# 2006 AMC 12A Problems/Problem 6

The following problem is from both the 2006 AMC 12A #6 and 2006 AMC 10A #7, so both problems redirect to this page.

## Problem

The $8\times18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$? $[asy] unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,0)--(12,0)--(12,-4)); label("A",(0,4),NW); label("B",(18,4),NE); label("C",(18,-4),SE); label("D",(0,-4),SW); label("y",(3,4),S); label("y",(15,-4),N); label("18",(9,4),N); label("18",(9,-4),S); label("8",(0,0),W); label("8",(18,0),E); dot((0,4)); dot((18,4)); dot((18,-4)); dot((0,-4));[/asy]$ $\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10$

## Solutions

### Solution 1

Since the two hexagons are going to be repositioned to form a square without overlap, the area will remain the same. The rectangle's area is $18\cdot8=144$. This means the square will have four sides of length 12. The only way to do this is shown below. $[asy] size(175); pair A,B,C,D,E,F,G,H; A=(0,8); B=(12,12); C=(12,4); D=(0,0); E=(0,12); F=(12,0); G=(6,4); H=(6,8); draw(A--E--B--C--G--H--A--D--F--C); label("A",A,W); label("B",B,NE); label("C",(12.6,4)); label("D",D,SW); label("12",E--B,N); label("12",D--F,S); label("4",E--A,W); label("4",(12.4,-1.75),E); label("8",A--D,W); label("8",(12.4,4),E); label("y",A--H,S); label("y",G--C,N); [/asy]$

As you can see from the diagram, the line segment denoted as $y$ is half the length of the side of the square, which leads to $y = \frac{12}{2} = 6 \Longrightarrow \mathrm{(A)}$.

### Solution 2 (Cheap)

Because the two hexagons are congruent, we know that the perpendicular line to A is half of BC, or $4$. Next, we plug the answer choices in to see which one works. Trying $A$, we get the area of one hexagon is $72$ , as desired, so the answer is $A$ .

~coolmath2017

### Solution 1 (Shortcut)

As solution 1 says, the two hexagons are going to be repositioned to form a square without overlap. Thus we create this square out of the original rectangle. $[asy] size(175); pair A,B,C,D,E,F,G,H; A=(0,8); B=(12,12); C=(12,4); D=(0,0); E=(0,12); F=(12,0); G=(6,4); H=(6,8); draw(A--E--B--C--G--H--A--D--F--C); label("y",A--H,S); label("y",G--C,N); [/asy]$

As you can see from the diagram, the length y fits into the previously blank side, so we know that it is equal to y. $[asy] unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,0)--(12,0)--(12,-4)); label("y",(9,-2),NW); label("A",(0,4),NW); label("B",(18,4),NE); label("C",(18,-4),SE); label("D",(0,-4),SW); label("y",(3,4),S); label("y",(15,-4),N); label("18",(9,4),N); label("18",(9,-4),S); label("8",(0,0),W); label("8",(18,0),E); dot((0,4)); dot((18,4)); dot((18,-4)); dot((0,-4));[/asy]$

From there we can say $3y = 18$ so $y = \frac{18}{3} = 6 \Longrightarrow \mathrm{(A)}$.

~Ezraft

## See also

 2006 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 7 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 AMC 12 Problems and Solutions
 2006 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 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 AMC 10 Problems and Solutions

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