AIME 2022 I, Problem 14

Given $\triangle ABC$ and a point $P$ on one of its sides, call line $\ell$ the $\textit{splitting line}$ 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 $\triangle ABC.$

Solution: Denote $BC = a$, $CA = b$, $AB = c$. Let the splitting line of $\triangle ABC$ through $M$ (resp. $N$) crosses $\triangle ABC$ at another point $X$ (resp. $Y$). WLOG, we assume $c \leq b$. $\textbf{Case 1}$: $a \leq c \leq b$. We extend segment $AB$ to $D$, such that $BD = a$. We extend segment $AC$ to $E$, such that $CE = a$. In this case, $X$ is the midpoint of $AE$, and $Y$ is the midpoint of $AD$. Because $M$ and $X$ are the midpoints of $AB$ and $AE$, respectively, $MX \parallel BE$. Because $N$ and $Y$ are the midpoints of $AC$ and $AD$, respectively, $NY \parallel CD$. Because $CB = CE$, $\angle CBE =\angle CEB = \frac{\angle ACB}{2}$. Because $BC = BD$, $\angle BCD = \angle BDC = \frac{\angle ABC}{2}$. Let $BE$ and $CD$ intersect at $O$. Because $MX \parallel BE$ and $NY \parallel CD$, the angle formed between lines $MX$ and $NY$ is congruent to $\angle BOD$. Hence, $\angle BOD = 30^\circ$ or $150^\circ$. We have \begin{align} \angle BOD & = \angle CBE + \angle BCD \\ & = \frac{\angle ACB}{2} + \frac{\angle ABC}{2} \\ & = 90^\circ - \frac{\angle A}{2} . \end{align} Hence, we must have $\angle BOD = 30^\circ$, not $150^\circ$. Hence, $\angle A = 120^\circ$. This implies $a > b$ and $a >c$. This contradicts the condition specified for this case. Therefore, this case is infeasible. $\textbf{Case 2}$: $c \leq a \leq b$. We extend segment $CB$ to $D$, such that $BD = c$. We extend segment $AC$ to $E$, such that $CE = a$. In this case, $X$ is the midpoint of $AE$, and $Y$ is the midpoint of $CD$. Because $M$ and $X$ are the midpoints of $AB$ and $AE$, respectively, $MX \parallel BE$. Because $N$ and $Y$ are the midpoints of $AC$ and $CD$, respectively, $NY \parallel AD$. Because $CB = CE$, $\angle CBE =\angle CEB = \frac{\angle ACB}{2}$. Because $BA = BD$, $\angle BAD = \angle BDA = \frac{\angle ABC}{2}$. Let $O$ be a point of $AC$, such that $BO \parallel AD$. Hence, $\angle OBC = \angle BDA = \frac{B}{2}$. Because $MX \parallel BE$ and $NY \parallel AD$ and $AD \parallel BO$, the angle formed between lines $MX$ and $NY$ is congruent to $\angle OBE$. Hence, $\angle OBE = 30^\circ$ or $150^\circ$. We have \begin{align} \angle OBE & = \angle OBC + \angle CBE \\ & = \frac{\angle ABC}{2} + \frac{\angle ACB}{2} \\ & = 90^\circ - \frac{\angle A}{2} . \end{align} Hence, we must have $\angle OBE = 30^\circ$, not $150^\circ$. Hence, $\angle A = 120^\circ$. This implies $a > b$ and $a >c$. This contradicts the condition specified for this case. Therefore, this case is infeasible. $\textbf{Case 3}$: $c \leq b \leq a$. We extend segment $CB$ to $D$, such that $BD = c$. We extend segment $BC$ to $E$, such that $CE = b$. In this case, $X$ is the midpoint of $BE$, and $Y$ is the midpoint of $CD$. Because $M$ and $X$ are the midpoints of $AB$ and $BE$, respectively, $MX \parallel AE$. Because $N$ and $Y$ are the midpoints of $AC$ and $CD$, respectively, $NY \parallel AD$. Because $CA = CE$, $\angle CAE =\angle CEB = \frac{\angle ACB}{2}$. Because $BA = BD$, $\angle BAD = \angle BDA = \frac{\angle ABC}{2}$. Because $MX \parallel AE$ and $NY \parallel AD$, the angle formed between lines $MX$ and $NY$ is congruent to $\angle DAE$. Hence, $\angle DAE = 30^\circ$ or $150^\circ$. We have \begin{align} \angle DAE & = \angle BAD + \angle CAE + \angle BAC \\ & = \frac{\angle ABC}{2} + \frac{\angle ACB}{2} + \angle BAC \\ & = 90^\circ + \frac{\angle BAC}{2} . \end{align} Hence, we must have $\angle OBE = 150^\circ$, not $30^\circ$. Hence, $\angle BAC = 120^\circ$. In $\triangle ABC$, by applying the law of cosines, we have \begin{align} a^2 & = b^2 + c^2 - 2bc \cos \angle BAC\\ & =  b^2 + c^2 - 2bc \cos 120^\circ \\ & = b^2 + c^2 + bc . \end{align} Because $a = 219$, we have \[ b^2 + c^2 + bc  = 219^2 . \] Now, we find integer solution(s) of this equation with $c \leq b$. Multiplying this equation by 4, we get \[ \left( 2 c + b \right)^2 + 3 b^2 = 438^2 . \hspace{1cm} (1) \] Denote $d = 2 c + b$. Because $c \leq b$, $b < d \leq 3 b$. Because $438^2 - 3 b^2 \equiv 0 \pmod{3}$, $d^2 \equiv 0 \pmod{3}$. Thus, $d \equiv 0 \pmod{3}$. This implies $d^2 \equiv 0 \pmod{9}$. We also have $438^2 \equiv 0 \pmod{9}$. Hence, $3 b^2 \equiv 0 \pmod{9}$. This implies $b \equiv 0 \pmod{3}$. Denote $b = 3 p$ and $d = 3 q$. Hence, $p < q \leq 3 p$. Hence, Equation (1) can be written as \[ q^2 + 3 p^2 = 146^2 . \hspace{1cm} (2) \] Now, we solve this equation. First, we find an upper bound of $q$. We have $q^2 + 3 p^2 \geq q^2 + 3 \left( \frac{q}{3} \right)^2 = \frac{4 q^2}{3}$. Hence, $\frac{4 q^2}{3} \leq 146^2$. Hence, $q \leq 73 \sqrt{3} < 73 \cdot 1.8 = 131.4$. Because $q$ is an integer, we must have $q \leq 131$. Second, we find a lower bound of $q$. We have $q^2 + 3 p^2 < q^2 + 3 q^2 = 4 q^2$. Hence, $4 q^2 > 146^2$. Hence, $q > 73$. Because $q$ is an integer, we must have $q \geq 74$. Now, we find the integer solutions of $p$ and $q$ that satisfy Equation (2) with $74 \leq q \leq 131$. First, modulo 9, \begin{align} q^2 & \equiv 146^2 - 3 p^2 \\ & \equiv 4 - 3 \cdot ( 0 \mbox{ or } 1 ) \\ & \equiv 4 \mbox{ or } 1 . \end{align} Hence $q \equiv \pm 1, \pm 2 \pmod{9}$. Second, modulo 5, \begin{align} q^2 & \equiv 146^2 - 3 p^2 \\ & \equiv 1 + 2 p^2 \\ & \equiv 1 + 2 \cdot ( 0 \mbox{ or } 1 \mbox{ or } -1 ) \\ & \equiv 1 \mbox{ or } 3 \mbox{ or } - 1 . \end{align} Because $q^2 \equiv 0 \mbox{ or } 1 \mbox{ or } - 1$, we must have $q^2 \equiv 1 \mbox{ or } - 1$. Hence, $5 \nmid q$. Third, modulo 7, \begin{align} q^2 & \equiv 146^2 - 3 p^2 \\ & \equiv 1 - 3 \cdot ( 0 \mbox{ or } 1 \mbox{ or } 5 \mbox{ or } 2 ) \\ & \equiv 1 \mbox{ or } 2 \mbox{ or } 3 \mbox{ or } 5 . \end{align} Because $q^2 \equiv 0 \mbox{ or } 1 \mbox{ or } 2 \mbox{ or } 4 \pmod{ 7 }$, we must have $q^2 \equiv 1 \mbox{ or } 2 \pmod{7}$. Hence, $q \equiv 1, 3, 4, 6 \pmod{7}$. Given all conditions above, the possible $q$ are 74, 83, 88, 92, 97, 101, 106, 109, 116, 118, 127. By testing all these numbers, we find that the only solution is $q = 97$. This implies $p = 63$. Hence, $b = 3p = 189$ and $d = 3q = 291$. Hence, $c = \frac{d - b}{2} = 51$. Therefore, the perimeter of $\triangle ABC$ is $b + c + a = 189 + 51 + 219 = \fbox{459}$. ~Steven Chen (www.professorchenedu.com)

Answer: 459