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

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==Problem==
 
==Problem==
 
Let <math>ABC</math> be a triangle inscribed in circle <math>\omega</math>. Let the tangents to <math>\omega</math> at <math>B</math> and <math>C</math> intersect at point <math>D</math>, and let <math>\overline{AD}</math> intersect <math>\omega</math> at <math>P</math>. If <math>AB=5</math>, <math>BC=9</math>, and <math>AC=10</math>, <math>AP</math> can be written as the form <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime integers. Find <math>m + n</math>.
 
Let <math>ABC</math> be a triangle inscribed in circle <math>\omega</math>. Let the tangents to <math>\omega</math> at <math>B</math> and <math>C</math> intersect at point <math>D</math>, and let <math>\overline{AD}</math> intersect <math>\omega</math> at <math>P</math>. If <math>AB=5</math>, <math>BC=9</math>, and <math>AC=10</math>, <math>AP</math> can be written as the form <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime integers. Find <math>m + n</math>.
 +
 +
==Diagram==
 +
<asy>
 +
import olympiad;
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unitsize(15);
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pair A, B, C, D, E, F, P, O;
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C = origin; A = (10,0); B = (7.8, 4.4899);
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draw(A--B--C--cycle); draw(A..B..C..cycle, red+dotted);
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O = circumcenter(A, B, C);
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E = rotate(90,B) * (O);
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F = rotate(90,C) * (O);
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D = IP(B..E + (B-E)*4, C..F + (C-F)*-3);
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draw(B--D--C--D--A);
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P = IP(D..A, A..B..C);
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dot(A); dot(B); dot(C); dot(D); dot(P);
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label("$A$", A, dir(335));
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label("$B$", B, dir(65));
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label("$C$", C, dir(200));
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label("$D$", D, dir(135));
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label("$P$", P, dir(235));
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</asy>
  
 
==Solution 1==
 
==Solution 1==
−
From the tangency condition we have <math>\let\angle BCP = \let\angle CBP = \let\angle A</math>. With LoC we have <math>\cos(A) = \frac{25+100-81}{2*5*10} = \frac{11}{25}</math> and <math>\cos(B) = \frac{81+25-100}{2*9*5} = \frac{1}{15}</math>. Then, <math>CD = \frac{\frac{9}{2}}{\cos(A)} = \frac{225}{22}</math>. Using LoC we can find <math>AD</math>: <math>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}</math>. Thus, <math>AD = \frac{5^2*13}{22}</math>. By Power of a Point, <math>DP*AD = CD^2</math> so <math>DP*\frac{5^2*13}{22} = (\frac{225}{22})^2</math> which gives <math>DP = \frac{5^2*9^2}{13*22}</math>. Finally, we have <math>AP = AD - DP = \frac{5^2*13}{22} - \frac{5^2*9^2}{13*22} = \frac{100}{13} \rightarrow \boxed{113}</math>.
+
From the tangency condition we have <math>\let\angle BCD = \let\angle CBD = \let\angle A</math>. With LoC we have <math>\cos(A) = \frac{25+100-81}{2*5*10} = \frac{11}{25}</math> and <math>\cos(B) = \frac{81+25-100}{2*9*5} = \frac{1}{15}</math>. Then, <math>CD = \frac{\frac{9}{2}}{\cos(A)} = \frac{225}{22}</math>. Using LoC we can find <math>AD</math>: <math>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}</math>. Thus, <math>AD = \frac{5^2*13}{22}</math>. By Power of a Point, <math>DP*AD = CD^2</math> so <math>DP*\frac{5^2*13}{22} = (\frac{225}{22})^2</math> which gives <math>DP = \frac{5^2*9^2}{13*22}</math>. Finally, we have <math>AP = AD - DP = \frac{5^2*13}{22} - \frac{5^2*9^2}{13*22} = \frac{100}{13} \rightarrow \boxed{113}</math>.
  
 
~angie.
 
~angie.
  
 
==Solution 2==
 
==Solution 2==
 +
[[File:2024 AIME I problem 10.png|300px|right]]
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We know <math>AP</math> is the symmedian,(see [[ Symmedians, Lemoine point | Symmedian and tangents ]])
  
−
We know <math>AP</math> is the symmedian, which implies <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>. Thus, we have <math>\frac{AP}{AC}=\frac{AB}{AM}, AP=\frac{100}{13}\implies \boxed{113}</math>
+
which implies <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>.  
 +
 
 +
Thus, we have <math>\frac{AP}{AC}=\frac{AB}{AM}, AP=\frac{100}{13}\implies \boxed{113}</math>
  
 
~Bluesoul
 
~Bluesoul
Line 96: Line 132:
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 +
==Solution 6==
 +
 +
Note that since P is a symmedian, <math>(AP;BC)</math> are harmonic. As a result, <math>\frac{BP}{PC} = \frac{BA}{CA}</math>. As a result, <math>2(BP) = PC</math>. Call <math>BP = x</math>. Then, <math>PC = 2x</math>. Since <math>cos A = \frac{11}{25}</math>, <math>cos BPC = - \frac{11}{25}</math>. Use LOC to find <math>x = \frac{45}{13}</math>. Finish with Ptolemy on ABPC, and finish to get <math>\frac{100}{13}</math>.
  
 
==See also==
 
==See also==

Latest revision as of 14:03, 5 October 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

From the tangency condition we have $\let\angle BCD = \let\angle CBD = \let\angle A$. 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 $AP = AD - DP = \frac{5^2*13}{22} - \frac{5^2*9^2}{13*22} = \frac{100}{13} \rightarrow \boxed{113}$.

~angie.

Solution 2

2024 AIME I problem 10.png

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

which implies $\triangle{ABP}\sim \triangle{AMC}$ 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 $\frac{100}{13}$.

See also

2024 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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