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

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==Problem==
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Given <math>\triangle ABC</math> and a point <math>P</math> on one of its sides, call line <math>\ell</math> the splitting line 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>.
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==The Geometry Part - Solution 1==
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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>.
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- MortemEtInteritum
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==The Geometry Part - Solution 2==
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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>.
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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>.
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- kevinmathz
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==The NT Part==
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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>.
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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>.
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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
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\begin{align*}
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73^2 &= N((8-\omega)^2) \\
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&= N(64 - 16\omega + \omega^2) \\
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&= N(64 - 16\omega + (-1-\omega)) \\
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&= N(63 - 17\omega),
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\end{align*}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.
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- MarkBcc168

Revision as of 21:43, 17 February 2022

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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