Easier Version of an Old HMMT Problem

Algebra Level 5

Let a b c a\geq b\geq c be real numbers with a + b + c > 0 a+b+c>0 such that a 2 b c + a b 2 c + a b c 2 + 21 = a + b + c , a 2 b + a 2 c + b 2 c + b 2 a + c 2 a + c 2 b + 3 a b c = 3 , a 2 b 2 c + a b 2 c 2 + a 2 b c 2 = 7 + a b + b c + c a . \begin{aligned}a^2bc+ab^2c+abc^2+21&=a+b+c,\\a^2b+a^2c+b^2c+b^2a+c^2a+c^2b+3abc&=-3,\\a^2b^2c+ab^2c^2+a^2bc^2&=7+ab+bc+ca.\end{aligned} If a 2 a^2 can be written as m + n p \dfrac{m+\sqrt n}p for positive integers m , n , m,n, and p p , find m + n + p m+n+p .


The answer is 22.

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1 solution

David Altizio
Apr 22, 2014

After rearranging, the three equations factor to

( a + b + c ) ( 1 a b c ) = 21 , (a+b+c)(1-abc)=21, ( a + b + c ) ( a b + a c + b c ) = 3 , (a+b+c)(ab+ac+bc)=-3, ( a b + a c + b c ) ( 1 a b c ) = 7. (ab+ac+bc)(1-abc)=-7.

Solving this system and remembering to take into account a + b + c > 0 a+b+c>0 gives a + b + c = 3 , a b + a c + b c = 1 , a b c = 6 a+b+c=3,ab+ac+bc=-1,abc=-6 , which means that a , b , c a,b,c are the roots of the cubic t 3 3 t 2 t + 6 t^3-3t^2-t+6 . By Rational Root Theorem, t = 2 t=2 is a root, and factoring this out gives ( t 2 ) ( t 2 t 3 ) = 0 (t-2)(t^2-t-3)=0 . The solutions to the latter quadratic are t = 1 ± 13 2 t=\tfrac{1\pm\sqrt{13}}2 , and since 1 + 13 2 > 2 > 1 13 2 \tfrac{1+\sqrt{13}}2>2>\tfrac{1-\sqrt{13}}2 , a = 1 + 13 2 a=\tfrac{1+\sqrt{13}}2 . This means that a 2 = ( 1 + 13 2 ) 2 = 7 + 13 2 a^2=\left(\dfrac{1+\sqrt{13}}2\right)^2=\dfrac{7+\sqrt{13}}2 and the requested answer is 7 + 13 + 2 = 22 7+13+2=\boxed{22} .

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