2023 AIME I Problems/Problem 15
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.