Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.1 AIME, harder Prove it Soviet Union

Problem:

A square is divided into $n$ parallel strips (parallel to the bottom side of the square). The width of each strip is integral. The total width of the strips with odd width equals the total width of the strips with even width. A diagonal of the square is drawn which divides each strip into a left part and a right part. Show that the sum of the areas of the left parts of the odd strips equals the sum of the areas of the right parts of the even strips.

Solution

Solution:

Let $LO$ be the total area of the left parts of the odd strips, $LE$ the total area of the left parts of the even strips, and $RE$ the total area of the right parts of the even strips. Since the diagonal bisects the square, $LO + LE = A / 2$ , where $A$ is the area of the square. Also $RE + LE = A / 2$ (because the total width of the even strips equals the total width of the odd strips). Subtracting, $LO = RE$ , as required.

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