The length of each side of a convex quadrilateral $ABCD$ is a positive integer. If the sum of the lengths of any three sides is divisible by the length of the remaining side then prove that some two sides of the quadrilateral have the same length.
Problem
Source: RMO (Mumbai Region) 2015 P8
Tags: inequalities, number theory, geometry, geometry proposed
libra8
06.12.2015 16:31
assume a< b< c< d. then as d divides a+b+c, and a+b+c < 3d, and a+b+c > d (triangle inequality), a+b+c = 2d. perform similarly with c, b, a.
AnonymousBunny
06.12.2015 16:31
Does this work?
Let the lengths be $a, b, c, d$ with $b, c, d <a$. Note that each number must divide $a+b+c+d$. Also $b+c+d$ is less than $a+a+a= 3a$ and more than $a$ (because the quadrilateral is convex), so $b+c+d= 2a$, so $b, c, d$ are all divisors of $a+b+c+d= 3a$. However the sum of the three largest divisors of $3a$ less than $a$ is at most $\dfrac{3a}{4} + \dfrac{3a}{5} + \dfrac{3a}{6} < 2a$, contradiction.
edit: sniped by few seconds
libra8
06.12.2015 16:32
yeah. same.
aditya21
06.12.2015 16:45
yeah that's a nice and easy problem.
trimo
03.11.2019 09:06
Same as Q. No 62 in the book "Maths Problem Book 1" by Kin Y. Li.
guptaamitu1
13.10.2020 21:29
AnonymousBunny wrote: Does this work?
Let the lengths be $a, b, c, d$ with $b, c, d <a$. Note that each number must divide $a+b+c+d$. Also $b+c+d$ is less than $a+a+a= 3a$ and more than $a$ (because the quadrilateral is convex), so $b+c+d= 2a$, so $b, c, d$ are all divisors of $a+b+c+d= 3a$. However the sum of the three largest divisors of $3a$ less than $a$ is at most $\dfrac{3a}{4} + \dfrac{3a}{5} + \dfrac{3a}{6} < 2a$, contradiction.
edit: sniped by few seconds This solution is wrong as there are could be some multiples of 3a which are less than a and have more larger values, for example, 20a/21 etc
sanyalarnab
20.04.2022 07:02
ok ig there is no complete solutions. So here it goes: Assume the sides are $a,b,c,d$ and that they are distinct. Then we can get an ordering. WLOG we assume $$a<b<c<d$$. Let $b+c+d=am_1$ $a+c+d=bm_2$ $b+a+d=cm_3$ $b+c+a=dm_4$ From the quadrilateral inequality we get $m_4>1$ Also $dm_4 < d+d+d$ Hence we get $1<m_4<3 \implies m_4=2$ Also $a(m_1+1)=b(m_2+1)=c(m_3+1)=3d$ $\implies m_1>m_2>m_3>2$ $\implies m_1 \ge 5,m_2 \ge 4, m_3 \ge 3$ Hence $3(a+b+c+d)=am_1+bm_2+cm_3+dm_4 \ge 5a+4b+3c+2d$ $\implies d \ge 2a+b$....(*) Also $(d+a+b)+(a+b+c) = cm_3+dm_4 \ge 3c+2d$ $\implies d \le 2a+2b-2c < 2a+2c-2c=2a < 2a+b$ which contradicts (*). Hence our assumption for $a,b,c,d$ to be distinct was wrong, completing the proof.