2012 AMC 8 Problems/Problem 17

Problem

A square with integer side length is cut into 10 squares, all of which have integer side length and at least 8 of which have area 1. What is the smallest possible value of the length of the side of the original square?

Solution

The first answer choice , can be eliminated since there must be squares with integer side lengths. We then test the next smallest sidelength which is . The square with area can be partitioned into squares with area and two squares with area , which satisfies all the conditions of the problem. Therefore, the smallest possible value of the length of the side of the original square is .

[asy] import olympiad; size(6cm);

// Outer 4×4 square draw((0,0)--(4,0)--(4,4)--(0,4)--cycle, linewidth(1.2));

// Split into four 2×2 squares draw((2,0)--(2,4)); draw((0,2)--(4,2));

// Split the two diagonal 2×2 squares into 1×1 squares // Bottom-left (0,0)-(2,2) draw((1,0)--(1,2)); draw((0,1)--(2,1));

// Top-right (2,2)-(4,4) draw((3,2)--(3,4)); draw((2,3)--(4,3));

// Labels label("", (2,4.25)); label("", (-0.25,2)); label("2×2", (1,3)); // top-left label("2×2", (3,1)); // bottom-right label("1×1", (1,1)); label("1×1", (3,3)); [/asy]

See Also

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.