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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>? <!-- [[Image:2006 AMC 12A Problem 6.png]] -->
+
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);
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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 ==
+
== 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>
  
−
[[Image:2006 AMC 12A Problem 6 - Solution.png]]
+
<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 <math>y</math> is half the length of the side of the square, which leads to <math> y = \frac{12}{2} = \boxed{\textbf{(A) }6}</math>.
 +
 
 +
=== 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 <math>y</math> fits into the previously blank side, so we know that it is equal to <math>y</math>.
 +
 
 +
<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} = \boxed{\textbf{(A) }6}</math>.
 +
 
 +
~Ezraft
 +
 
 +
== Solution 3 (Cheap) ==
 +
Because the two hexagons are congruent, we know that the perpendicular line to <math>A</math> is half of <math>BC</math>, 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>\boxed{\textbf{(A) }6}</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</math><math> = \frac{12}{2} = 6 \Longrightarrow \mathrm{(A)}</math>.
+
~coolmath2017
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2006|ab=A|num-b=5|num-a=7}}
 
{{AMC12 box|year=2006|ab=A|num-b=5|num-a=7}}
 
{{AMC10 box|year=2006|ab=A|num-b=6|num-a=8}}
 
{{AMC10 box|year=2006|ab=A|num-b=6|num-a=8}}
 +
{{MAA Notice}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

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

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. AMC logo.png

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