Lengths of the sides of quadrilateral

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SUMMARY

The discussion centers on a mathematical proof regarding the lengths of the sides of a quadrilateral, specifically that if the lengths are positive integers and each side divides the sum of the other three sides, then at least two sides must be of equal length. The proof utilizes properties of divisibility and integer relationships to establish this conclusion definitively. Key terms include "positive integers," "divisibility," and "quadrilateral." The conclusion is reached through logical deduction and mathematical reasoning.

PREREQUISITES
  • Understanding of basic number theory, particularly divisibility rules.
  • Familiarity with properties of quadrilaterals in geometry.
  • Knowledge of mathematical proof techniques, including direct proof and contradiction.
  • Ability to work with positive integers and their relationships.
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  • Study the properties of quadrilaterals and their side relationships.
  • Learn about divisibility rules in number theory.
  • Explore mathematical proof techniques, focusing on direct proofs and proofs by contradiction.
  • Investigate integer partitions and their implications in geometry.
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Mathematicians, geometry enthusiasts, and students studying number theory or mathematical proofs will benefit from this discussion.

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The lengths of the sides of a quadrilateral are positive integers. The length of each side divides the sum of the other three lengths. Prove that two of the sides have the same length.
 
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If the quadrilateral has sides $a,b,c,d$ and perimeter $s = a+b+c+d$ then $a < b+c+d$. Also, $a$ divides $b+c+d$, so $b+c+d$ is at least twice $a$. But $b+c+d = s-a$, so $a$ also divides $s$, and $s$ must be at least $3$ times $a$, say $s=ka$ with $k\geqslant3$. Similarly, $s = lb$, $s=mc$ and $s=nd$, with all of $l,m,n$ greater than or equal to $3$.

Then $a = \dfrac1ks$, $b = \dfrac1ls$, $c = \dfrac1ms$, $d = \dfrac1ns$. Therefore $s = a+b+c+d = \dfrac1ks + \dfrac1ls + \dfrac1ms + \dfrac1ns$ and so $$\frac1k + \frac1l + \frac1m + \frac1n = 1.$$ But under the given conditions, if $a,b,c,d$ are all different then so are $k,l,m,n$, and the largest possible value for $\dfrac1k + \dfrac1l + \dfrac1m + \dfrac1n$ is $\dfrac13 +\dfrac14 + \dfrac15 + \dfrac16 = \dfrac{57}{60}$, which is less than $1$. So those conditions cannot be satisfied and therefore at least two of the four sides must have the same length.
 

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