2025 AMC 12A Problems/Problem 16: Difference between revisions

Revision as of 18:17, 8 November 2025

The following problem is from both the 2025 AMC 10A #23 and 2025 AMC 12A #16, so both problems redirect to this page.

Problem

Triangle $\triangle ABC$ has side lengths $AB = 80$ , $BC = 45$ , and $AC = 75$ . The bisector $\angle B$ 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

Let $CD \perp AB$ with foot $D$ . Right triangles $ACD$ and $BCD$ give $AC^2 = AD^2+CD^2$ , $BC^2 = BD^2+CD^2$ , $AC^2-BC^2 = AD^2-BD^2 =(AD-BD)(AD+BD)$ .

Since $AD+BD = AB = 80$ and $AC^2-BC^2 = 75^2-45^2 = 3600$ , substituting we get $AD-BD = \frac{3600}{80} = 45$ , $AD = \frac{125}{2}$ , $BD = \frac{35}{2}$ . $CD^2 = AC^2-AD^2 = 75^2-\left(\frac{125}{2}\right)^2 = \frac{6875}{4}$ , $CD = \frac{25\sqrt{11}}{2}$ .

By Angle Bisector Theorem, $\frac{DP}{PC} = \frac{DB}{BC} = \frac{\frac{35}{2}}{45} = \frac{7}{18}$ , $PC = CD-DP$ thus, $18DP = 7(CD-DP)$ , $25DP = 7CD$ , $DP = \left(\frac{7}{25}\right)CD = \left(\frac{7}{25}\right)\left(\frac{25\sqrt{11}}{2}\right) = \frac{7\sqrt{11}}{2}$ . $BP^2 = BD^2+DP^2 = \left(\frac{35}{2}\right)^2+\left(\frac{7\sqrt{11}}{2}\right)^2 = \frac{1225}{4}+\frac{49(11)}{4} = \frac{1764}{4} = 441$ , thus $BP = \boxed{\text{(D) }21}.$

~pigwash

Solution 2 (Law of Cosines)

Scale this down to a $9-15-16$ triangle (we will multiply the result by $5$ in the end). Note that $9^2+16^2-18*16*\cos(\angle B) = 15^2$ , so $112 = 18 * 16 * \cos(\angle B)$ , which simplifies to $\frac{7}{18} = \cos(\angle B)$ . Then $\cos(\frac{\angle B}{2}) = \sqrt{\frac{1+\frac{7}{18}}{2}} = \sqrt{\frac{25}{36}} = \frac{5}{6}$ (positive root since the angle is acute). Therefore, we have $\frac{BD}{BP} = \frac{5}{6}$ , assuming that $D$ is the foot of the altitude. There are many ways to proceed from here to find $BD$ . Note that by Heron's formula, the area of the scaled-down triangle is $\sqrt{(20)(4)(5)(11)} = 20\sqrt{11}$ . Therefore, $CD = \frac{5}{2}\sqrt{11}$ . Using Pythagorean Theorem, we get $BD= \sqrt{81-\frac{25 \cdot 11}{4}} = \sqrt{\frac{324-275}{4}} = \frac{7}{2}$ . Therefore, we get $\frac{\frac{7}{2}}{BP} = \frac{5}{6}$ , so $BP = \frac{21}{5}$ , and we scale up by $5$ to get $\boxed{21}$ . ~ScoutViolet

Solution 3 (Stewarts)

Let the foot of the altitude coming from $C$ on segment $AB$ be $D$ . Using the fact that $CD$ is a common leg in right triangles $\triangle CDB$ and $\triangle CDA$ , we have $45^2 - BD^2 = 75^2 - (80-BD)^2.$ Expanding gives $45^2=75^2-80^2+160BD,$ so $BD=\frac{35}{2}.$ Let the foot of the angle bisector from $B$ to $AC$ be point $E$ . Since $BE$ is the angle bisector of $\angle CBA$ , we can use the angle bisector theorem. This gives $\frac{45}{CE}=\frac{80}{75-CE},$ so $CE=27$ and $AE=75-27=48$ . Now we can use Stewart’s Theorem to find $\overline{BE}$ . We have $(48)(45)^2+(27)(80)^2=(27)(48)(75)+75BE^2.$ To simplify this expression, just divide by the greatest common divisor and solve from there. In the end, we get $BE=48$ . Let $BP=x$ , so $PE=48-x$ . Draw the altitude from $E$ down to $AB$ . Let the foot of this altitude be $F$ . Since $EF || CD$ , we have $\triangle AFE \sim \triangle ADC$ . Hence, we can write the equation $\frac{48}{75}=\frac{AF}{\frac{125}{2}}.$ Solving gives $AF=40$ , so $FD=\frac{125}{2}-40=\frac{45}{2}$ . Since $PD || EF$ , we also have $\triangle BDP \sim \triangle BFE$ , so we have $\frac{x}{48}=\frac{\frac{35}{2}}{40}.$ Solving for $x$ gives $\boxed{x=21}$ or $\boxed{\text{D}}.$

~evanhliu2009

Solution 4 (Rulerbash)

When we have a problem as such, involving a simple diagram with minimal instructions, I use a method I named "rulerbash". Rulerbash should only be used in specific cases and as a last resort, mainly in the event of a time crunch or a difficult problem. Start with the longest side, drawing a line with a length of $8$ cm ( $AB$ ). Then, using a compass, draw 2 circles centered around points $A$ and $B$ , $7.5$ and $4.5 cm$ radiuses respectfully. At the point of intersection of these 2 circles, we have point $C$ , completing a perfectly scaled drawing of $\triangle ABC$ . (Note the circles are not necessary with a bit of trial and error with the side lengths, they simply offer a way to get it done first try).

unitsize(1cm);

pair A=(0,0), B=(8,0);
real rA=7.5, rB=4.5;
path cA=circle(A,rA), cB=circle(B,rB);
pair[] pts = intersectionpoints(cA,cB);
pair C = pts[0];

draw(A--B);
draw(cA);
draw(cB);
draw(A--C--B--cycle);

dot(A); dot(B); dot(C);
label("$A$",A,S);
label("$B$",B,S);
label("$C$",C,N);
label("$8\text{ cm}$",(A+B)/2,S);
label("$7.5\text{ cm}$",A--C,NE,fontsize(9));
label("$4.5\text{ cm}$",B--C,NW,fontsize(9));
 (Error making remote request. Unknown error_msg)

~curryswish, shreyan.chethan

Video Solution 1 by OmegaLearn

https://youtu.be/DI-q_-cYMVU

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2025 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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