Difference between revisions of "2023 AIME II Problems/Problem 12"

Revision as of 17:16, 17 February 2023

Problem

In $\triangle ABC$ with side lengths $AB = 13,$ $BC = 14,$ and $CA = 15,$ let $M$ be the midpoint of $\overline{BC}.$ Let $P$ be the point on the circumcircle of $\triangle ABC$ such that $M$ is on $\overline{AP}.$ There exists a unique point $Q$ on segment $\overline{AM}$ such that $\angle PBQ = \angle PCQ.$ Then $AQ$ can be written as $\frac{m}{\sqrt{n}},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution

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

Because $A$ , $B$ , $C$ , and $P$ 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)

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 $\tan{MAL_1}=\frac{7-5}{12}=\frac{1}{6}$ , $\tan{CAL_1}=\frac{9}{12}=\frac{3}{4}$ , and $\tan{L_1AB}=\frac{5}{12}$ . 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 $99+148=\boxed{247}$ . ~sigma

Solution 3 (simplest)

It is clear that $BQCP$ is a parallelogram. By Stewart's Theorem, $AM=\sqrt{148}$ , POP tells $PM=\frac{49}{\sqrt{148}}$

As $QM=PM, AQ=AM-PM=\frac{99}{\sqrt{148}}$ leads to $\boxed{247}$

~bluesoul

@@ Line 7: / Line 7: @@
 Because <math>M</math> is the midpoint of <math>BC</math>, following from the Steward's theorem, <math>AM = 2 \sqrt{37}</math>.
-Because <math>A</math>, <math>B</math>, <math>C</math>, <math>P</math> are concyclic, <math>\angle BPA = \angle C</math>, <math>\angle CPA = \angle B</math>.
+Because <math>A</math>, <math>B</math>, <math>C</math>, and <math>P</math> are concyclic, <math>\angle BPA = \angle C</math>, <math>\angle CPA = \angle B</math>.
 Denote <math>\theta = \angle PBQ</math>.

2023 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search