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

(Solution 2 (Cheap))
(Solution 2 (Cheap))
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== 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
  
 
== See also ==
 
== See also ==

Revision as of 19:37, 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 $8\times18$ 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 $y = \frac{12}{2} = 6 \Longrightarrow \mathrm{(A)}$.

Solution 2 (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 $A$ . ~coolmath2017

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

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