2022 AIME I Problems/Problem 14

Problem

Given $\triangle ABC$ and a point $P$ on one of its sides, call line $\ell$ the splitting line of $\triangle ABC$ through $P$ if $\ell$ passes through $P$ and divides $\triangle ABC$ into two polygons of equal perimeter. Let $\triangle ABC$ be a triangle where $BC = 219$ and $AB$ and $AC$ are positive integers. Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{AC}$ , respectively, and suppose that the splitting lines of $\triangle ABC$ through $M$ and $N$ intersect at $30^{\circ}$ . Find the perimeter of $\triangle ABC$ .

The Geometry Part - Solution 1

Consider the splitting line through $M$ . Extend $D$ on ray $BC$ such that $CD=CA$ . Then the splitting line bisects segment $BD$ , so in particular it is the midline of triangle $ABD$ and thus it is parallel to $AD$ . But since triangle $ACD$ is isosceles, we can easily see $AD$ is parallel to the angle bisector of $C$ , so the splitting line is also parallel to this bisector, and similar for the splitting line through $N$ . Some simple angle chasing reveals the condition is now equivalent to $\angle A=120^\circ$ .

- MortemEtInteritum

The Geometry Part - Solution 2

Let $PM$ and $QN$ be the splitting lines. Reflect $B$ across $Q$ to be $B'$ and $C$ across $P$ to be $C'$ . Take $S_B$ and $S_C$ , which are spiral similarity centers on the other side of $BC$ as $A$ such that $\triangle S_BB'C \sim \triangle S_BBA$ and $\triangle S_CC'B \sim \triangle S_CCA$ . This gets that because $\angle S_BCB = \angle S_BCB' = \angle S_BAB$ and $\angle S_CBC = \angle S_CBC' = \angle S_CAC$ , then $S_B$ and $S_C$ are on $\triangle ABC$ 's circumcircle. Now, we know that $\triangle S_BBB' \sim \triangle S_BAC$ and $\triangle S_CCC' \sim \triangle S_CAB$ so because $BA=B'C$ and $CA=C'B$ , then $S_BB=SBB'$ and $S_CC=S_CC'$ and $S_BQ \perp BC$ and $S_CP \perp BC$ .

We also notice that because $Q$ and $N$ correspond on $\triangle S_BBB'$ and $\triangle S_BAC$ , and because $P$ and $M$ correspond on $\triangle S_CCC'$ and $\triangle S_CAB$ , then the angle formed by $NQ$ and $BA$ is equal to the angle formed by $B'C$ and $NQ$ which is equal to $\angle BS_BQ = \angle QS_BB'$ . Thus, $\angle CBA=2\angle CQN$ . Similarly, $\angle BCA = 2\angle QPM$ and so $\angle CBA + \angle BCA = 2\angle PQN + 2\angle QPM = 60^{\circ}$ and $\angle A = 120^{\circ}$ .

- kevinmathz

The NT Part

We now need to solve $a^2+ab+b^2 = 3^2\cdot 73^2$ . A quick $(\bmod 9)$ check gives that $3\mid a$ and $3\mid b$ . Thus, it's equivalent to solve $x^2+xy+y^2 = 73^2$ .

Let $\omega$ be one root of $\omega^2+\omega+1=0$ . Then, recall that $\mathbb Z[\omega]$ is the ring of integers of $\mathbb Q[\sqrt{-3}]$ and is a unique factorization domain. Notice that $N(x-y\omega) = (x-y\omega)(x-y\omega^2) = x^2-xy+y^2$ . Therefore, it suffices to find an element of $\mathbb Z[\omega]$ with the norm $73^2$ .

To do so, we factor $73$ in $\mathbb Z[\omega]$ . Since it's $1\pmod 3$ , it must split. A quick inspection gives $73 = (8-\omega)(8-\omega^2)$ . Thus, $N(8-\omega) = 73$ , so \begin{align*} 73^2 &= N((8-\omega)^2) \\ &= N(64 - 16\omega + \omega^2) \\ &= N(64 - 16\omega + (-1-\omega)) \\ &= N(63 - 17\omega), \end{align*}giving the solution $x=63$ and $y=17$ , yielding $a=189$ and $b=51$ , so the sum is $\boxed{459}$ . Since $8-\omega$ and $8-\omega^2$ are primes in $\mathbb Z[\omega]$ , the solution must divide $73^2$ . One can then easily check that this is the unique solution.

- MarkBcc168

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2022 AIME I Problems/Problem 14

Contents

Problem

The Geometry Part - Solution 1

The Geometry Part - Solution 2

The NT Part