Difference between revisions of "1987 AIME Problems/Problem 14"
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MRENTHUSIASM (talk | contribs) (→Solution 3.1 (Exponential Form)) |
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Two solutions follow from here: | Two solutions follow from here: | ||
− | === Solution 3.1 ( | + | === Solution 3.1 (Polar Form) === |
− | We rewrite <math>N</math> to the | + | We rewrite <math>N</math> to the polar form <cmath>N=r(\cos\theta+i\sin\theta)=r\mathrm{ \ cis \ }\theta,</cmath> where <math>r</math> is the magnitude of <math>N</math> such that <math>r\geq0,</math> and <math>\theta</math> is the argument of <math>N</math> such that <math>0\leq\theta<2\pi.</math> |
+ | |||
+ | By <b>De Moivre's Theorem</b>, we have <cmath>r^4\mathrm{cis}(4\theta)=18^2(-1),</cmath> from which | ||
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li><math>r^4=18^2,</math> so <math>r=3\sqrt2.</math></li><p> | ||
+ | <li><math>\begin{cases} | ||
+ | \begin{aligned} | ||
+ | \cos(4\theta) &= -1 \ | ||
+ | \sin(4\theta) &= 0 | ||
+ | \end{aligned}, | ||
+ | \end{cases}</math> so <math>\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}.</math></li><p> | ||
+ | </ol> | ||
+ | By the <b>Factor Theorem</b>, we get | ||
+ | <cmath>\begin{align*} | ||
+ | N^4+18^2&=\biggl(N-3\sqrt2\mathrm{ \ cis \ }\frac{\pi}{4}\biggr)\biggl(N-3\sqrt2\mathrm{ \ cis \ }\frac{3\pi}{4}\biggr)\biggl(N-3\sqrt2\mathrm{ \ cis \ }\frac{5\pi}{4}\biggr)\biggl(N-3\sqrt2\mathrm{ \ cis \ }\frac{7\pi}{4}\biggr) \ | ||
+ | &=\biggl[\biggl(N-3\sqrt2\mathrm{ \ cis \ }\frac{\pi}{4}\biggr)\biggl(N-3\sqrt2\mathrm{ \ cis \ }\frac{7\pi}{4}\biggr)\biggr]\biggl[\biggl(N-3\sqrt2\mathrm{ \ cis \ }\frac{3\pi}{4}\biggr)\biggl(N-3\sqrt2\mathrm{ \ cis \ }\frac{5\pi}{4}\biggr)\biggr] \ | ||
+ | &=\left[(N-(3+3i))(N-(3-3i))\right]\left[(N-(-3+3i))(N-(-3-3i))\right] \ | ||
+ | &=\left[((N-3)-3i)((N-3)+3i)\right]\left[((N+3)-3i)((N+3)+3i)\right] \ | ||
+ | &=\left[(N-3)^2+9\right]\left[(N+3)^2+9\right]. | ||
+ | \end{align*}</cmath> | ||
+ | We continue with the last paragraph of Solution 2 to get the answer <math>\boxed{337}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
=== Solution 3.2 (Rectangular Form) === | === Solution 3.2 (Rectangular Form) === |
Revision as of 14:45, 2 July 2021
Contents
[hide]Problem
Compute
Solution 1 (Sophie Germain Identity)
The Sophie Germain Identity states that can be factored as . Each of the terms is in the form of . Using Sophie Germain, we get that so the original expression becomes
Almost all of the terms cancel out! We are left with .
Solution 2 (Completing the Square and Difference of Squares)
In both the numerator and the denominator, each factor is of the form for some positive integer
We factor by completing the square, then applying the difference of squares: The original expression now becomes ~MRENTHUSIASM
Solution 3 (Complex Numbers)
In both the numerator and the denominator, each factor is of the form for some positive integer
We factor by solving the equation or
Two solutions follow from here:
Solution 3.1 (Polar Form)
We rewrite to the polar form where is the magnitude of such that and is the argument of such that
By De Moivre's Theorem, we have from which
- so
- so
By the Factor Theorem, we get We continue with the last paragraph of Solution 2 to get the answer
~MRENTHUSIASM
Solution 3.2 (Rectangular Form)
Video Solution
https://youtu.be/ZWqHxc0i7ro?t=1023
~ pi_is_3.14
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.