Mathematical Reflections 2013, Issue 3 - Problem J265

Problem:
Let $a,b,c$ be real numbers such that $$5(a+b+c)-2(ab+bc+ca)=9.$$
Prove that any two of the equalities $$|3a-4b|=|5c-6|, \qquad |3b-4c|=|5a-6|, \qquad |3c-4a|=|5b-6|$$
imply the third.

Proposed by Titu Andreescu.

Solution:
By symmetry, we can suppose that the first two equalities are given. Since $|x|=|y|$ if and only if $x^2=y^2$ for real numbers $x,y$, then
$$9a^2-24ab+16b^2=25c^2-60c+36, \qquad 9b^2-24bc+16c^2=25a^2-60a+36.$$ Summing up the two equalities and reordering, we have
$$25b^2-24(ab+bc)+36=16a^2-60(a+c)+9c^2+108.$$
Since $60(a+c)-24(ab+bc)=108-60b+24ca$, we get $$25b^2+(108-60b+24ca)+36=16a^2+9c^2+108,$$ and reordering we get $$(5b-6)^2=(3c-4a)^2,$$ i.e. $|5b-6|=|3c-4a|$, which is the third equality.