Difference between revisions of "2022 AIME I Problems/Problem 14"

Latest revision as of 04:07, 4 May 2025

Problem

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 $\triangle ABC.$

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

Solution (Geometry + Number Theory)

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 $b + c + a = 189 + 51 + 219 = \boxed{\textbf{(459) }}$ .

~Steven Chen (www.professorchenedu.com)

Solution (Number Theory Part)

We wish to solve the Diophantine equation $a^2+ab+b^2=3^2 \cdot 73^2$ . It can be shown that $3|a$ and $3|b$ , so we make the substitution $a=3x$ and $b=3y$ to obtain $x^2+xy+y^2=73^2$ as our new equation to solve for.

Notice that $r^2+r+1=(r-\omega)(r-{\omega}^2)$ , where $\omega=e^{i\frac{2\pi}{3}}$ . Thus, $x^2+xy+y^2 = y^2((x/y)^2+(x/y)+1) = y^2 (\frac{x}{y}-\omega)(\frac{x}{y}-{\omega}^2) = (x-y\omega)(x-y{\omega}^2).$

Note that $8^2+1^2+8 \cdot 1=73$ . Thus, $(8-\omega)(8-{\omega}^2)=73$ . Squaring both sides yields $\begin{align} (8-\omega)^2(8-{\omega}^2)^2&=73^2\\ (63-17\omega)(63-17{\omega}^2)&=73^2. \end{align}$ Thus, by $(2)$ , $(63, 17)$ is a solution to $x^2+xy+y^2=73^2$ . This implies that $a=189$ and $b=51$ , so our final answer is $189+51+219=\boxed{459}$ .

~ Leo.Euler

Solution(Visual geometry)

We look at upper and middle diagrams and get $\angle BAC = 120^\circ$ .

Next we use only the lower Diagram. Let $I$ be incenter $\triangle ABC$ , E be midpoint of biggest arc $\overset{\Large\frown} {BC}.$ Then bisector $AI$ cross circumcircle $\triangle ABC$ at point $E$ . Quadrilateral $ABEC$ is cyclic, so $\angle BEC = 180^\circ - \angle ABC = 60^\circ \implies BE = CE = IE = BC.$ $AE \cdot BC = AB \cdot CE + AC \cdot BE \implies AE = AB + AC$ $\implies AI +EI = AB + AC, \hspace{10mm} AI = AB+ AC – BC$ is integer. $AI = \frac {2AB \cdot AC \cdot cos \angle CAI}{AB+AC + BC} = \frac {AB \cdot AC}{AB+AC + BC} =$ $= AB + AC – BC \implies AC^2 + AB^2 + AB \cdot AC = BC^2.$ A quick $(\mod9)$ check gives that $3\mid AC$ and $3\mid AB$ . $AI \le A_0I_0 = EA_0 – EI_0 = \frac{2 BC}{\sqrt{3}} – BC = \frac {2 - \sqrt{3}}{\sqrt{3}} BC = 33.88.$ Denote $a= \frac {BC}{3}= 73, b = \frac {AC}{3}, c = \frac {AB}{3}, l = \frac {AI}{3} \le 11.$

We have equations in integers $\frac{bc}{a+b+c} = b + c – a = l \le 11.$

The solution $(b > c)$ is $b = \frac{a + l +\sqrt{a^2 – 6al – 3l^2}}{2}, c = \frac{a + l -\sqrt{a^2 – 6al – 3l^2}}{2}.$ Suppose, $a^2 – 6al – 3l^2 = (a – 3l – t)^2 \implies \frac {12l^2}{t} + t + 6l= 2a = 146.$

Now we check all possible $t = {2,3,4,6,12, ml}.$

Case $t = 2 \implies 6l^2 + 6l = 146 – 2 \implies l^2 + l = 24 \implies \O$

Case $t = 3 \implies 4l^2 + 6l = 146 – 3 = 143\implies \O.$

Case $t = 4 \implies 3l^2 + 6l = 146 – 4 =142 \implies \O.$

Case $t = 6 \implies 2l^2 + 6l = 146 – 6 = 140 \implies l = 7, b = 63, c = 17.$

Case $t = 12 \implies l^2 + 6l = 146 – 12 = 134 \implies \O.$

Case $t = ml \implies \frac{12l}{m} + 6l + ml = 146 \implies \frac{12}{m} + 6 + m = \frac{73 \cdot 2}{l}\implies \O.$

vladimir.shelomovskii@gmail.com, vvsss

Solution (The Geometry Part Using Menelaus)

Let the 2 splitting lines crossing $N$ and $M$ cross $BC$ at $G$ and $H$ . Extend $AB$ and $NG$ so that the two lines intersect at $E$ and extend $AC$ and $MH$ so that the two lines intersect at $F$ . Let $a$ , $b$ , $c$ be the corresponding side lengths for $\angle A$ , $\angle B$ , $\angle C$ . According to the question, $MH$ and $NG$ creates two sections of $\triangle ABC$ of equal perimeter, we could list out the equations:

$BG+C=a-BG$

$CH+b=a-BH$

$BG+GH+CH=a$

This results in:

$BG=\frac{a-c}{2}$

$CH=\frac{a-b}{2}$

$GH=\frac{b+c}{2}$

Now, we can apply the Menelaus Theorem to $\triangle ABC$ . Let $BE = x$ : $\frac{BE}{AE}\cdot\frac{AN}{CN}\cdot\frac{CG}{BG} = 1$ $\frac{x}{x+c}\cdot1\cdot\frac{\frac{a+c}{2}}{\frac{a-c}{2}} = 1$ $\frac{x}{x+c}\cdot\frac{a+c}{a-c} = 1$ $\frac{x+c}{x}=\frac{a+c}{a-c}$ $\frac{c}{x}=\frac{2c}{a-c}$ $BE=x=\frac{a-c}{2}=BG$ Similarly, $CF=CH$ .

Therefore, $\angle BGE = \angle E$ and $\angle CHF = \angle F$ . Because the splitting lines of $\triangle ABC$ through $M$ and $N$ intersect at $30^\circ$ , $\angle NOF$ = $30^\circ$ . $\angle BGE+\angle CHF=\angle BGE+\angle CHF=\angle NOF=30^\circ$ .

We conclude that $\angle A=360^\circ-\angle E-\angle F-\angle EOF(reflex)=360^\circ-30^\circ-210^\circ=120^\circ$

Apply the Law of Cosines on $\triangle ABC$

$c^2+a^2-2ac\cos 120^\circ=219^2$

$a^2+c^2+ac=219^2=47961$

To find the final answer, you would only need to solve this equation, which steps could be found in previous solutions.

~cassphe

Video Solution

https://youtu.be/T6zq1e1RZdg

~MathProblemSolvingSkills.com

Video Solution

https://www.youtube.com/watch?v=kkous52vPps&t=3023s

~Steven Chen (wwww.professorchenedu.com)

Animated Video Solution

https://youtu.be/o-aDdxdnTWY

~Star League (https://starleague.us)

2022 AIME I (Problems • Answer Key • Resources)
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Difference between revisions of "2022 AIME I Problems/Problem 14"

Latest revision as of 04:07, 4 May 2025

Contents

Problem

The Geometry Part - Solution 1

The Geometry Part - Solution 2

The NT Part

Solution (Geometry + Number Theory)

Solution (Number Theory Part)

Solution(Visual geometry)

Solution (The Geometry Part Using Menelaus)

Video Solution

Video Solution

Animated Video Solution

See Also

@@ Line 1: / Line 1: @@
-.
+==Problem==
+Given <math>\triangle ABC</math> and a point <math>P</math> on one of its sides, call line <math>\ell</math> the <math>\textit{splitting line}</math> of <math>\triangle ABC</math> through <math>P</math> if <math>\ell</math> passes through <math>P</math> and divides <math>\triangle ABC</math> into two polygons of equal perimeter. Let <math>\triangle ABC</math> be a triangle where <math>BC = 219</math> and <math>AB</math> and <math>AC</math> are positive integers. Let <math>M</math> and <math>N</math> be the midpoints of <math>\overline{AB}</math> and <math>\overline{AC},</math> respectively, and suppose that the splitting lines of <math>\triangle ABC</math> through <math>M</math> and <math>N</math> intersect at <math>30^\circ.</math> Find the perimeter of <math>\triangle ABC.</math>
+==The Geometry Part - Solution 1==
+Consider the splitting line through <math>M</math>. Extend <math>D</math> on ray <math>BC</math> such that <math>CD=CA</math>. Then the splitting line bisects segment <math>BD</math>, so in particular it is the midline of triangle <math>ABD</math> and thus it is parallel to <math>AD</math>. But since triangle <math>ACD</math> is isosceles, we can easily see <math>AD</math> is parallel to the angle bisector of <math>C</math>, so the splitting line is also parallel to this bisector, and similar for the splitting line through <math>N</math>. Some simple angle chasing reveals the condition is now equivalent to <math>\angle A=120^\circ</math>.
+- MortemEtInteritum
+==The Geometry Part - Solution 2==
+Let <math>PM</math> and <math>QN</math> be the splitting lines. Reflect <math>B</math> across <math>Q</math> to be <math>B'</math> and <math>C</math> across <math>P</math> to be <math>C'</math>. Take <math>S_B</math> and <math>S_C</math>, which are spiral similarity centers on the other side of <math>BC</math> as <math>A</math> such that <math>\triangle S_BB'C \sim \triangle S_BBA</math> and <math>\triangle S_CC'B \sim \triangle S_CCA</math>. This gets that because <math>\angle S_BCB = \angle S_BCB' = \angle S_BAB</math> and <math>\angle S_CBC = \angle S_CBC' = \angle S_CAC</math>, then <math>S_B</math> and <math>S_C</math> are on <math>\triangle ABC</math>'s circumcircle. Now, we know that <math>\triangle S_BBB' \sim \triangle S_BAC</math> and <math>\triangle S_CCC' \sim \triangle S_CAB</math> so because <math>BA=B'C</math> and <math>CA=C'B</math>, then <math>S_BB=SBB'</math> and <math>S_CC=S_CC'</math> and <math>S_BQ \perp BC</math> and <math>S_CP \perp BC</math>.
+We also notice that because <math>Q</math> and <math>N</math> correspond on <math>\triangle S_BBB'</math> and <math>\triangle S_BAC</math>, and because <math>P</math> and <math>M</math> correspond on <math>\triangle S_CCC' </math> and <math>\triangle S_CAB</math>, then the angle formed by <math>NQ</math> and <math>BA</math> is equal to the angle formed by <math>B'C</math> and <math>NQ</math> which is equal to <math>\angle BS_BQ = \angle QS_BB'</math>. Thus, <math>\angle CBA=2\angle CQN</math>. Similarly, <math>\angle BCA = 2\angle QPM</math> and so <math>\angle CBA + \angle BCA = 2\angle PQN + 2\angle QPM = 60^{\circ}</math> and <math>\angle A = 120^{\circ}</math>.
+- kevinmathz
+==The NT Part==
+We now need to solve <math>a^2+ab+b^2 = 3^2\cdot 73^2</math>. A quick <math>(\bmod 9)</math> check gives that <math>3\mid a</math> and <math>3\mid b</math>. Thus, it's equivalent to solve <math>x^2+xy+y^2 = 73^2</math>.
+Let <math>\omega</math> be one root of <math>\omega^2+\omega+1=0</math>. Then, recall that <math>\mathbb Z[\omega]</math> is the ring of integers of <math>\mathbb Q[\sqrt{-3}]</math> and is a unique factorization domain. Notice that <math>N(x-y\omega) = (x-y\omega)(x-y\omega^2) = x^2+xy+y^2</math>. Therefore, it suffices to find an element of <math>\mathbb Z[\omega]</math> with the norm <math>73^2</math>.
+To do so, we factor <math>73</math> in <math>\mathbb Z[\omega]</math>. Since it's <math>1\pmod 3</math>, it must split. A quick inspection gives <math>73 = (8-\omega)(8-\omega^2)</math>. Thus, <math>N(8-\omega) = 73</math>, so
+<cmath>\begin{align*}
+^2 &= N((8-\omega)^2) \\
+&= N(64 - 16\omega + \omega^2) \\
+&= N(64 - 16\omega + (-1-\omega)) \\
+&= N(63 - 17\omega),
+\end{align*}</cmath>
+giving the solution <math>x=63</math> and <math>y=17</math>, yielding <math>a=189</math> and <math>b=51</math>, so the sum is <math>\boxed{459}</math>. Since <math>8-\omega</math> and <math>8-\omega^2</math> are primes in <math>\mathbb Z[\omega]</math>, the solution must divide <math>73^2</math>. One can then easily check that this is the unique solution.
+- MarkBcc168
+==Solution (Geometry + Number Theory)==
+Denote <math>BC = a</math>, <math>CA = b</math>, <math>AB = c</math>.
+Let the splitting line of <math>\triangle ABC</math> through <math>M</math> (resp. <math>N</math>) crosses <math>\triangle ABC</math> at another point <math>X</math> (resp. <math>Y</math>).
+WLOG, we assume <math>c \leq b</math>.
+<math>\textbf{Case 1}</math>: <math>a \leq c \leq b</math>.
+We extend segment <math>AB</math> to <math>D</math>, such that <math>BD = a</math>.
+We extend segment <math>AC</math> to <math>E</math>, such that <math>CE = a</math>.
+In this case, <math>X</math> is the midpoint of <math>AE</math>, and <math>Y</math> is the midpoint of <math>AD</math>.
+Because <math>M</math> and <math>X</math> are the midpoints of <math>AB</math> and <math>AE</math>, respectively, <math>MX \parallel BE</math>.
+Because <math>N</math> and <math>Y</math> are the midpoints of <math>AC</math> and <math>AD</math>, respectively, <math>NY \parallel CD</math>.
+Because <math>CB = CE</math>, <math>\angle CBE =\angle CEB = \frac{\angle ACB}{2}</math>.
+Because <math>BC = BD</math>, <math>\angle BCD = \angle BDC = \frac{\angle ABC}{2}</math>.
+Let <math>BE</math> and <math>CD</math> intersect at <math>O</math>.
+Because <math>MX \parallel BE</math> and <math>NY \parallel CD</math>, the angle formed between lines <math>MX</math> and <math>NY</math> is congruent to <math>\angle BOD</math>. Hence, <math>\angle BOD = 30^\circ</math> or <math>150^\circ</math>.
+We have
+<cmath>
+\begin{align*}
+\angle BOD & = \angle CBE + \angle BCD \\
+& = \frac{\angle ACB}{2} + \frac{\angle ABC}{2} \\
+& = 90^\circ - \frac{\angle A}{2} .
+\end{align*}
+</cmath>
+Hence, we must have <math>\angle BOD = 30^\circ</math>, not <math>150^\circ</math>.
+Hence, <math>\angle A = 120^\circ</math>.
+This implies <math>a > b</math> and <math>a >c</math>. This contradicts the condition specified for this case.
+Therefore, this case is infeasible.
+<math>\textbf{Case 2}</math>: <math>c \leq a \leq b</math>.
+We extend segment <math>CB</math> to <math>D</math>, such that <math>BD = c</math>.
+We extend segment <math>AC</math> to <math>E</math>, such that <math>CE = a</math>.
+In this case, <math>X</math> is the midpoint of <math>AE</math>, and <math>Y</math> is the midpoint of <math>CD</math>.
+Because <math>M</math> and <math>X</math> are the midpoints of <math>AB</math> and <math>AE</math>, respectively, <math>MX \parallel BE</math>.
+Because <math>N</math> and <math>Y</math> are the midpoints of <math>AC</math> and <math>CD</math>, respectively, <math>NY \parallel AD</math>.
+Because <math>CB = CE</math>, <math>\angle CBE =\angle CEB = \frac{\angle ACB}{2}</math>.
+Because <math>BA = BD</math>, <math>\angle BAD = \angle BDA = \frac{\angle ABC}{2}</math>.
+Let <math>O</math> be a point of <math>AC</math>, such that <math>BO \parallel AD</math>.
+Hence, <math>\angle OBC = \angle BDA = \frac{B}{2}</math>.
+Because <math>MX \parallel BE</math> and <math>NY \parallel AD</math> and <math>AD \parallel BO</math>, the angle formed between lines <math>MX</math> and <math>NY</math> is congruent to <math>\angle OBE</math>. Hence, <math>\angle OBE = 30^\circ</math> or <math>150^\circ</math>.
+We have
+<cmath>
+\begin{align*}
+\angle OBE & = \angle OBC + \angle CBE \\
+& = \frac{\angle ABC}{2} + \frac{\angle ACB}{2} \\
+& = 90^\circ - \frac{\angle A}{2} .
+\end{align*}
+</cmath>
+Hence, we must have <math>\angle OBE = 30^\circ</math>, not <math>150^\circ</math>.
+Hence, <math>\angle A = 120^\circ</math>.
+This implies <math>a > b</math> and <math>a >c</math>. This contradicts the condition specified for this case.
+Therefore, this case is infeasible.
+<math>\textbf{Case 3}</math>: <math>c \leq b \leq a</math>.
+We extend segment <math>CB</math> to <math>D</math>, such that <math>BD = c</math>.
+We extend segment <math>BC</math> to <math>E</math>, such that <math>CE = b</math>.
+In this case, <math>X</math> is the midpoint of <math>BE</math>, and <math>Y</math> is the midpoint of <math>CD</math>.
+Because <math>M</math> and <math>X</math> are the midpoints of <math>AB</math> and <math>BE</math>, respectively, <math>MX \parallel AE</math>.
+Because <math>N</math> and <math>Y</math> are the midpoints of <math>AC</math> and <math>CD</math>, respectively, <math>NY \parallel AD</math>.
+Because <math>CA = CE</math>, <math>\angle CAE =\angle CEB = \frac{\angle ACB}{2}</math>.
+Because <math>BA = BD</math>, <math>\angle BAD = \angle BDA = \frac{\angle ABC}{2}</math>.
+Because <math>MX \parallel AE</math> and <math>NY \parallel AD</math>, the angle formed between lines <math>MX</math> and <math>NY</math> is congruent to <math>\angle DAE</math>. Hence, <math>\angle DAE = 30^\circ</math> or <math>150^\circ</math>.
+We have
+<cmath>
+\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*}
+</cmath>
+Hence, we must have <math>\angle OBE = 150^\circ</math>, not <math>30^\circ</math>.
+Hence, <math>\angle BAC = 120^\circ</math>.
+In <math>\triangle ABC</math>, by applying the law of cosines, we have
+<cmath>
+\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*}
+</cmath>
+Because <math>a = 219</math>, we have
+<cmath>
+\[
+b^2 + c^2 + bc  = 219^2 .
+\]
+</cmath>
+Now, we find integer solution(s) of this equation with <math>c \leq b</math>.
+Multiplying this equation by 4, we get
+<cmath>
+\[
+\left( 2 c + b \right)^2 + 3 b^2 = 438^2 . \hspace{1cm} (1)
+\]
+</cmath>
+Denote <math>d = 2 c + b</math>. Because <math>c \leq b</math>, <math>b < d \leq 3 b</math>.
+Because <math>438^2 - 3 b^2 \equiv 0 \pmod{3}</math>, <math>d^2 \equiv 0 \pmod{3}</math>.
+Thus, <math>d \equiv 0 \pmod{3}</math>.
+This implies <math>d^2 \equiv 0 \pmod{9}</math>.
+We also have <math>438^2 \equiv 0 \pmod{9}</math>.
+Hence, <math>3 b^2 \equiv 0 \pmod{9}</math>.
+This implies <math>b \equiv 0 \pmod{3}</math>.
+Denote <math>b = 3 p</math> and <math>d = 3 q</math>. Hence, <math>p < q \leq 3 p</math>.
+Hence, Equation (1) can be written as
+<cmath>
+\[
+q^2 + 3 p^2 = 146^2 . \hspace{1cm} (2)
+\]
+</cmath>
+Now, we solve this equation.
+First, we find an upper bound of <math>q</math>.
+We have <math>q^2 + 3 p^2 \geq q^2 + 3 \left( \frac{q}{3} \right)^2 = \frac{4 q^2}{3}</math>.
+Hence, <math>\frac{4 q^2}{3} \leq 146^2</math>.
+Hence, <math>q \leq 73 \sqrt{3} < 73 \cdot 1.8 = 131.4</math>.
+Because <math>q</math> is an integer, we must have <math>q \leq 131</math>.
+Second, we find a lower bound of <math>q</math>.
+We have <math>q^2 + 3 p^2 < q^2 + 3 q^2 = 4 q^2</math>.
+Hence, <math>4 q^2 > 146^2</math>.
+Hence, <math>q > 73</math>.
+Because <math>q</math> is an integer, we must have <math>q \geq 74</math>.
+Now, we find the integer solutions of <math>p</math> and <math>q</math> that satisfy Equation (2) with <math>74 \leq q \leq 131</math>.
+First, modulo 9,
+<cmath>
+\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*}
+</cmath>
+Hence <math>q \equiv \pm 1, \pm 2 \pmod{9}</math>.
+Second, modulo 5,
+<cmath>
+\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*}
+</cmath>
+Because <math>q^2 \equiv 0 \mbox{ or } 1 \mbox{ or } - 1</math>, we must have <math>q^2 \equiv 1 \mbox{ or } - 1</math>.
+Hence, <math>5 \nmid q</math>.
+Third, modulo 7,
+<cmath>
+\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*}
+</cmath>
+Because <math>q^2 \equiv 0 \mbox{ or } 1 \mbox{ or } 2 \mbox{ or } 4 \pmod{ 7 }</math>, we must have <math>q^2 \equiv 1 \mbox{ or } 2 \pmod{7}</math>.
+Hence, <math>q \equiv 1, 3, 4, 6 \pmod{7}</math>.
+Given all conditions above, the possible <math>q</math> are 74, 83, 88, 92, 97, 101, 106, 109, 116, 118, 127.
+By testing all these numbers, we find that the only solution is <math>q = 97</math>.
+This implies <math>p = 63</math>.
+Hence, <math>b = 3p = 189</math> and <math>d = 3q = 291</math>.
+Hence, <math>c = \frac{d - b}{2} = 51</math>.
+Therefore, the perimeter of <math>\triangle ABC</math> is <math>b + c + a = 189 + 51 + 219 = \boxed{\textbf{(459) }}</math>.
+~Steven Chen (www.professorchenedu.com)
+==Solution (Number Theory Part)==
+We wish to solve the Diophantine equation <math>a^2+ab+b^2=3^2 \cdot 73^2</math>. It can be shown that <math>3|a</math> and <math>3|b</math>, so we make the substitution <math>a=3x</math> and <math>b=3y</math> to obtain <math>x^2+xy+y^2=73^2</math> as our new equation to solve for.
+Notice that <math>r^2+r+1=(r-\omega)(r-{\omega}^2)</math>, where <math>\omega=e^{i\frac{2\pi}{3}}</math>. Thus,
+<cmath>x^2+xy+y^2 = y^2((x/y)^2+(x/y)+1) = y^2 (\frac{x}{y}-\omega)(\frac{x}{y}-{\omega}^2) = (x-y\omega)(x-y{\omega}^2).</cmath>
+Note that <math>8^2+1^2+8 \cdot 1=73</math>. Thus, <math>(8-\omega)(8-{\omega}^2)=73</math>. Squaring both sides yields
+<cmath>\begin{align}
+(8-\omega)^2(8-{\omega}^2)^2&=73^2\\
+(63-17\omega)(63-17{\omega}^2)&=73^2.
+\end{align}</cmath>
+Thus, by <math>(2)</math>, <math>(63, 17)</math> is a solution to <math>x^2+xy+y^2=73^2</math>. This implies that <math>a=189</math> and <math>b=51</math>, so our final answer is <math>189+51+219=\boxed{459}</math>.
+~ Leo.Euler
+==Solution(Visual geometry)==
+[[File:AIME-I-2022-14a.png|400px|right]]
+[[File:AIME-I-2022-14b.png|400px|right]]
+[[File:AIME-I-2022-14c.png|400px|right]]
+We look at  upper and middle  diagrams and get <math>\angle BAC = 120^\circ</math>.
+Next we use only the lower Diagram. Let  <math>I</math> be incenter <math>\triangle ABC</math>, E be midpoint of biggest arc <math>\overset{\Large\frown} {BC}.</math>
+Then bisector <math>AI</math> cross circumcircle <math>\triangle ABC</math> at point <math>E</math>. Quadrilateral  <math>ABEC</math> is cyclic, so
+<cmath> \angle BEC = 180^\circ - \angle ABC = 60^\circ \implies BE = CE = IE = BC.</cmath>
+<cmath>AE \cdot BC = AB \cdot CE + AC \cdot  BE \implies AE = AB + AC</cmath>
+<math>\implies AI +EI = AB + AC, \hspace{10mm} AI = AB+ AC – BC</math> is integer.
+<cmath>AI = \frac {2AB \cdot AC \cdot  cos \angle CAI}{AB+AC + BC} =  \frac {AB \cdot AC}{AB+AC + BC} =</cmath>
+<cmath>= AB + AC – BC \implies AC^2 + AB^2 + AB \cdot AC = BC^2.</cmath>
+A quick <math>(\mod9)</math> check gives that <math>3\mid AC</math> and <math>3\mid AB</math>.
+<cmath>AI \le A_0I_0 = EA_0 – EI_0 = \frac{2 BC}{\sqrt{3}} – BC = \frac {2 - \sqrt{3}}{\sqrt{3}} BC = 33.88.</cmath>
+Denote <math>a= \frac {BC}{3}= 73, b = \frac {AC}{3}, c = \frac {AB}{3}, l = \frac {AI}{3} \le 11.</math>
+We have equations in integers
+<math>\frac{bc}{a+b+c} = b + c – a = l \le 11.</math>
+The solution <math>(b > c)</math> is
+<cmath>b = \frac{a + l +\sqrt{a^2 – 6al – 3l^2}}{2},
+c = \frac{a + l -\sqrt{a^2 – 6al – 3l^2}}{2}.</cmath>
+Suppose,  <math>a^2 – 6al – 3l^2 = (a – 3l – t)^2 \implies \frac {12l^2}{t} + t + 6l= 2a = 146.</math>
+Now we check all possible <math>t = {2,3,4,6,12, ml}.</math>
+Case  <math>t = 2 \implies 6l^2 + 6l = 146 – 2 \implies l^2 + l = 24 \implies \O </math>
+Case  <math>t = 3 \implies 4l^2 + 6l = 146 – 3 = 143\implies \O.</math>
+Case <math>t = 4 \implies 3l^2 + 6l = 146 – 4 =142 \implies \O.</math>
+Case <math>t = 6 \implies 2l^2 + 6l = 146 – 6 = 140 \implies l = 7, b = 63, c = 17.</math>
+Case <math>t = 12 \implies l^2 + 6l = 146 – 12 = 134 \implies \O.</math>
+Case <math>t = ml \implies \frac{12l}{m} + 6l + ml = 146 \implies \frac{12}{m} + 6 + m = \frac{73 \cdot 2}{l}\implies \O.</math>
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+==Solution (The Geometry Part Using Menelaus)==
+[[File:2022AIME1.png|500px]]
+Let the 2 splitting lines crossing <math>N</math> and <math>M</math> cross <math>BC</math> at <math>G</math> and <math>H</math>. Extend <math>AB</math> and <math>NG</math> so that the two lines intersect at <math>E</math> and extend <math>AC</math> and <math>MH</math> so that the two lines intersect at <math>F</math>. Let <math>a</math>,<math>b</math>,<math>c</math> be the corresponding side lengths for <math>\angle A</math>,<math>\angle B</math>,<math>\angle C</math>. According to the question, <math>MH</math> and <math>NG</math> creates two sections of <math>\triangle ABC</math> of equal perimeter, we could list out the equations:
+<math>BG+C=a-BG</math>
+<math>CH+b=a-BH</math>
+<math>BG+GH+CH=a</math>
+This results in:
+<math>BG=\frac{a-c}{2}</math>
+<math>CH=\frac{a-b}{2}</math>
+<math>GH=\frac{b+c}{2}</math>
+Now, we can apply the Menelaus Theorem to <math>\triangle ABC</math>.
+Let <math>BE = x</math>:
+<cmath>\frac{BE}{AE}\cdot\frac{AN}{CN}\cdot\frac{CG}{BG} = 1</cmath>
+<cmath>\frac{x}{x+c}\cdot1\cdot\frac{\frac{a+c}{2}}{\frac{a-c}{2}} = 1</cmath>
+<cmath>\frac{x}{x+c}\cdot\frac{a+c}{a-c} = 1</cmath>
+<cmath>\frac{x+c}{x}=\frac{a+c}{a-c}</cmath>
+<cmath>\frac{c}{x}=\frac{2c}{a-c}</cmath>
+<cmath>BE=x=\frac{a-c}{2}=BG</cmath>
+Similarly, <math>CF=CH</math>.
+Therefore, <math>\angle BGE = \angle E</math> and <math>\angle CHF = \angle F</math>. Because the splitting lines of <math>\triangle ABC</math> through <math>M</math> and <math>N</math> intersect at <math>30^\circ</math>, <math>\angle NOF</math> = <math>30^\circ</math>. <math>\angle BGE+\angle CHF=\angle BGE+\angle CHF=\angle NOF=30^\circ</math>.
+We conclude that <math>\angle A=360^\circ-\angle E-\angle F-\angle EOF(reflex)=360^\circ-30^\circ-210^\circ=120^\circ</math>
+Apply the Law of Cosines on <math>\triangle ABC</math>
+<math>c^2+a^2-2ac\cos 120^\circ=219^2</math>
+<math>a^2+c^2+ac=219^2=47961</math>
+To find the final answer, you would only need to solve this equation, which steps could be found in previous solutions.
+~cassphe
+==Video Solution==
+https://youtu.be/T6zq1e1RZdg
+~MathProblemSolvingSkills.com
+==Video Solution==
+https://www.youtube.com/watch?v=kkous52vPps&t=3023s
+~Steven Chen (wwww.professorchenedu.com)
+==Animated Video Solution==
+https://youtu.be/o-aDdxdnTWY
+~Star League (https://starleague.us)
+==See Also==
+{{AIME box|year=2022|n=I|num-b=13|num-a=15}}
+[[Category:Intermediate Geometry Problems]]
+[[Category:Intermediate Number Theory Problems]]
+{{MAA Notice}}