Competition problem — Algebra — Maths Olympiad Prep

Library / /96 of 220

Algebra Difficulty 5.8 AIME, harder Prove it Ukraine

Square $ABCD$ of size $2019 \times 2019$ is divided by two lines into four rectangles with integer side lengths, and some of the rectangles might be squares. Turns out, the area of the rectangle containing vertex $A$ equals the perimeter of the rectangle containing vertex $B$ . What is the area of the smallest of the four rectangles (or squares)?

Solution

Let us denote the sides of the rectangle containing vertex $A$ as $a$ and $b$ (Fig. 21). Then the sides of the rectangle containing vertex $B$ are equal to $a$ and $2019-b$ . The equality of the area of the first rectangle and perimeter of the second yields:
$ab = 2 \cdot (a + 2019 - b) \Rightarrow ab - 2a + 2b = 4038 \Rightarrow \\ (a+2)(b-2) = 4034 = 2017 \cdot 2.$

Since, on the left-hand side, one of the multipliers is greater than $2$ , and $2017$ is prime, the only solution is: $a+2=2017$ and $b-2=2$ . Hence, $a=2015$ and $b=4$ . Clearly, the smallest rectangle of the four would be the square with the area $4 \cdot 4 = 16$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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