Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.6 AIME Prove it United States

Problem:
Find the area of the region of the $x y$ -plane defined by the inequality $|x|+|y|+|x+y| \leq 1$ .

Solution

Solution:
To graph this region we divide the $x y$ -plane into six sectors depending on which of $x, y, x+y$ are $\geq 0$ , or $\leq 0$ . The inequality simplifies in each case:

Sector	Inequality	Simplified inequality
$x \geq 0, y \geq 0, x+y \geq 0$	$x+y+(x+y) \leq 1$	$x+y \leq 1 / 2$
$x \geq 0, y \leq 0, x+y \geq 0$	$x-y+(x+y) \leq 1$	$x \leq 1 / 2$
$x \geq 0, y \leq 0, x+y \leq 0$	$x-y-(x+y) \leq 1$	$y \geq-1 / 2$
$x \leq 0, y \geq 0, x+y \geq 0$	$-x+y+(x+y) \leq 1$	$y \leq 1 / 2$
$x \leq 0, y \geq 0, x+y \leq 0$	$-x+y-(x+y) \leq 1$	$x \geq-1 / 2$
$x \leq 0, y \leq 0, x+y \leq 0$	$-x-y-(x+y) \leq 1$	$x+y \geq-1 / 2$

We then draw the region; we get a hexagon as shown. The hexagon intersects each region in an isosceles right triangle of area

1 / 8

, so the total area is

6 \cdot 1 / 8=3 / 4

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