Difference between revisions of "2023 AIME I Problems/Problem 15"
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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{ | + | <cmath> |
+ | \begin{align*} | ||
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{align*} |
+ | </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 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: | ||
+ | <cmath> | ||
\begin{align*} | \begin{align*} | ||
{\rm Re} \left( z^3 \right) + {\rm Im} \left( z^3 \right) & > p \hspace{1cm} (3) \\ | {\rm Re} \left( z^3 \right) + {\rm Im} \left( z^3 \right) & > p \hspace{1cm} (3) \\ | ||
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p + {\rm Im} \left( z^3 \right) & > {\rm Re} \left( z^3 \right) \hspace{1cm} (5) | p + {\rm Im} \left( z^3 \right) & > {\rm Re} \left( z^3 \right) \hspace{1cm} (5) | ||
\end{align*} | \end{align*} | ||
+ | </cmath> | ||
We notice that <math>| z^3 | = p^{3/2}</math>, and <math>{\rm Re} \left( z^3 \right)</math>, <math>{\rm Im} \left( z^3 \right)</math>, and <math>| z^3 |</math> form a right triangle. Thus, <math>{\rm Re} z^3 + {\rm Im} z^3 > p^{3/2}</math>. | We notice that <math>| z^3 | = p^{3/2}</math>, and <math>{\rm Re} \left( z^3 \right)</math>, <math>{\rm Im} \left( z^3 \right)</math>, and <math>| z^3 |</math> form a right triangle. Thus, <math>{\rm Re} z^3 + {\rm Im} z^3 > p^{3/2}</math>. | ||
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Conditions (4) and (5) can be written in the joint form as | Conditions (4) and (5) can be written in the joint form as | ||
− | + | <cmath> | |
\left| {\rm Re} \left( z^3 \right) - {\rm Im} \left( z^3 \right) \right| < p . \hspace{1cm} (4) | \left| {\rm Re} \left( z^3 \right) - {\rm Im} \left( z^3 \right) \right| < p . \hspace{1cm} (4) | ||
− | + | </cmath> | |
+ | |||
We have | We have | ||
+ | <cmath> | ||
\begin{align*} | \begin{align*} | ||
{\rm Re} \left( z^3 \right) - {\rm Im} \left( z^3 \right) | {\rm Re} \left( z^3 \right) - {\rm Im} \left( z^3 \right) | ||
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& = \left( a + b \right) \left( a^2 - 4 ab + b^2 \right) | & = \left( a + b \right) \left( a^2 - 4 ab + b^2 \right) | ||
\end{align*} | \end{align*} | ||
+ | </cmath> | ||
and <math>p = a^2 + b^2</math>. | and <math>p = a^2 + b^2</math>. | ||
Thus, (5) can be written as | Thus, (5) can be written as | ||
− | + | <cmath> | |
\left| \left( a + b \right) \left( a^2 - 4 ab + b^2 \right) \right| | \left| \left( a + b \right) \left( a^2 - 4 ab + b^2 \right) \right| | ||
< a^2 + b^2 . \hspace{1cm} (6) | < a^2 + b^2 . \hspace{1cm} (6) | ||
− | + | </cmath> | |
Therefore, we need to jointly solve (1), (2), (6). | Therefore, we need to jointly solve (1), (2), (6). | ||
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Thus, (1) and (2) are reduced to | Thus, (1) and (2) are reduced to | ||
− | + | <cmath> | |
a > \sqrt{3} b . \hspace{1cm} (7) | a > \sqrt{3} b . \hspace{1cm} (7) | ||
− | + | </cmath> | |
Let <math>a = \lambda b</math>. Plugging this into (6), we get | Let <math>a = \lambda b</math>. Plugging this into (6), we get | ||
+ | <cmath> | ||
\begin{align*} | \begin{align*} | ||
\left| \left( \left( \lambda - 2 \right)^2 - 3 \right) \right| | \left| \left( \left( \lambda - 2 \right)^2 - 3 \right) \right| | ||
< \frac{1}{b} \frac{\lambda^2 + 1}{\lambda + 1} . \hspace{1cm} (8) | < \frac{1}{b} \frac{\lambda^2 + 1}{\lambda + 1} . \hspace{1cm} (8) | ||
\end{align*} | \end{align*} | ||
+ | </cmath> | ||
Because <math>p= a^2 + b^2</math> is a prime, <math>a</math> and <math>b</math> are relatively prime. | Because <math>p= a^2 + b^2</math> is a prime, <math>a</math> and <math>b</math> are relatively prime. | ||
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Thus, <math>\sqrt{3} < \lambda \leq \frac{30}{9}</math>. | Thus, <math>\sqrt{3} < \lambda \leq \frac{30}{9}</math>. | ||
Thus, the R.H.S. of (8) has the following upper bound | Thus, the R.H.S. of (8) has the following upper bound | ||
+ | <cmath> | ||
\begin{align*} | \begin{align*} | ||
\frac{1}{b} \frac{\lambda^2 + 1}{\lambda + 1} | \frac{1}{b} \frac{\lambda^2 + 1}{\lambda + 1} | ||
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& < \frac{10}{27} . | & < \frac{10}{27} . | ||
\end{align*} | \end{align*} | ||
+ | </cmath> | ||
Hence, to satisfy (8), a necessary condition is | Hence, to satisfy (8), a necessary condition is | ||
+ | <cmath> | ||
\begin{align*} | \begin{align*} | ||
\left| \left( \left( \lambda - 2 \right)^2 - 3 \right) \right| | \left| \left( \left( \lambda - 2 \right)^2 - 3 \right) \right| | ||
< \frac{10}{27} . | < \frac{10}{27} . | ||
\end{align*} | \end{align*} | ||
+ | </cmath> | ||
However, this cannot be satisfied for <math>\sqrt{3} < \lambda \leq \frac{30}{9}</math>. | However, this cannot be satisfied for <math>\sqrt{3} < \lambda \leq \frac{30}{9}</math>. |
Revision as of 13:31, 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,
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.
\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)