2022 AIME I Problems/Problem 14
Contents
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
\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
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)
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.