Difference between revisions of "2022 AIME I Problems/Problem 14"
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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>. | 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 | + | 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 | 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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==Solution(Visual geometry)== | ==Solution(Visual geometry)== | ||
− | [[File:AIME 2022 | + | [[File:AIME-I-2022-14a.png|400px|right]] |
− | [[File:AIME 2022 | + | [[File:AIME-I-2022-14b.png|400px|right]] |
[[File:AIME-I-2022-14c.png|400px|right]] | [[File:AIME-I-2022-14c.png|400px|right]] | ||
We look at upper and middle diagrams and get <math>\angle BAC = 120^\circ</math>. | We look at upper and middle diagrams and get <math>\angle BAC = 120^\circ</math>. |
Latest revision as of 02:56, 4 January 2023
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
Case
Case
Case
Case
Case
Case
vladimir.shelomovskii@gmail.com, vvsss
Video Solution
~MathProblemSolvingSkills.com
Video Solution
https://www.youtube.com/watch?v=kkous52vPps&t=3023s
~Steven Chen (wwww.professorchenedu.com)
Animated Video Solution
~Star League (https://starleague.us)
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.