Difference between revisions of "2006 AMC 12A Problems/Problem 6"

Latest revision as of 17:29, 7 May 2024

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]$

$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10$

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 $y = \frac{12}{2} = \boxed{\textbf{(A) }6}$ .

Solution 2 (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 $y = \frac{18}{3} = \boxed{\textbf{(A) }6}$ .

~Ezraft

Solution 3 (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 $\boxed{\textbf{(A) }6}$ .

~coolmath2017

2006 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 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 by Problem 6	Followed by Problem 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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