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2023 AIME II Problems/Problem 12

Revision as of 18:58, 16 February 2023 by Sigmapie (talk | contribs)

Solution

Because $M$ is the midpoint of $BC$, following from the Steward's theorem, $AM = 2 \sqrt{37}$.

Because $A$, $B$, $C$, $P$ are concyclic, $\angle BPA = \angle C$, $\angle CPA = \angle B$.

Denote $\theta = \angle PBQ$.

In $\triangle BPQ$, following from the law of sines, \[ \frac{BQ}{\sin \angle BPA} = \frac{PQ}{\angle PBQ} . \]

Thus, \[ \frac{BQ}{\sin C} = \frac{PQ}{\sin \theta} . \hspace{1cm} (1) \]

In $\triangle CPQ$, following from the law of sines, \[ \frac{CQ}{\sin \angle CPA} = \frac{PQ}{\angle PCQ} . \]

Thus, \[ \frac{CQ}{\sin B} = \frac{PQ}{\sin \theta} . \hspace{1cm} (2) \]

Taking $\frac{(1)}{(2)}$, we get \[ \frac{BQ}{\sin C} = \frac{CQ}{\sin B} . \]

In $\triangle ABC$, following from the law of sines, \[ \frac{AB}{\sin C} = \frac{AC}{\sin B} . \hspace{1cm} (3) \]

Thus, Equations (2) and (3) imply \begin{align*} \frac{BQ}{CQ} & = \frac{AB}{AC} \\ & = \frac{13}{15} . \hspace{1cm} (4) \end{align*}


Next, we compute $BQ$ and $CQ$.

We have \begin{align*} BQ^2 & = AB^2 + AQ^2 - 2 AB\cdot AQ \cos \angle BAQ \\ & = AB^2 + AQ^2 - 2 AB\cdot AQ \cos \angle BAM \\ & = AB^2 + AQ^2 - 2 AB\cdot AQ \cdot \frac{AB^2 + AM^2 - BM^2}{2 AB \cdot AM} \\ & = AB^2 + AQ^2 - AQ \cdot \frac{AB^2 + AM^2 - BM^2}{AM} \\ & = 169 + AQ^2 - \frac{268}{2 \sqrt{37}} AQ . \hspace{1cm} (5) \end{align*}

We have \begin{align*} CQ^2 & = AC^2 + AQ^2 - 2 AC\cdot AQ \cos \angle CAQ \\ & = AC^2 + AQ^2 - 2 AC\cdot AQ \cos \angle CAM \\ & = AC^2 + AQ^2 - 2 AC\cdot AQ \cdot \frac{AC^2 + AM^2 - CM^2}{2 AC \cdot AM} \\ & = AC^2 + AQ^2 - AQ \cdot \frac{AC^2 + AM^2 - CM^2}{AM} \\ & = 225 + AQ^2 - \frac{324}{2 \sqrt{37}} AQ . \hspace{1cm} (6) \end{align*}

Taking (5) and (6) into (4), we get $AQ = \frac{99}{\sqrt{148}}$. Therefore, the answer is $99 + 148 = \boxed{\textbf{(247) }}$.

Solution 2 (

Define $L_1$ to be the foot of the altitude from $A$ to $BC$. Furthermore, define $L_2$ to be the foot of the altitude from $Q$ to $BC$. From here, one can find $AL_1=12$, either using the 13-14-15 triangle or by calculating the area of $ABC$ two ways. Then, we find $BL_1=5$ and $L_1C = 9$ using Pythagorean theorem. Let $QL_2=x$. By AA similarity, $\triangle{AL_1M}$ and $\triangle{QL_2M}$ are similar. By similarity ratios, \[\frac{AL_1}{L_1M}=\frac{QL_2}{L_2M}\] \[\frac{12}{2}=\frac{x}{L_2M}\] \[L_2M = \frac{x}{6}\] Thus, $BL_2=BM-L_2M=7-\frac{x}{6}$. Similarly, $CL_2=7+\frac{x}{6}$. Now, we angle chase from our requirement to obtain new information. \[\angle{PBQ}=\angle{PCQ}\] \[\angle{QCM}+\angle{PCM}=\angle{QBM}+\angle{PBM}\] \[\angle{QCL_2}+\angle{PCM}=\angle{QBL_2}+\angle{PBM}\] \[\angle{PCM}-\angle{PBM}=\angle{QBL_2}-\angle{QCL_2}\] \[\angle{MAB}-\angle{MAC}=\angle{QBL_2}-\angle{QCL_2}\text{(By inscribed angle theorem)}\] \[(\angle{MAL_1}+\angle{L_1AB})-(\angle{CAL_1}-\angle{MAL_1})=\angle{QBL_2}-\angle{QCL_2}\] \[2\angle{MAL_1}+\angle{L_1AB}-\angle{CAL_1}=\angle{QBL_2}-\angle{QCL_2}\] Take the tangent of both sides to obtain \[\tan(2\angle{MAL_1}+\angle{L_1AB}-\angle{CAL_1})=\tan(\angle{QBL_2}-\angle{QCL_2})\] By the definition of the tangent function on right triangles, we have $\tan^{-1}{MAL_1}=\frac{7-5}{12}=\frac{1}{6}$, $\tan^{-1}{CAL_1}=\frac{9}{12}=\frac{3}{4}$, and $\tan^{-1}{L_1AB}=\frac{5}{12}$. By abusing the tangent angle addition formula, we can find that \[\tan(2\angle{MAL_1}+\angle{L_1AB}-\angle{CAL_1})=\frac{196}{2397}\] By substituting $\tan{\angle{QBL_2}}=\frac{6x}{42-x}$, $\tan{\angle{QCL_2}}=\frac{6x}{42+x}$ and using tangent angle subtraction formula we find that \[x=\frac{147}{37}\] Finally, using similarity formulas, we can find \[\frac{AQ}{AM}=\frac{12-x}{x}\]. Plugging in $x=\frac{147}{37}$ and $AM=\sqrt{148}$, we find that \[AQ=\frac{99}{\sqrt{148}}\] Thus, our final answer is $99+148=\boxed{247}$. ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

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