Difference between revisions of "2022 AIME I Problems/Problem 14"
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− | . | + | ==Problem== |
+ | 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>. | ||
+ | |||
+ | ==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 | ||
+ | \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 <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 |
Revision as of 20:43, 17 February 2022
Contents
[hide]Problem
Given and a point on one of its sides, call line the splitting line of through if passes through and divides into two polygons of equal perimeter. Let be a triangle where and and are positive integers. Let and be the midpoints of and , respectively, and suppose that the splitting lines of through and intersect at . Find the perimeter of .
The Geometry Part - Solution 1
Consider the splitting line through . Extend on ray such that . Then the splitting line bisects segment , so in particular it is the midline of triangle and thus it is parallel to . But since triangle is isosceles, we can easily see is parallel to the angle bisector of , so the splitting line is also parallel to this bisector, and similar for the splitting line through . Some simple angle chasing reveals the condition is now equivalent to .
- MortemEtInteritum
The Geometry Part - Solution 2
Let and be the splitting lines. Reflect across to be and across to be . Take and , which are spiral similarity centers on the other side of as such that and . This gets that because and , then and are on 's circumcircle. Now, we know that and so because and , then and and and .
We also notice that because and correspond on and , and because and correspond on and , then the angle formed by and is equal to the angle formed by and which is equal to . Thus, . Similarly, and so and .
- kevinmathz
The NT Part
We now need to solve . A quick check gives that and . Thus, it's equivalent to solve .
Let be one root of . Then, recall that is the ring of integers of and is a unique factorization domain. Notice that . Therefore, it suffices to find an element of with the norm .
To do so, we factor in . Since it's , it must split. A quick inspection gives . Thus, , so
- MarkBcc168