Difference between revisions of "1987 AIME Problems/Problem 14"

Revision as of 15:13, 31 August 2021

Problem

Compute $\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.$

Solution 1 (Sophie Germain Identity)

The Sophie Germain Identity states that $a^4 + 4b^4$ can be factored as $\left(a^2 + 2b^2 - 2ab\right)\left(a^2 + 2b^2 + 2ab\right).$ Each of the terms is in the form of $x^4 + 324.$ Using Sophie Germain, we get that $\begin{align*} x^4 + 324 &= x^4 + 4\cdot 3^4 \\ &= \left(x^2 + 2 \cdot 3^2 - 2\cdot 3\cdot x\right)\left(x^2 + 2 \cdot 3^2 + 2\cdot 3\cdot x\right) \\ &= (x(x-6) + 18)(x(x+6)+18), \end{align*}$ so the original expression becomes $\frac{[(10(10-6)+18)(10(10+6)+18)][(22(22-6)+18)(22(22+6)+18)]\cdots[(58(58-6)+18)(58(58+6)+18)]}{[(4(4-6)+18)(4(4+6)+18)][(16(16-6)+18)(16(16+6)+18)]\cdots[(52(52-6)+18)(52(52+6)+18)]},$ which simplifies to $\frac{(10(4)+18)(10(16)+18)(22(16)+18)(22(28)+18)\cdots(58(52)+18)(58(64)+18)}{(4(-2)+18)(4(10)+18)(16(10)+18)(16(22)+18)\cdots(52(46)+18)(52(58)+18)}.$ Almost all of the terms cancel out! We are left with $\frac{58(64)+18}{4(-2)+18} = \frac{3730}{10} = \boxed{373}.$

~Azjps (Solution)

~MRENTHUSIASM (Minor Reformatting)

Solution 2 (Completing the Square and Difference of Squares)

In both the numerator and the denominator, each factor is of the form $N^4+324=N^4+18^2$ for some positive integer $N.$

We factor $N^4+18^2$ by completing the square, then applying the difference of squares: $\begin{align*} N^4+18^2&=\left(N^4+36N^2+18^2\right)-36N^2 \\ &=\left(N^2+18\right)^2-(6N)^2 \\ &=\left(N^2-6N+18\right)\left(N^2+6N+18\right) \\ &=\left((N-3)^2+9\right)\left((N+3)^2+9\right). \end{align*}$ The original expression now becomes $\frac{\left[(7^2+9)(13^2+9)\right]\left[(19^2+9)(25^2+9)\right]\left[(31^2+9)(37^2+9)\right]\left[(43^2+9)(49^2+9)\right]\left[(55^2+9)(61^2+9)\right]}{\left[(1^2+9)(7^2+9)\right]\left[(13^2+9)(19^2+9)\right]\left[(25^2+9)(31^2+9)\right]\left[(37^2+9)(43^2+9)\right]\left[(49^2+9)(55^2+9)\right]}=\frac{61^2+9}{1^2+9}=\boxed{373}.$ ~MRENTHUSIASM

Solution 3 (Complex Numbers)

In both the numerator and the denominator, each factor is of the form $N^4+324=N^4+18^2$ for some positive integer $N.$

We factor $N^4+18^2$ by solving the equation $N^4+18^2=0,$ or $N^4=-18^2.$

Two solutions follow from here:

Solution 3.1 (Polar Form)

We rewrite $N$ to the polar form $N=r(\cos\theta+i\sin\theta)=r\mathrm{ \ cis \ }\theta,$ where $r$ is the magnitude of $N$ such that $r\geq0,$ and $\theta$ is the argument of $N$ such that $0\leq\theta<2\pi.$

By De Moivre's Theorem, we have $N^4=r^4\mathrm{ \ cis \ }(4\theta)=18^2(-1),$ from which

$r^4=18^2,$ so $r=3\sqrt2.$

$\begin{cases} \begin{aligned} \cos{(4\theta)} &= -1 \\ \sin{(4\theta)} &= 0 \end{aligned}, \end{cases}$ so $\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}.$

By the Factor Theorem, we get $\begin{align*} N^4+18^2&=\biggl(N-3\sqrt2\mathrm{ \ cis \ }\frac{\pi}{4}\biggr)\biggl(N-3\sqrt2\mathrm{ \ cis \ }\frac{3\pi}{4}\biggr)\biggl(N-3\sqrt2\mathrm{ \ cis \ }\frac{5\pi}{4}\biggr)\biggl(N-3\sqrt2\mathrm{ \ cis \ }\frac{7\pi}{4}\biggr) \\ &=\biggl[\biggl(N-3\sqrt2\mathrm{ \ cis \ }\frac{\pi}{4}\biggr)\biggl(N-3\sqrt2\mathrm{ \ cis \ }\frac{7\pi}{4}\biggr)\biggr]\biggl[\biggl(N-3\sqrt2\mathrm{ \ cis \ }\frac{3\pi}{4}\biggr)\biggl(N-3\sqrt2\mathrm{ \ cis \ }\frac{5\pi}{4}\biggr)\biggr] \\ &=\left[(N-(3+3i))(N-(3-3i))\right]\left[(N-(-3+3i))(N-(-3-3i))\right] \\ &=\left[((N-3)-3i)((N-3)+3i)\right]\left[((N+3)-3i)((N+3)+3i)\right] \\ &=\left[(N-3)^2+9\right]\left[(N+3)^2+9\right]. \end{align*}$ We continue with the last paragraph of Solution 2 to get the answer $\boxed{373}.$

~MRENTHUSIASM

Solution 3.2 (Rectangular Form)

We rewrite $N$ to the rectangular form $N=a+bi$ for some real numbers $a$ and $b.$

Note that $N^2=\pm18i,$ so there are two cases:

$N^2=18i$
We have $\begin{align*} (a+bi)^2&=18i \\ a^2-b^2+2abi&=18i. \end{align*}$ We need $\begin{cases} \begin{aligned} a^2-b^2 &= 0 \\ 2ab &= 18 \end{aligned}, \end{cases}$ from which $(a,b)=(3,3),(-3,-3),$ or $N=3+3i,-3-3i.$

$N^2=-18i$
We have $\begin{align*} (a+bi)^2&=-18i \\ a^2-b^2+2abi&=-18i. \end{align*}$ We need $\begin{cases} \begin{aligned} a^2-b^2 &= 0 \\ 2ab &= -18 \end{aligned}, \end{cases}$ from which $(a,b)=(3,-3),(-3,3),$ or $N=3-3i,-3+3i.$

By the Factor Theorem, we get $\begin{align*} N^4+18^2&=(N-(3+3i))(N-(-3-3i))(N-(3-3i))(N-(-3+3i)) \\ &=\left[(N-(3+3i))(N-(3-3i))\right]\left[(N-(-3+3i))(N-(-3-3i))\right] \\ &=\left[((N-3)-3i)((N-3)+3i)\right]\left[((N+3)-3i)((N+3)+3i)\right] \\ &=\left[(N-3)^2+9\right]\left[(N+3)^2+9\right]. \end{align*}$ We continue with the last paragraph of Solution 2 to get the answer $\boxed{373}.$

~MRENTHUSIASM

Video Solutions

https://youtu.be/ZWqHxc0i7ro?t=1023

~ pi_is_3.14

https://www.youtube.com/watch?v=yoOWcx2Otcw

~Michael Penn

@@ Line 41: / Line 41: @@
 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>N^4=r^4\mathrm{cis}(4\theta)=18^2(-1),</cmath> from which
+By <b>De Moivre's Theorem</b>, we have <cmath>N^4=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 \\
+\cos{(4\theta)} &= -1 \\
-\sin(4\theta) &= 0
+\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>

1987 AIME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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