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Algebra Difficulty 4.4 AIME Find the answer United States

Problem:
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?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:
Let $x$ , $y$ , $z$ be the integers. We have
$\begin{aligned} & \frac{x+y}{2} + z = 42 \\ & \frac{y+z}{2} + x = 13 \\ & \frac{x+z}{2} + y = 37 \end{aligned}$
Adding these three equations yields $2(x+y+z) = 92$ , so $x+y+z = 46$ .

Now, from $\frac{x+y}{2} + z = 42$ :
$\frac{x+y}{2} + z = 42 \\ \Rightarrow x+y = 2(42 - z) \\$
But $x+y+z = 46$ , so $x+y = 46 - z$ .

Set equal:
$46 - z = 2(42 - z) \\ 46 - z = 84 - 2z \\ 2z - z = 84 - 46 \\ z = 38$

Now $x+y = 46 - 38 = 8$ .

From $\frac{x+z}{2} + y = 37$ :
$\frac{x+z}{2} + y = 37 \\ x+z = 2(37 - y) \\$
But $x+z = (x+y+z) - y = 46 - y$ .

Set equal:
$46 - y = 2(37 - y) \\ 46 - y = 74 - 2y \\ 2y - y = 74 - 46 \\ y = 28$

Now $x = 46 - 28 - 38 = -20$ .

Therefore, the three integers are $-20$ , $28$ , $38$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.