AIME 2024 I · 第 10 题

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 Apollonius' 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 tangent 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 $\frac{100}{13}$.

Solution 7 (1 min solve)

Note that $AD$ is the A-symmedian in $\triangle ABC$. Let $M$ be the midpoint of $BC$. It is a well known property that $\triangle ABP \sim \triangle AMC$. Therefore, using the median length formula,

$$AM = \frac{\sqrt{200+50-81}}{2} = \frac{13}{2},$$ so

$$\frac{\frac{13}{2}}{10} = \frac{5}{AP} \implies AP = \frac{100}{13},$$ so our answer is $\boxed{113}$.

~Yiyj1

Solution 8 (Inversion)

Take a $\sqrt{bc}$ inversion, and note that the circumcircle of $\triangle ABC$ gets mapped to $\overline{BC}$. Therefore, $P$ is mapped to a point on $\overline{BC}$. Also, the tangents map to circles that pass through $A$ and are tangent to $\overline{BC}$ at $B$ or $C$. The radical axis of the two circles therefore passes through the midpoint of $\overline{BC}$, and thus $P$ is mapped to the midpoint of $\overline{BC}$. By Stewart's, $AM=\frac{13}{2}$, so $AP=\frac{AB*AC}{AM}=\frac{100}{13}$. Therefore, the answer is $\boxed{113}$.

~MATHLOVERSSD