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[SOLVED] Lengths of the sides of quadrilateral

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MHB POTW Director
Staff member
Feb 14, 2012
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.


MHB Oldtimer
Staff member
Feb 7, 2012
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.