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

m (Minor typo)
m (Problem: replace with <asy>)
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{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #6]] and [[2006 AMC 10A Problems/Problem 7|2006 AMC 10A #7]]}}
 
{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #6]] and [[2006 AMC 10A Problems/Problem 7|2006 AMC 10A #7]]}}
 
== Problem ==
 
== Problem ==
The <math>8\times18</math> [[rectangle]] <math>ABCD</math> 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 <math>y</math>?
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The <math>8\times18</math> [[rectangle]] <math>ABCD</math> 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 <math>y</math>? <!-- [[Image:2006 AMC 12A Problem 6.png]] -->
 
+
<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));
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label("$A$",(0,4),NW);
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label("$B$",(18,4),NE);
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label("$C$",(18,-4),SE);
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label("$D$",(0,-4),SW);
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label("$y$",(3,4),S);
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label("$y$",(15,-4),N);
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label("$18$",(9,4),N);
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label("$18$",(9,-4),S);
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label("$8$",(0,0),W);
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label("$8$",(18,0),E);
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dot((0,4));
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dot((18,4));
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dot((18,-4));
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dot((0,-4));</asy>
 
<math>\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10</math>
 
<math>\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10</math>
 
[[Image:2006 AMC 12A Problem 6.png]]
 
  
 
== Solution ==
 
== Solution ==

Revision as of 22:57, 25 August 2011

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

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.

2006 AMC 12A Problem 6 - Solution.png

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)}$.

See also

2006 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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