2022 AIME I Problems/Problem 14
Contents
Problem
Given and a point on one of its sides, call line the 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 giving the solution and , yielding and , so the sum is . Since and are primes in , the solution must divide . One can then easily check that this is the unique solution.
- MarkBcc168
Solution (Geometry + Number Theory)
Denote , , .
Let the splitting line of through (resp. ) crosses at another point (resp. ).
WLOG, we assume .
: .
We extend segment to , such that . We extend segment to , such that .
In this case, is the midpoint of , and is the midpoint of .
Because and are the midpoints of and , respectively, . Because and are the midpoints of and , respectively, .
Because , . Because , .
Let and intersect at . Because and , the angle formed between lines and is congruent to . Hence, or .
We have
Hence, we must have , not . Hence, .
This implies and . This contradicts the condition specified for this case.
Therefore, this case is infeasible.
: .
We extend segment to , such that . We extend segment to , such that .
In this case, is the midpoint of , and is the midpoint of .
Because and are the midpoints of and , respectively, . Because and are the midpoints of and , respectively, .
Because , . Because , .
Let be a point of , such that . Hence, .
Because and and , the angle formed between lines and is congruent to . Hence, or .
We have
Hence, we must have , not . Hence, .
This implies and . This contradicts the condition specified for this case.
Therefore, this case is infeasible.
: .
We extend segment to , such that . We extend segment to , such that .
In this case, is the midpoint of , and is the midpoint of .
Because and are the midpoints of and , respectively, . Because and are the midpoints of and , respectively, .
Because , . Because , .
Because and , the angle formed between lines and is congruent to . Hence, or .
We have
Hence, we must have , not . Hence, .
In , by applying the law of cosines, we have
Because , we have
Now, we find integer solution(s) of this equation with .
Multiplying this equation by 4, we get
Denote . Because , .
Because , . Thus, . This implies .
We also have . Hence, . This implies .
Denote and . Hence, . Hence, Equation (1) can be written as
Now, we solve this equation.
First, we find an upper bound of .
We have . Hence, . Hence, . Because is an integer, we must have .
Second, we find a lower bound of .
We have . Hence, . Hence, . Because is an integer, we must have .
Now, we find the integer solutions of and that satisfy Equation (2) with .
First, modulo 9,
Hence .
Second, modulo 5,
Because , we must have . Hence, .
Third, modulo 7,
Because , we must have . Hence, .
Given all conditions above, the possible are 74, 83, 88, 92, 97, 101, 106, 109, 116, 118, 127.
By testing all these numbers, we find that the only solution is . This implies .
Hence, and . Hence, .
Therefore, the perimeter of is .
~Steven Chen (www.professorchenedu.com)
Solution (Number Theory Part)
We wish to solve the Diophantine equation . It can be shown that and , so we make the substitution and to obtain as our new equation to solve for.
Notice that , where . Thus,
Note that . Thus, . Squaring both sides yields Thus, by , is a solution to . This implies that and , so our final answer is .
~ Leo.Euler
Solution(Visual geometry)
We look at upper and middle diagrams and get .
Next we use only the lower Diagram. Let be incenter , E be midpoint of biggest arc Then bisector cross circumcircle at point . Quadrilateral is cyclic, so is integer. A quick check gives that and . Denote
We have equations in integers
The solution is Suppose,
Now we check all possible
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vladimir.shelomovskii@gmail.com, vvsss
Video Solution
https://www.youtube.com/watch?v=kkous52vPps&t=3023s
~Steven Chen (wwww.professorchenedu.com)
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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