Difference between revisions of "2023 AIME I Problems/Problem 15"
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==Answer: 349== | ==Answer: 349== | ||
− | Suppose <math>z=a+bi</math>; notice that <math>\arg(z^{3})\approx 45^{\circ}</math>, so by De Moivre’s theorem <math>\arg(z)\approx 15^{\circ}</math> and <math>\tfrac{b}{a}\approx tan(15^{\circ})=2-\sqrt{3}</math>. Now just try pairs <math>(a, b)=(t, \left(2-\sqrt{3}\right)t)</math> going down from <math>t=\lfloor\sqrt{1000}\rfloor</math>, writing down the value of <math>a^{2}+b^{2}</math> on the right; and eventually we arrive at <math>(a, b)=(18, 5)</math> the first time <math>a^{2}+b^{2}</math> is prime. Therefore, <math>p=\boxed{349}</math>. | + | Suppose <math>z=a+bi</math>; notice that <math>\arg(z^{3})\approx 45^{\circ}</math>, so by De Moivre’s theorem <math>\arg(z)\approx 15^{\circ}</math> and <math>\tfrac{b}{a}\approx\tan(15^{\circ})=2-\sqrt{3}</math>. Now just try pairs <math>(a, b)=(t, \left(2-\sqrt{3}\right)t)</math> going down from <math>t=\lfloor\sqrt{1000}\rfloor</math>, writing down the value of <math>a^{2}+b^{2}</math> on the right; and eventually we arrive at <math>(a, b)=(18, 5)</math> the first time <math>a^{2}+b^{2}</math> is prime. Therefore, <math>p=\boxed{349}</math>. |
==Solution== | ==Solution== |
Revision as of 13:34, 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}
$$ (Error compiling LaTeX. Unknown error_msg)
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)