately. I used $5:1$ cm. Use the ruler to measure the distance between $B$ and $P$ for $BP=\boxed{\textbf{(D)}~21}$ .

~dareckolo and friends ~Ben Dong (typo fix, originally said distance between $B$ and $C$ )

Video Solution (In 3 Mins)

https://youtu.be/nAimLnvSTwQ?si=cRm14r4GyfgxRq5L ~ Pi Academy

2025 AMC 10A (Problems • Answer Key • Resources)
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