2022 AIME I Problems/Problem 14
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