Maths Olympiad Prep

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Algebra Difficulty 5.5 AIME, harder Find the answer

## Task B-1.1.

If xyz=abcx y z=a b c, what is bxxy+ab+bx+cyyz+bc+cy+azzx+ca+az\frac{b x}{x y+a b+b x}+\frac{c y}{y z+b c+c y}+\frac{a z}{z x+c a+a z}?

A number or a short expression. Spacing and $ signs are ignored.

Solution

## Solution.

First, extend the first fraction by zz, and the second by aa.

bxxy+ab+bxzz+cyyz+bc+cyaa+azzx+ca+az=bxzxyz+abz+bxz+cayayz+abc+acy+azzx+ca+az \begin{aligned} & \frac{b x}{x y+a b+b x} \cdot \frac{z}{z}+\frac{c y}{y z+b c+c y} \cdot \frac{a}{a}+\frac{a z}{z x+c a+a z}= \\ & \frac{b x z}{x y z+a b z+b x z}+\frac{c a y}{a y z+a b c+a c y}+\frac{a z}{z x+c a+a z} \end{aligned}

Replace the expression xyzxyz in the denominator of the first fraction with abcabc, and the expression abcabc in the denominator of the second fraction with xyzxyz.

bxzabc+abz+bxz+acyayz+xyz+acy+azzx+ca+az=bxzb(ac+az+xz)+acyy(az+xz+ac)+azzx+ca+az=xz+ac+azac+xz+az=1. \begin{gathered} \frac{b x z}{a b c+a b z+b x z}+\frac{a c y}{a y z+x y z+a c y}+\frac{a z}{z x+c a+a z}= \\ \frac{b x z}{b(a c+a z+x z)}+\frac{a c y}{y(a z+x z+a c)}+\frac{a z}{z x+c a+a z}=\frac{x z+a c+a z}{a c+x z+a z}=1 . \end{gathered}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.