Difference between revisions of "2024 AIME I Problems/Problem 10"

Latest revision as of 19:23, 22 December 2024

Problem

Let $ABC$ be a triangle inscribed in circle $\omega$ . Let the tangents to $\omega$ at $B$ and $C$ intersect at point $D$ , and let $\overline{AD}$ intersect $\omega$ at $P$ . If $AB=5$ , $BC=9$ , and $AC=10$ , $AP$ can be written as the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime integers. Find $m + n$ .

Diagram

$[asy] import olympiad; unitsize(15); pair A, B, C, D, E, F, P, O; C = origin; A = (10,0); B = (7.8, 4.4899); draw(A--B--C--cycle); draw(A..B..C..cycle, red+dotted); O = circumcenter(A, B, C); E = rotate(90,B) * (O); F = rotate(90,C) * (O); D = IP(B..E + (B-E)*4, C..F + (C-F)*-3); draw(B--D--C--D--A); P = IP(D..A, A..B..C); dot(A); dot(B); dot(C); dot(D); dot(P); label("$A$", A, dir(335)); label("$B$", B, dir(65)); label("$C$", C, dir(200)); label("$D$", D, dir(135)); label("$P$", P, dir(235)); [/asy]$

Solution 1

We have $\let\angle BCD = \let\angle CBD = \let\angle A$ from the tangency condition. With LoC we have $\cos(A) = \frac{25+100-81}{2*5*10} = \frac{11}{25}$ and $\cos(B) = \frac{81+25-100}{2*9*5} = \frac{1}{15}$ . Then, $CD = \frac{\frac{9}{2}}{\cos(A)} = \frac{225}{22}$ . Using LoC we can find $AD$ : $AD^2 = AC^2 + CD^2 - 2(AC)(CD)\cos(A+C) = 10^2+(\frac{225}{22})^2 + 2(10)\frac{225}{22}\cos(B) = 100 + \frac{225^2}{22^2} + 2(10)\frac{225}{22}*\frac{1}{15} = \frac{5^4*13^2}{484}$ . Thus, $AD = \frac{5^2*13}{22}$ . By Power of a Point, $DP*AD = CD^2$ so $DP*\frac{5^2*13}{22} = (\frac{225}{22})^2$ which gives $DP = \frac{5^2*9^2}{13*22}$ . Finally, we have $AP = AD - DP = \frac{5^2*13}{22} - \frac{5^2*9^2}{13*22} = \frac{100}{13}$ . So the answer is $\boxed{113}$ .

~angie.

Solution 2

We know $AP$ is the symmedian (see Symmedian and tangents ) ,

which implies that $\triangle{ABP}\sim \triangle{AMC}$ where $M$ is the midpoint of $BC$ .

By Appolonius' theorem, $AM=\frac{13}{2}$ .

Thus, we have $\frac{AP}{AC}=\frac{AB}{AM}, AP=\frac{100}{13}\implies \boxed{113}$

~Bluesoul

Solution 3

Extend sides $\overline{AB}$ and $\overline{AC}$ to points $E$ and $F$ , respectively, such that $B$ and $C$ are the feet of the altitudes in $\triangle AEF$ . Denote the feet of the altitude from $A$ to $\overline{EF}$ as $X$ , and let $H$ denote the orthocenter of $\triangle AEF$ . Call $M$ the midpoint of segment $\overline{EF}$ . By the Three Tangents Lemma, we have that $MB$ and $MC$ are both tangents to $(ABC)$ $\implies$ $M = D$ , and since $M$ is the midpoint of $\overline{EF}$ , $MF = MB$ . Additionally, by angle chasing, we get that: $\angle ABC \cong \angle AHC \cong \angle EHX$ Also, $\angle EHX = 90 ^\circ - \angle HEF = 90 ^\circ - (90 ^\circ - \angle AFE) = \angle AFE$ Furthermore, $AB = AF \cdot \cos(A)$ From this, we see that $\triangle ABC \sim \triangle AFE$ with a scale factor of $\cos(A)$ . By the Law of Cosines, $\cos(A) = \frac{10^2 + 5^2 - 9^2}{2 \cdot 10 \cdot 5} = \frac{11}{25}$ Thus, we can find that the side lengths of $\triangle AEF$ are $\frac{250}{11}, \frac{125}{11}, \frac{225}{11}$ . Then, by Stewart's theorem, $AM = \frac{13 \cdot 25}{22}$ . By Power of a Point, $\overline{MB} \cdot \overline{MB} = \overline{MA} \cdot \overline{MP}$ $\frac{225}{22} \cdot \frac{225}{22} = \overline{MP} \cdot \frac{13 \cdot 25}{22} \implies \overline{MP} = \frac{225 \cdot 9}{22 \cdot 13}$ Thus, $AP = AM - MP = \frac{13 \cdot 25}{22} - \frac{225 \cdot 9}{22 \cdot 13} = \frac{100}{13}$ Therefore, the answer is $\boxed{113}$ .

~mathwiz_1207

Solution 4 (LoC spam)

Connect lines $\overline{PB}$ and $\overline{PC}$ . From the angle by tanget formula, we have $\angle PBD = \angle DAB$ . Therefore by AA similarity, $\triangle PBD \sim \triangle BAD$ . Let $\overline{BP} = x$ . Using ratios, we have $\frac{x}{5}=\frac{BD}{AD}.$ Similarly, using angle by tangent, we have $\angle PCD = \angle DAC$ , and by AA similarity, $\triangle CPD \sim \triangle ACD$ . By ratios, we have $\frac{PC}{10}=\frac{CD}{AD}.$ However, because $\overline{BD}=\overline{CD}$ , we have $\frac{x}{5}=\frac{PC}{10},$ so $\overline{PC}=2x.$ Now using Law of Cosines on $\angle BAC$ in triangle $\triangle ABC$ , we have $9^2=5^2+10^2-100\cos(\angle BAC).$ Solving, we find $\cos(\angle BAC)=\frac{11}{25}$ . Now we can solve for $x$ . Using Law of Cosines on $\triangle BPC,$ we have \begin{align*} 81&=x^2+4x^2-4x^2\cos(180-\angle BAC) \\ &= 5x^2+4x^2\cos(BAC). \\ \end{align*}

Solving, we get $x=\frac{45}{13}.$ Now we have a system of equations using Law of Cosines on $\triangle BPA$ and $\triangle CPA$ , $AP^2=5^2+\left(\frac{45}{13}\right)^2 -(10) \left(\frac{45}{13} \right)\cos(ABP)$ $AP^2=10^2+4 \left(\frac{45}{13} \right)^2 + (40) \left(\frac{45}{13} \right)\cos(ABP).$

Solving, we find $\overline{AP}=\frac{100}{13}$ , so our desired answer is $100+13=\boxed{113}$ .

~evanhliu2009

Solution 5

Following from the law of cosines, we can easily get $\cos A = \frac{11}{25}$ , $\cos B = \frac{1}{15}$ , $\cos C = \frac{13}{15}$ .

Hence, $\sin A = \frac{6 \sqrt{14}}{25}$ , $\cos 2C = \frac{113}{225}$ , $\sin 2C = \frac{52 \sqrt{14}}{225}$ . Thus, $\cos \left( A + 2C \right) = - \frac{5}{9}$ .

Denote by $R$ the circumradius of $\triangle ABC$ . In $\triangle ABC$ , following from the law of sines, we have $R = \frac{BC}{2 \sin A} = \frac{75}{4 \sqrt{14}}$ .

Because $BD$ and $CD$ are tangents to the circumcircle $ABC$ , $\triangle OBD \cong \triangle OCD$ and $\angle OBD = 90^\circ$ . Thus, $OD = \frac{OB}{\cos \angle BOD} = \frac{R}{\cos A}$ .

In $\triangle AOD$ , we have $OA = R$ and $\angle AOD = \angle BOD + \angle AOB = A + 2C$ . Thus, following from the law of cosines, we have

\begin{align*} AD & = \sqrt{OA^2 + OD^2 - 2 OA \cdot OD \cos \angle AOD} \\ & = \frac{26 \sqrt{14}}{33} R. \end{align*}

Following from the law of cosines,

\begin{align*} \cos \angle OAD & = \frac{AD^2 + OA^2 - OD^2}{2 AD \cdot OA} \\ & = \frac{8 \sqrt{14}}{39} . \end{align*}

Therefore,

\begin{align*} AP & = 2 OA \cos \angle OAD \\ & = \frac{100}{13} . \end{align*}

Therefore, the answer is $100 + 13 = \boxed{\textbf{(113) }}$ .

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution 1 by OmegaLearn.org

https://youtu.be/heryP002bp8

Video Solution

https://youtu.be/RawwQmVYyaw

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 6

Note that since P is a symmedian, $(AP;BC)$ are harmonic. As a result, $\frac{BP}{PC} = \frac{BA}{CA}$ . As a result, $2(BP) = PC$ . Call $BP = x$ . Then, $PC = 2x$ . Since $cos A = \frac{11}{25}$ , $cos BPC = - \frac{11}{25}$ . Use LOC to find $x = \frac{45}{13}$ . Finish with Ptolemy on ABPC, and finish to get $\frac{100}{13}$ .

@@ Line 43: / Line 43: @@
 which implies that <math>\triangle{ABP}\sim \triangle{AMC}</math> where <math>M</math> is the midpoint of <math>BC</math>.
-By Appolonius theorem, <math>AM=\frac{13}{2}</math>.
+By Appolonius' theorem, <math>AM=\frac{13}{2}</math>.
 Thus, we have <math>\frac{AP}{AC}=\frac{AB}{AM}, AP=\frac{100}{13}\implies \boxed{113}</math>

2024 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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