Solving the Equation ( frac{1}{a} frac{1}{b} frac{1}{c} frac{1}{d} frac{1}{4} ) and ( abcd 99 )

In this article, we will explore how to solve a system of equations: ( frac{1}{a} frac{1}{b} frac{1}{c} frac{1}{d} frac{1}{4} ) and ( abcd 99 ). We will use algebraic manipulation, Vieta's formulas, and the AM-HM inequality to find the solution.

Understanding the Problem

We are given two equations:

( frac{1}{a} frac{1}{b} frac{1}{c} frac{1}{d} frac{1}{4} ) ( abcd 99 )

Let's break down each step to solve this problem.

Solving the Equations

First, we rewrite the first equation in a more convenient form:

( frac{1}{a} frac{1}{b} frac{1}{c} frac{1}{d} frac{1}{4} )

This can be expressed as:

( frac{bcd acd abd abc}{abcd} frac{1}{4} )

Cross-multiplying, we get:

( 4bcd acd abd abc abcd )

Rearranging gives us:

( abcd - 4bcd acd abd abc 0 )

Defining Useful Summations

Let's define two important summations:

( S_1 a b c d 99 ) ( S_2 ab ac ad bc bd cd )

Using Vieta's formulas, we know the following:

The sum of the roots ( S_1 a b c d ). The sum of the products of the roots taken two at a time ( S_2 ab ac ad bc bd cd ). The product of the roots ( P abcd ).

We can express ( abcd ) in terms of ( S_2 ):

( abcd 4S_2 )

Applying the AM-HM Inequality

To further simplify, we use the AM-HM inequality:

( frac{S_1}{4} geq frac{4}{frac{1}{a} frac{1}{b} frac{1}{c} frac{1}{d}} 4 cdot 4 16 )

This implies that:

( S_1 geq 16 cdot 4 64 )

Since ( S_1 99 ) is greater than 64, the equality in the AM-HM inequality holds, which means:

( a b c d )

Setting ( a b c d x ), from ( a b c d 99 ),

( 4x 99 )

( x frac{99}{4} 24.75 )

Substituting ( x ) back into the product:

( abcd x^4 left( frac{99}{4} right)^4 )

Calculating ( left( frac{99}{4} right)^4 ):

( left( frac{99}{4} right)^4 frac{99^4}{4^4} frac{96059601}{256} 375000 )

Brute Force Solution in J Programming Language

We can use the J programming language to generate all unique combinations of 4 integers from 1 to 70, allowing duplicates, and find those that sum to 99 and whose inverses sum to 1/4:

Generate all unique combinations of 4 integers from the integer set 1 to 70, allowing duplicates:

n ~./:~
1088430

So there are 1088430 unique sets of 4 integers with duplicates from the set of integers 1 to 70.

Find all the sets of 4 integers in n that sum to 99 and whose inverses sum to 1/4:

m n~99/
7 20 30 42
12 12 15 60

So there are two sets of 4 integers that meet the criteria: 7 20 30 42 and 12 12 15 60.

The product of each of these sets is:

/ 7 20 30 42 12 12 15 60
176400 129600

Therefore, the solutions are 176400 and 129600.

Should duplicates be prohibited, the solution is 176400.

Thus, the values of ( abcd ) are:

[ boxed{375000} ]