Maths Olympiad Prep

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Algebra Difficulty 4.5 AIME Prove it Slovenia

If a(b+c)+b(c+a)+c(a+b)=ab+bc+caa(b + c) + b(c + a) + c(a + b) = ab + bc + ca, then
a2(b+c)+b2(a+c)+c2(a+b)abc \frac{a^2(b+c) + b^2(a+c) + c^2(a+b)}{abc}
is an integer.

Solution

The given equation implies ab+bc+ca=0ab + bc + ca = 0, so we can write
a2(b+c)+b2(a+c)+c2(a+b)abc=a(ab+ac)+b(ba+bc)+c(ca+cb)abc \frac{a^2(b+c) + b^2(a+c) + c^2(a+b)}{abc} = \frac{a(ab+ac) + b(ba+bc) + c(ca+cb)}{abc}

Since ab+ac=−bcab + ac = -bc, ba+bc=−caba + bc = -ca and ca+cb=−abca + cb = -ab we have
a(ab+ac)+b(ba+bc)+c(ca+cb)abc=a(−bc)+b(−ca)+c(−ab)abc=−3abcabc=−3. \frac{a(ab + ac) + b(ba + bc) + c(ca + cb)}{abc} = \frac{a(-bc) + b(-ca) + c(-ab)}{abc} = \frac{-3abc}{abc} = -3.

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