MHB Lengths of the sides of quadrilateral

  • Thread starter Thread starter anemone
  • Start date Start date
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
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.
 
Mathematics news on Phys.org
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.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.

Similar threads

Replies
4
Views
2K
Replies
38
Views
3K
Replies
1
Views
1K
Replies
3
Views
2K
Replies
30
Views
5K
Replies
6
Views
1K
Replies
1
Views
1K
Back
Top