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Difference between revisions of "2024 AIME I Problems/Problem 10"
m (→‎Solution 3)
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==Solution 3==
 
==Solution 3==
  
−
Extend sides <math>\overline{AB}</math> and <math>\overline{AC}</math> to points <math>E</math> and <math>F</math>, respectively, such that <math>B</math> and <math>C</math> are the feet of the altitudes in <math>\triangle AEF</math>. Denote the feet of the altitude from <math>A</math> to <math>\overline{EF}</math> as <math>X</math>, and let <math>H</math> denote the orthocenter of <math>\triangle AEF</math>. Call <math>M</math> the midpoint of segment <math>\overline{EF}</math>. By the Three Tangents Lemma, we have that <math>MB</math> and <math>MC</math> are both tangents to <math>(ABC)</math> <math>\implies</math> <math>M = P</math>, and since <math>M</math> is the midpoint of <math>\overline{EF}</math>, <math>MF = MB</math>. Additionally, by angle chasing, we get that:  
+
Extend sides <math>\overline{AB}</math> and <math>\overline{AC}</math> to points <math>E</math> and <math>F</math>, respectively, such that <math>B</math> and <math>C</math> are the feet of the altitudes in <math>\triangle AEF</math>. Denote the feet of the altitude from <math>A</math> to <math>\overline{EF}</math> as <math>X</math>, and let <math>H</math> denote the orthocenter of <math>\triangle AEF</math>. Call <math>M</math> the midpoint of segment <math>\overline{EF}</math>. By the Three Tangents Lemma, we have that <math>MB</math> and <math>MC</math> are both tangents to <math>(ABC)</math> <math>\implies</math> <math>M = D</math>, and since <math>M</math> is the midpoint of <math>\overline{EF}</math>, <math>MF = MB</math>. Additionally, by angle chasing, we get that:  
 
<cmath>\angle ABC \cong \angle AHC \cong \angle EHX</cmath>
 
<cmath>\angle ABC \cong \angle AHC \cong \angle EHX</cmath>
 
Also,  
 
Also,  
Line 25: Line 25:
 
Thus, we can find that the side lengths of <math>\triangle AEF</math> are <math>\frac{250}{11}, \frac{125}{11}, \frac{225}{11}</math>. Then, by Stewart's theorem, <math>AM = \frac{13 \cdot 25}{22}</math>. By Power of a Point,  
 
Thus, we can find that the side lengths of <math>\triangle AEF</math> are <math>\frac{250}{11}, \frac{125}{11}, \frac{225}{11}</math>. Then, by Stewart's theorem, <math>AM = \frac{13 \cdot 25}{22}</math>. By Power of a Point,  
 
<cmath>\overline{MB} \cdot \overline{MB} = \overline{MA} \cdot \overline{MD}</cmath>
 
<cmath>\overline{MB} \cdot \overline{MB} = \overline{MA} \cdot \overline{MD}</cmath>
−
<cmath>\frac{225}{22} \cdot \frac{225}{22} = \overline{MD} \cdot \frac{13 \cdot 25}{22} \implies \overline{MD} = \frac{225 \cdot 9}{22 \cdot 13}</cmath>
+
<cmath>\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}</cmath>
 
Thus,  
 
Thus,  
−
<cmath>AD = AM - MD = \frac{13 \cdot 25}{22} - \frac{225 \cdot 9}{22 \cdot 13} = \frac{100}{13}</cmath>
+
<cmath>AP = AM - MP = \frac{13 \cdot 25}{22} - \frac{225 \cdot 9}{22 \cdot 13} = \frac{100}{13}</cmath>
 
Therefore, the answer is <math>\boxed{113}</math>.
 
Therefore, the answer is <math>\boxed{113}</math>.
  

Revision as of 19:55, 3 February 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 $P$. Find $AP$, if $AB=5$, $BC=9$, and $AC=10$.

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

Well know $AP$ is the symmedian, 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, \[\overline{MB} \cdot \overline{MB} = \overline{MA} \cdot \overline{MD}\] \[\frac{225}{22} \cdot \frac{225}{22} = \overline{MP} \cdot \frac{13 \cdot 25}{22} \implies \overline{MP} = \frac{225 \cdot 9}{22 \cdot 13}\] 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(BPC). \\ \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

Video Solution 1 by OmegaLearn.org

https://youtu.be/heryP002bp8


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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