Difference between revisions of "2006 AMC 12A Problems/Problem 6"
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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>? | + | 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>? |
+ | |||
<asy> | <asy> | ||
unitsize(3mm); | unitsize(3mm); | ||
Line 22: | Line 24: | ||
dot((18,-4)); | dot((18,-4)); | ||
dot((0,-4));</asy> | 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> | ||
− | == Solution 1 == | + | == Solutions == |
+ | === Solution 1 === | ||
Since the two [[hexagon]]s are going to be repositioned to form a [[square (geometry) | square]] without overlap, the [[area]] will remain the same. The rectangle's area is <math>18\cdot8=144</math>. This means the square will have four sides of length 12. The only way to do this is shown below.<br> | Since the two [[hexagon]]s are going to be repositioned to form a [[square (geometry) | square]] without overlap, the [[area]] will remain the same. The rectangle's area is <math>18\cdot8=144</math>. This means the square will have four sides of length 12. The only way to do this is shown below.<br> | ||
Line 48: | Line 52: | ||
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 (Cheap) == | + | === Solution 2 (Cheap) === |
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. Trying <math>A</math>, we get the area of one hexagon is <math>72</math> , as desired, so the answer is <math>A</math> . | 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. Trying <math>A</math>, we get the area of one hexagon is <math>72</math> , as desired, so the answer is <math>A</math> . | ||
+ | |||
~coolmath2017 | ~coolmath2017 | ||
+ | |||
+ | === Solution 1 (Shortcut)=== | ||
+ | |||
+ | As solution 1 says, the two [[hexagon]]s are going to be repositioned to form a [[square (geometry) | 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 <math>3y = 18</math> so <math>y = \frac{18}{3} = 6 \Longrightarrow \mathrm{(A)}</math>. | ||
+ | |||
+ | ~Ezraft | ||
== See also == | == See also == |
Latest revision as of 21:00, 26 January 2021
- The following problem is from both the 2006 AMC 12A #6 and 2006 AMC 10A #7, so both problems redirect to this page.
Contents
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 ?
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 . This means the square will have four sides of length 12. The only way to do this is shown below.
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. Trying , we get the area of one hexagon is , as desired, so the answer is .
~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.
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.
From there we can say so .
~Ezraft
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.