2023 AIME II Problems/Problem 12

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 $99 + 148 = \boxed{\textbf{(247) }}$ .

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

Page

Toolbox

Search

2023 AIME II Problems/Problem 12

Solution