# Difference between revisions of "1984 AIME Problems/Problem 15"

Scorpius119 (talk | contribs) (solution added) |
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== Solution == | == Solution == | ||

− | {{ | + | For each of the values <math>t=4,16,36,64</math>, we have the equation |

+ | |||

+ | <math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25)</math> | ||

+ | |||

+ | <math>=(t-1)(t-9)(t-25)(t-49)-(t-4)(t-16)(t-36)(t-64)</math> | ||

+ | |||

+ | However, each side of the equation is a polynomial in <math>t</math> of degree at most 3, and they have 4 common roots. Therefore, the polynomials must be equal. | ||

+ | |||

+ | |||

+ | Now we can plug in <math>t=1</math> into the polynomial equation. Most terms drop, and we end up with | ||

+ | |||

+ | <math>x^2(-8)(-24)(-48)=-(-3)(-15)(-35)(-63)</math> | ||

+ | |||

+ | so that | ||

+ | |||

+ | <math>x^2=\frac{3\cdot 15\cdot 35\cdot 63}{8\cdot 24\cdot 48}=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}</math> | ||

+ | |||

+ | Similarly, we can plug in <math>t=9,25,49</math> and get | ||

+ | |||

+ | <math>y^2=\frac{5\cdot 7\cdot 27\cdot 55}{8\cdot 16\cdot 40}=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}}</math> | ||

+ | |||

+ | <math>z^2=\frac{21\cdot 9\cdot 11\cdot 39}{24\cdot 16\cdot 24}=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}}</math> | ||

+ | |||

+ | <math>w^2=\frac{45\cdot 33\cdot 13\cdot 15}{48\cdot 40\cdot 24}=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}}</math> | ||

+ | |||

+ | |||

+ | Now add them up... | ||

+ | |||

+ | <math>z^2+w^2=\frac{3^2\cdot 11\cdot 13(7+5)}{2^{10}}=\frac{3^3\cdot 11\cdot 13}{2^8}</math> | ||

+ | |||

+ | <math>x^2+y^2=\frac{3^2\cdot 5\cdot 7(5\cdot 7+3\cdot 11)}{2^{10}}=\frac{3^2\cdot 5\cdot 7\cdot 17}{2^8}</math> | ||

+ | |||

+ | with a sum of | ||

+ | |||

+ | <math>\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=36</math> | ||

== See also == | == See also == | ||

* [[1984 AIME Problems/Problem 14 | Previous problem]] | * [[1984 AIME Problems/Problem 14 | Previous problem]] | ||

* [[1984 AIME Problems]] | * [[1984 AIME Problems]] |

## Revision as of 19:26, 26 March 2007

## Problem

Determine if

## Solution

For each of the values , we have the equation

However, each side of the equation is a polynomial in of degree at most 3, and they have 4 common roots. Therefore, the polynomials must be equal.

Now we can plug in into the polynomial equation. Most terms drop, and we end up with

so that

Similarly, we can plug in and get

Now add them up...

with a sum of