Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
(→‎Video Solution)
Line 101: Line 101:
  
 
~ pi_is_3.14
 
~ pi_is_3.14
 +
https://www.youtube.com/watch?v=yoOWcx2Otcw - Michael Penn
  
 
== See also ==
 
== See also ==

Revision as of 14:27, 19 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 $(a^2 + 2b^2 - 2ab)(a^2 + 2b^2 + 2ab)$. Each of the terms is in the form of $x^4 + 324$. Using Sophie Germain, we get that \[x^4 + 4\cdot 3^4 = (x^2 + 2 \cdot 3^2 - 2\cdot 3\cdot x)(x^2 + 2 \cdot 3^2 + 2\cdot 3\cdot x) = (x(x-6) + 18)(x(x+6)+18),\] so the original expression becomes

$\frac{[(10(10-6)+18)(10(10+6)+18)][(22(22-6)+18)(22(22+6)+18)]\cdots[(58(58-6)+18)(58(58+6)+18)]}{[(4(4-6)+18)(4(4+6)+18)][(16(16-6)+18)(16(16+6)+18)]\cdots[(52(52-6)+18)(52(52+6)+18)]}$

$= \frac{(10(4)+18)(10(16)+18)(22(16)+18)(22(28)+18)\cdots(58(52)+18)(58(64)+18)}{(4(-2)+18)(4(10)+18)(16(10)+18)(16(22)+18)\cdots(52(46)+18)(52(58)+18)}$

Almost all of the terms cancel out! We are left with $\frac{58(64)+18}{4(-2)+18} = \frac{3730}{10} = \boxed{373}$.

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

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

  2. $\begin{cases} \begin{aligned} \cos(4\theta) &= -1 \\ \sin(4\theta) &= 0 \end{aligned}, \end{cases}$ 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:

  1. $N^2=18i$

    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 $N=3+3i,-3-3i.$

  2. $N^2=-18i$

    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 Solution

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

~ pi_is_3.14 https://www.youtube.com/watch?v=yoOWcx2Otcw - Michael Penn

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_14&oldid=160572"