Difference between revisions of "2023 AIME I Problems/Problem 15"
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==Problem== | ==Problem== | ||
− | + | Find the largest prime number <math>p<1000</math> for which there exists a complex number <math>z</math> satisfying | |
* the real and imaginary part of <math>z</math> are both integers; | * the real and imaginary part of <math>z</math> are both integers; | ||
* <math>|z|=\sqrt{p},</math> and | * <math>|z|=\sqrt{p},</math> and | ||
− | * there exists a triangle whose three side lengths are <math>p,</math> the real part of <math>z^{3},</math> and the imaginary part of <math>z^{3}.</math | + | * there exists a triangle whose three side lengths are <math>p,</math> the real part of <math>z^{3},</math> and the imaginary part of <math>z^{3}.</math> |
==Solution== | ==Solution== |
Latest revision as of 01:49, 21 August 2024
Contents
Problem
Find the largest prime number for which there exists a complex number
satisfying
- the real and imaginary part of
are both integers;
and
- there exists a triangle whose three side lengths are
the real part of
and the imaginary part of
Solution
Assume that . Then,
Note that by the Triangle Inequality,
Thus, we know
Without loss of generality, assume
(as otherwise, consider
). If
, then
`Thus, this means
or
. Also note that the roots of
are
, so thus if
,
Note that
so
, and
. If
, then
. Note that
, and
, so
or
. However, then
, absurd.
If , by similar logic, we have that
, so
. However, once again,
. If
, by the same logic,
, so
, where we run into the same problem. Thus
indeed.
If , note that
We note that
works. Thus, we just need to make sure that if
,
. But this is easy, as
absurd. Thus, the answer is
.
Solution 2
Denote . Thus,
.
Thus,
Because ,
,
are three sides of a triangle, we have
and
.
Thus,
Because ,
,
are three sides of a triangle, we have the following triangle inequalities:
We notice that , and
,
, and
form a right triangle. Thus,
.
Because
,
.
Therefore, (3) holds.
Conditions (4) and (5) can be written in the joint form as
We have
and
.
Thus, (5) can be written as
Therefore, we need to jointly solve (1), (2), (6).
From (1) and (2), we have either , or
.
In (6), by symmetry, without loss of generality, we assume
.
Thus, (1) and (2) are reduced to
Let . Plugging this into (6), we get
Because is a prime,
and
are relatively prime.
Therefore, we can use (7), (8), , and
and
are relatively prime to solve the problem.
To facilitate efficient search, we apply the following criteria:
\begin{enumerate*}
\item To satisfy (7) and , we have
.
In the outer layer, we search for
in a decreasing order.
In the inner layer, for each given
, we search for
.
\item Given
, we search for
in the range
.
\item We can prove that for
, there is no feasible
.
The proof is as follows.
For , to satisfy
, we have
.
Thus,
.
Thus, the R.H.S. of (8) has the following upper bound
Hence, to satisfy (8), a necessary condition is
However, this cannot be satisfied for .
Therefore, there is no feasible solution for
.
Therefore, we only need to consider
.
\item We eliminate that are not relatively prime to
.
\item We use the following criteria to quickly eliminate that make
a composite number.
- For
, we eliminate
satisfying
.
- For
(resp.
), we eliminate
satisfying
(resp.
).
\item For the remaining , check whether (8) and the condition that
is prime are both satisfied.
The first feasible solution is and
.
Thus,
.
\item For the remaining search, given , we only search for
.
Following the above search criteria, we find the final answer as and
.
Thus, the largest prime
is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~MathProblemSolvingSkills.com
Animated Video Solution
~Star League (https://starleague.us)
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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