Difference between revisions of "2023 AIME I Problems/Problem 15"
(→Solution) |
(→Solution) |
||
Line 19: | Line 19: | ||
Because <math>p</math>, <math>{\rm Re} \left( z^3 \right)</math>, <math>{\rm Im} \left( z^3 \right)</math> are three sides of a triangle, we have <math>{\rm Re} \left( z^3 \right) > 0</math> and <math>{\rm Im} \left( z^3 \right) > 0</math>. | Because <math>p</math>, <math>{\rm Re} \left( z^3 \right)</math>, <math>{\rm Im} \left( z^3 \right)</math> are three sides of a triangle, we have <math>{\rm Re} \left( z^3 \right) > 0</math> and <math>{\rm Im} \left( z^3 \right) > 0</math>. | ||
Thus, | Thus, | ||
− | \begin{ | + | \begin{eqnarray*} |
a \left( a^2 - 3 b^2 \right) & > 0 , \hspace{1cm} (1) \\ | a \left( a^2 - 3 b^2 \right) & > 0 , \hspace{1cm} (1) \\ | ||
b \left( - b^2 + 3 a^2 \right) & > 0. \hspace{1cm} (2) | b \left( - b^2 + 3 a^2 \right) & > 0. \hspace{1cm} (2) | ||
− | \end{ | + | \end{eqnarray*} |
Because <math>p</math>, <math>{\rm Re} \left( z^3 \right)</math>, <math>{\rm Im} \left( z^3 \right)</math> are three sides of a triangle, we have the following triangle inequalities: | Because <math>p</math>, <math>{\rm Re} \left( z^3 \right)</math>, <math>{\rm Im} \left( z^3 \right)</math> are three sides of a triangle, we have the following triangle inequalities: |
Revision as of 12:29, 8 February 2023
Problem 15
Find the largest prime number for which there exists a complex number satisfying
- the real and imaginary parts of are integers;
- , and
- there exists a triangle with side lengths , the real part of , and the imaginary part of .
Answer: 349
Suppose ; notice that , so by De Moivre’s theorem and . Now just try pairs going down from , writing down the value of on the right; and eventually we arrive at the first time is prime. Therefore, .
Solution
Denote . Thus, .
Thus,
Because , , are three sides of a triangle, we have and . Thus, \begin{eqnarray*} a \left( a^2 - 3 b^2 \right) & > 0 , \hspace{1cm} (1) \\ b \left( - b^2 + 3 a^2 \right) & > 0. \hspace{1cm} (2) \end{eqnarray*}
Because , , are three sides of a triangle, we have the following triangle inequalities: \begin{align*} {\rm Re} \left( z^3 \right) + {\rm Im} \left( z^3 \right) & > p \hspace{1cm} (3) \\ p + {\rm Re} \left( z^3 \right) & > {\rm Im} \left( z^3 \right) \hspace{1cm} (4) \\ p + {\rm Im} \left( z^3 \right) & > {\rm Re} \left( z^3 \right) \hspace{1cm} (5) \end{align*}
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 \[ \left| {\rm Re} \left( z^3 \right) - {\rm Im} \left( z^3 \right) \right| < p . \hspace{1cm} (4) \]
We have \begin{align*} {\rm Re} \left( z^3 \right) - {\rm Im} \left( z^3 \right) & = \left( a^3 - 3 a b^2 \right) - \left( - b^3 + 3 a^2 b \right) \\ & = \left( a + b \right) \left( a^2 - 4 ab + b^2 \right) \end{align*} and .
Thus, (5) can be written as \[ \left| \left( a + b \right) \left( a^2 - 4 ab + b^2 \right) \right| < a^2 + b^2 . \hspace{1cm} (6) \]
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 \[ a > \sqrt{3} b . \hspace{1cm} (7) \]
Let . Plugging this into (6), we get \begin{align*} \left| \left( \left( \lambda - 2 \right)^2 - 3 \right) \right| < \frac{1}{b} \frac{\lambda^2 + 1}{\lambda + 1} . \hspace{1cm} (8) \end{align*}
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 \begin{align*} \frac{1}{b} \frac{\lambda^2 + 1}{\lambda + 1} & < \frac{1}{b} \frac{\lambda^2 + \lambda}{\lambda + 1} \\ & = \frac{\lambda}{b} \\ & \leq \frac{\frac{30}{9}}{9} \\ & < \frac{10}{27} . \end{align*}
Hence, to satisfy (8), a necessary condition is \begin{align*} \left| \left( \left( \lambda - 2 \right)^2 - 3 \right) \right| < \frac{10}{27} . \end{align*}
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. \begin{enumerate} \item For , we eliminate satisfying . \item For (resp. ), we eliminate satisfying (resp. ). \end{enumerate}
\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 . \end{enumerate}
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)