Difference between revisions of "2023 AIME I Problems/Problem 15"
(→Solution) |
|||
Line 13: | Line 13: | ||
Thus, | Thus, | ||
− | + | <cmath> | |
z^3 = a \left( a^2 - 3 b^2 \right) + i b \left( - b^2 + 3 a^2 \right) . | z^3 = a \left( a^2 - 3 b^2 \right) + i b \left( - b^2 + 3 a^2 \right) . | ||
− | + | </cmath> | |
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>. |
Revision as of 13:28, 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{align*}
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{align*}
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)