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

Latest revision as of 22:15, 11 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 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

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 $AC^2-BC^2 = 75^2-45^2 = 3600$ , we get the equation $3600 = 80(AD-BD)$ . This equation simplifies to $45 = AD - BD$ . We can solve the system of equations $AD + BD = 80$ and $AD - BD = 45$ easily via elimination, and we get $AD = \frac{125}{2}$ , $BD = \frac{35}{2}$ . $CD^2 = AC^2-AD^2 = 75^2-\left(\frac{125}{2}\right)^2 = \frac{6875}{4}$ , $CD = \frac{25\sqrt{11}}{2}$ .

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 ~aldzandrtc(fixed logical jump)

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)

Preface: 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 $8\,\text{cm}$ ( $AB$ ). Then, using a compass, draw 2 circles centered around points $A$ and $B$ , $7.5\,\text{cm}$ and $4.5\,\text{cm}$ 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).

$[asy] unitsize(0.25cm); pair A=(0,0), B=(8,0); path cA=circle(A,7.5), cB=circle(B,4.5); pair C=intersectionpoint(cA,cB); draw(A--B); draw(cA); draw(cB); draw(A--C--B--cycle); dot(A); dot(B); dot(C); label("A",A,S, fontsize(8)); label("B",B,S, fontsize(8)); label("C",C,N, fontsize(8)); label("8 cm",(A+B)/2, S, fontsize(6)); label("7.5 cm",(A+C)/2 + (-0.85,0.8),fontsize(6)); label("4.5 cm",(B+C)/2 + (1.75,-0.1),fontsize(6)); [/asy]$

After this, dropping the altitude to $AB$ is simple with a ruler and careful placement, and angle bisector can be estimated quite accurately.

$[asy] unitsize(0.55cm); import olympiad; pair A=(0,0), B=(8,0); pair C=intersectionpoint(circle(A,7.5),circle(B,4.5)); draw(A--B--C--cycle); pair F=foot(C,A,B); draw(C--F); draw(rightanglemark(A,F,C,2)); pair U=unit(unit(A-B)+unit(C-B)); pair P=intersectionpoint(B--(B+100*U), C--F); draw(B--P,dashed); dot(A);dot(B);dot(C);dot(P); label("A",A,S,fontsize(12)); label("B",B,S,fontsize(12)); label("C",C,N,fontsize(12)); label("P",P,NE,fontsize(12)); [/asy]$

After all of this, we can reuse our ruler and measure $BP = 2.1\,\text{cm}$ , and using our scale of $80=8\,\text{cm}$ , our final answer is $\boxed{\text{(D) }21}.$

~shreyan.chethan (notes by curryswish)

Solution 5 (No Trig)

Let $D$ be the intersection of the altitude from $C$ to $AB$. To simplify calculations, 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 $DB$: $DB^2 + \frac{25 \cdot 11}{4} = 81 \quad \Rightarrow \quad 4DB^2 + 275 = 324 \quad \Rightarrow \quad DB^2 = \frac{49}{4},$ so $DB = \tfrac{7}{2}$. Next, we use the Angle Bisector Theorem to find $PD$. Let $x = PC$ and $y = PD$. Since $x + y = \tfrac{5\sqrt{11}}{2}$, we have $x = \tfrac{5\sqrt{11}}{2} - y$. From the given ratio, $\frac{9}{x} = \frac{7}{2y} \quad \Rightarrow \quad 18y = 7x.$ Substituting $x = \tfrac{5\sqrt{11}}{2} - y$, $18y = 7\left(\tfrac{5\sqrt{11}}{2} - y\right) \quad \Rightarrow \quad 25y = \tfrac{35\sqrt{11}}{2},$ so $y = \tfrac{7\sqrt{11}}{10}$.

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

Finally, multiplying the side lengths by $5$ again gives $BP = 21.$, or $\boxed{\text{D}}.$

~Voidling

Solution 6 (Similar Triangle)

$[asy] /* Figure drawn by reda*/ import geometry; unitsize(2.5); pair _A = (0, 0); pair _B = (80, 0); pair _C = (75*(75^2+80^2-45^2)/(2*75*80), 75*sqrt(1-((75^2+80^2-45^2)/(2*75*80))^2)); pair _D = (_C.x, 0); pair _E = (_A * 45 + _C * 80)/(45 + 80); pair _F = (_E.x, 0); pair _P = extension(_C, _D, _B, _E); draw(_B -- _A -- _C -- _B -- _E ^^ _C -- _D); draw(_E -- _F, dashed); /*dot(_A ^^ _B ^^ _C ^^ _D ^^ _E ^^ _F ^^ _P);*/ markrightangle(_B, _D, _P, 0.2*markangleradius()); markrightangle(_B, _F, _E, 0.2*markangleradius()); markangle(_P, _B, _D, radius = 0.25*markangleradius()); markangle(_P, _B, _D, radius = 0.3*markangleradius()); markangle(_C, _B, _P, radius = 0.25*markangleradius()); markangle(_C, _B, _P, radius = 0.3*markangleradius()); markangle(_B, _A, _C, radius = 0.25*markangleradius()); markangle(_B, _A, _C, radius = 0.3*markangleradius()); label("$A$", _A, S); label("$B$", _B, S); label("$C$", _C, N); label("$D$", _D, S); label("$E$", _E, NW); label("$F$", _F, S); label("$P$", _P, NE); label("$27$", (_C + _E)/2, N); label("$48$", (_A + _E)/2, N); label("$45$", (_B + _C)/2, E); label("$80$", (_A + _B)/2, 4S); [/asy]$

Due to Angle Bisector Theorem $\frac{CE}{AE} = \frac{BC}{BA} = \frac{45}{80} = \frac{9}{16}$ $CE = AC \times \frac{9}{25} = 27$ $AE = AC \times \frac{16}{25} = 48$ Notice that $\dfrac{CE}{CB} = \dfrac{CB}{CA} = \dfrac{3}{5}$ , $\triangle CBE \sim \triangle CAB$ , $\angle A = \angle CBE = \angle EBF$ , so $\triangle ABE$ is isosceles triangle, hence $AF = BF$ Notice that $AF:FD = AE:EC = 16:9$ , $DB = FB - FD$ , then $AF:DB = 16:7$

Notice that $\triangle PBD \sim \triangle EAF$ , $PB = AE \times \dfrac{BD}{AF} = 48 \times \dfrac{7}{16} = \boxed{\text{(D) }21}.$

~范_mandymath

Solution 7 (Pythagoras+angle bisector theorem)

Let $CD$ be the altitude to side $AB$ , and let $E$ be the point of intersection of the angle bisector of $\angle B$ . Let $P$ be the point of intersection of the angle bisector and the altitude.

By Pythagoras, $CD^2=BC^2-BD^2=AC^2-AD^2$

We let $BD = x$ , so $AD=80-x$

Now we have

$45^2-x^2=75^2-(80-x)^2$

(This looks complicated, but we can see that we are going to have two $-x^2$ terms on both sides, so it cancels out)

Simplifying and rearranging gives us $80-2x=45$ , so $x=\frac{35}{2}$

Therefore, $BD=\frac{35}{2}$ , and $AD=\frac{125}{2}$ . This gives us $CD=\frac{25}{2}\sqrt{11}$

Now, note that in triangle $BDC$ , $BP$ bisects $\angle DBC$

By the angle bisection theorem, we have $\frac{BD}{DP}=\frac{BC}{CP}$

which gives us $\frac{DP}{CP}=\frac{BD}{BC}=\frac{7}{18}$

We have $DP=\frac{7}{25}*CD=\frac{7}{2}\sqrt{11}$ and $CP=\frac{18}{25}*CD=\frac{18}{2}\sqrt{11}$

So

$BP^2=BD*BC-DP*CP=\frac{35}{2}*45-\frac{7}{2}*\frac{18}{2}*11=441$

Therefore, $BP=21$

(This is my first solution, pls point out any mistakes in math or LATEX). ~backtosq-1

Solution 8 (Pythagora's + Double Angle Identity)

Since all the lengths share a factor of $5$ , divide by $5$ and multiply it back at the end for smaller numbers. Drop the altitude from vertex $C$ to point $D$ on $AB$ . Then comparing the altitude by Pythagora's on both sides gives $9^2-x^2=15^2-(16-x)^2=15^2-16^2+32x-x^2=-31+32x-x^2$ Cancelling the $-x^2$ then solving gives us $x=\frac{112}{32}=\frac{7}{2}$ . Since $\triangle BDC$ is a right triangle with $\angle B=2\theta$ , we find $\cos(2\theta)=\frac{\frac{7}{2}}{9}=\frac{7}{18}$ .

Let $\theta=\angle PBA=\angle PBC$ . Because $\angle B$ is opposite to the leg $9<15$ and not the hypotenuse, we must have that $2\theta$ is acute, likewise for $\theta$ . Hence, $\cos \theta>0$ .

Thus, using cosine definition on $\triangle BDP$ and the double angle identity $\cos 2\theta=2\cos^2 \theta-1$ , we get $\cos \theta=\frac{7/2}{BP}=\sqrt{ \frac{1+\cos 2\theta}{2} }=\sqrt{ \frac{1+\frac{7}{18}}{2} }=\sqrt{ \frac{25}{36} }=\frac{5}{6}$ Finally, solving for $BP=\frac{7}{2\cos\theta}=\frac{7}{2}\cdot \frac{6}{5}=\frac{21}{5}$ . Scaling back to the original diagram by $5$ gives $\boxed{\textbf{(D)}~21}$ .

Video Solution (In 3 Mins) (Really Easy)

https://youtu.be/nAimLnvSTwQ?si=85o8QuW5HbDldWdU ~ Pi Academy

Video Solution 1 by OmegaLearn

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

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

@@ Line 216: / Line 216: @@
 Finally, solving for <imath>BP=\frac{7}{2\cos\theta}=\frac{7}{2}\cdot \frac{6}{5}=\frac{21}{5}</imath>. Scaling back to the original diagram by <imath>5</imath> gives <imath>\boxed{\textbf{(D)}~21}</imath>.
-== Video Solution (In 2 Mins) (Really Easy) ==
+== Video Solution (In 3 Mins) (Really Easy) ==
 https://youtu.be/nAimLnvSTwQ?si=85o8QuW5HbDldWdU ~ Pi Academy

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