Problem
NT-B2-M07-P018 A Harder Egyptian Fraction Search
Find all triples of positive integers \(x\le y\le z\) such that \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2}\).
Use the ordering to get \(x\le 6\). Values \(x=1,2\) are impossible. Then solve a two-fraction equation for each \(x=3,4,5,6\).
From \(x\le y\le z\), we get \(\frac{3}{x}\ge \frac{1}{2}\), so \(x\le 6\). If \(x=1\), the sum is already greater than \(\frac{1}{2}\); if \(x=2\), the remaining fractions would have to sum to \(0\), impossible.
If \(x=3\), then \(\frac{1}{y}+\frac{1}{z}=\frac{1}{6}\), so \((y-6)(z-6)=36\). With \(y\le z\), this gives \((y,z)=(7,42),(8,24),(9,18),(10,15),(12,12)\).
If \(x=4\), then \(\frac{1}{y}+\frac{1}{z}=\frac{1}{4}\), so \((y-4)(z-4)=16\). We get \((5,20),(6,12),(8,8)\).
If \(x=5\), then \(\frac{1}{y}+\frac{1}{z}=\frac{3}{10}\). Multiplying and completing a product gives \((3y-10)(3z-10)=100\). With \(y\ge 5\) and \(y\le z\), the only suitable choice is \(3y-10=5\), hence \((y,z)=(5,10)\).
If \(x=6\), then \(\frac{1}{y}+\frac{1}{z}=\frac{1}{3}\). Since \(y,z\ge 6\), the maximum is \(\frac{1}{3}\), so \(y=z=6\). The answer is \((3,7,42),(3,8,24),(3,9,18),(3,10,15),(3,12,12),(4,5,20),(4,6,12),(4,8,8),(5,5,10),(6,6,6)\).
This is a full olympiad-style branching problem; watch that students do not skip the case \(x=5\).