Maths Olympiad Prep

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Algebra Difficulty 5.0 AIME, harder Find the answer

James writes down three integers. Alex picks some two of those integers, takes the average of them, and adds the result to the third integer. If the possible final results Alex could get are 42, 13, and 37, what are the three integers James originally chose?

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

Solution

Let x,y,zx, y, z be the integers. We have x+y2+z=42y+z2+x=13x+z2+y=37\begin{aligned} & \frac{x+y}{2}+z=42 \\ & \frac{y+z}{2}+x=13 \\ & \frac{x+z}{2}+y=37 \end{aligned} Adding these three equations yields 2(x+y+z)=922(x+y+z)=92, so x+y2+z=23+z2=42\frac{x+y}{2}+z=23+\frac{z}{2}=42 so z=38z=38. Similarly, x=−20x=-20 and y=28y=28.

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