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Difference between revisions of "2012 AMC 8 Problems/Problem 17"

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==Solution==
 
==Solution==
 
The first answer choice <math> {\textbf{(A)}\ 3} </math>, can be eliminated since there must be <math> 10 </math> squares with integer side lengths. We then test the next smallest sidelength which is <math> 4 </math>. The square with area <math> 16 </math> can be partitioned into <math> 8 </math> squares with area <math> 1 </math> and two squares with area <math> 4 </math>, which satisfies all the conditions of the problem. Therefore, the smallest possible value of the length of the side of the original square is <math> \boxed{\textbf{(B)}\ 4} </math>.
 
The first answer choice <math> {\textbf{(A)}\ 3} </math>, can be eliminated since there must be <math> 10 </math> squares with integer side lengths. We then test the next smallest sidelength which is <math> 4 </math>. The square with area <math> 16 </math> can be partitioned into <math> 8 </math> squares with area <math> 1 </math> and two squares with area <math> 4 </math>, which satisfies all the conditions of the problem. Therefore, the smallest possible value of the length of the side of the original square is <math> \boxed{\textbf{(B)}\ 4} </math>.
 +
 +
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("<math>4</math>", (2,4.25));
 +
label("<math>4</math>", (-0.25,2));
 +
label("2×2", (1,3)); // top-left 2×2
 +
label("2×2", (3,1)); // bottom-right 2×2
 +
 +
label("1×1", (0.5,0.5), fontsize(9));
 +
label("1×1", (1.5,0.5), fontsize(9));
 +
label("1×1", (0.5,1.5), fontsize(9));
 +
label("1×1", (1.5,1.5), fontsize(9));
 +
 +
label("1×1", (2.5,2.5), fontsize(9));
 +
label("1×1", (3.5,2.5), fontsize(9));
 +
label("1×1", (2.5,3.5), fontsize(9));
 +
label("1×1", (3.5,3.5), fontsize(9));
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|num-b=16|num-a=18}}
 
{{AMC8 box|year=2012|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:10, 11 October 2025

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?

$\text{(A)}\hspace{.05in}3\qquad\text{(B)}\hspace{.05in}4\qquad\text{(C)}\hspace{.05in}5\qquad\text{(D)}\hspace{.05in}6\qquad\text{(E)}\hspace{.05in}7$

Solution

The first answer choice ${\textbf{(A)}\ 3}$, can be eliminated since there must be $10$ squares with integer side lengths. We then test the next smallest sidelength which is $4$. The square with area $16$ can be partitioned into $8$ squares with area $1$ and two squares with area $4$, which satisfies all the conditions of the problem. Therefore, the smallest possible value of the length of the side of the original square is $\boxed{\textbf{(B)}\ 4}$.

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("$4$", (2,4.25)); label("$4$", (-0.25,2)); label("2×2", (1,3)); // top-left 2×2 label("2×2", (3,1)); // bottom-right 2×2

label("1×1", (0.5,0.5), fontsize(9)); label("1×1", (1.5,0.5), fontsize(9)); label("1×1", (0.5,1.5), fontsize(9)); label("1×1", (1.5,1.5), fontsize(9));

label("1×1", (2.5,2.5), fontsize(9)); label("1×1", (3.5,2.5), fontsize(9)); label("1×1", (2.5,3.5), fontsize(9)); label("1×1", (3.5,3.5), fontsize(9));

See Also

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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

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