Difference between revisions of "2006 AMC 12A Problems/Problem 6"
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As you can see from the diagram, the [[line segment]] denoted as <math>y</math> is half the length of the side of the square, which leads to <math> y = \frac{12}{2} = 6 \Longrightarrow \mathrm{(A)}</math>. | As you can see from the diagram, the [[line segment]] denoted as <math>y</math> is half the length of the side of the square, which leads to <math> y = \frac{12}{2} = 6 \Longrightarrow \mathrm{(A)}</math>. | ||
| − | == Solution 2== | + | == Solution 2 (Cheap) == |
| − | Because the two hexagons are congruent, we know that | + | Because the two hexagons are congruent, we know that the perpendicular line to A is half of BC, or <math>4</math>. Next, we plug the answer choices in to see which one works. |
== See also == | == See also == | ||
Revision as of 18:35, 3 May 2020
- 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 
rectangle 
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 
?
![[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]](http://latex.artofproblemsolving.com/6/6/a/66a524a2244d311dbe57cc4d9aaffc1b5e83bb03.png)
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 
. 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]](http://latex.artofproblemsolving.com/1/7/9/1794d4dcd643f680a8bcabca8f5ff34ed2b54a70.png)
As you can see from the diagram, the line segment denoted as is half the length of the side of the square, which leads to 
.
Solution 2 (Cheap)
Because the two hexagons are congruent, we know that the perpendicular line to A is half of BC, or 
. Next, we plug the answer choices in to see which one works.
See also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
