Difference between revisions of "2020 AMC 12A Problems/Problem 6"
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label("$C$",(4.5,0.5)); | label("$C$",(4.5,0.5)); | ||
</asy> | </asy> | ||
| − | Since <math>A</math> and <math>C</math> are not on either line of symmetry, they each contribute <math>4</math> shaded unit squares to the resulting figure; since <math>B</math> is on one line of symmetry, it contributes <math>2</math> shaded unit squares to the resulting figure. | + | Since <math>A</math> and <math>C</math> are not on either line of symmetry, they each contribute <math>4</math> shaded unit squares to the resulting figure; since <math>B</math> is on one line of symmetry, it contributes <math>2</math> shaded unit squares to the resulting figure. The shaded unit squares contributed by <math>A,B,</math> and <math>C</math> are all distinct, so we need to shade at least <math>4+4+2-3=\boxed{\textbf{(D) } 7}</math> unit squares in additional, as shown below: |
<asy> | <asy> | ||
import olympiad; | import olympiad; | ||
Revision as of 18:38, 26 April 2021
Problem
In the plane figure shown below, 
of the unit squares have been shaded. What is the least number of additional unit squares that must be shaded so that the resulting figure has two lines of symmetry?
![[asy] import olympiad; unitsize(25); filldraw((1,3)--(1,4)--(2,4)--(2,3)--cycle, gray(0.7)); filldraw((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7)); filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7)); for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { pair A = (j,i); } } for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { if (j != 5) { draw((j,i)--(j+1,i)); } if (i != 4) { draw((j,i)--(j,i+1)); } } } [/asy]](http://latex.artofproblemsolving.com/b/f/0/bf0987ebf1bcdda763449e461f9bc1e7a648be9a.png)
Solution 1 (Graphical)
The two lines of symmetry must be horizontally and vertically through the middle. We can then fill the boxes in like so:
![[asy] import olympiad; unitsize(25); filldraw((1,3)--(1,4)--(2,4)--(2,3)--cycle, gray(0.7)); filldraw((0,3)--(0,4)--(1,4)--(1,3)--cycle, gray(0.9)); filldraw((3,3)--(3,4)--(4,4)--(4,3)--cycle, gray(0.9)); filldraw((4,3)--(4,4)--(5,4)--(5,3)--cycle, gray(0.9)); filldraw((2,2)--(2,3)--(3,3)--(3,2)--cycle, gray(0.9)); filldraw((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7)); filldraw((3,0)--(4,0)--(4,1)--(3,1)--cycle, gray(0.9)); filldraw((1,0)--(2,0)--(2,1)--(1,1)--cycle, gray(0.9)); filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7)); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray(0.9)); for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { pair A = (j,i); } } for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { if (j != 5) { draw((j,i)--(j+1,i)); } if (i != 4) { draw((j,i)--(j,i+1)); } } } [/asy]](http://latex.artofproblemsolving.com/e/a/e/eaee082a58a8eca3546d8932ab326a3c7d0eb780.png)
where the light gray boxes are the ones we have filled. Counting these, we get 
total boxes. ~ciceronii
Solution 2 (Analytical)
We label the three shaded unit squares 
and 
then construct the two lines of symmetry of the resulting figure, as shown below:
![[asy] import olympiad; unitsize(25); filldraw((1,3)--(1,4)--(2,4)--(2,3)--cycle, gray(0.7)); filldraw((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7)); filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7)); for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { pair A = (j,i); } } for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { if (j != 5) { draw((j,i)--(j+1,i)); } if (i != 4) { draw((j,i)--(j,i+1)); } } } draw((-1,2)--(6,2),dashed+linewidth(2)); draw((2.5,-1)--(2.5,5),dashed+linewidth(2)); label("$A$",(1.5,3.5)); label("$B$",(2.5,1.5)); label("$C$",(4.5,0.5)); [/asy]](http://latex.artofproblemsolving.com/4/5/7/4572ee018da607881bd77f0cd373454942e58a0c.png)
Since 
and 
are not on either line of symmetry, they each contribute 
shaded unit squares to the resulting figure; since 
is on one line of symmetry, it contributes 
shaded unit squares to the resulting figure. The shaded unit squares contributed by and are all distinct, so we need to shade at least 
unit squares in additional, as shown below:
![[asy] import olympiad; unitsize(25); filldraw((1,3)--(1,4)--(2,4)--(2,3)--cycle, gray(0.7)); filldraw((0,3)--(0,4)--(1,4)--(1,3)--cycle, gray(0.9)); filldraw((3,3)--(3,4)--(4,4)--(4,3)--cycle, gray(0.9)); filldraw((4,3)--(4,4)--(5,4)--(5,3)--cycle, gray(0.9)); filldraw((2,2)--(2,3)--(3,3)--(3,2)--cycle, gray(0.9)); filldraw((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7)); filldraw((3,0)--(4,0)--(4,1)--(3,1)--cycle, gray(0.9)); filldraw((1,0)--(2,0)--(2,1)--(1,1)--cycle, gray(0.9)); filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7)); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray(0.9)); for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { pair A = (j,i); } } for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { if (j != 5) { draw((j,i)--(j+1,i)); } if (i != 4) { draw((j,i)--(j,i+1)); } } } draw((-1,2)--(6,2),dashed+linewidth(2)); draw((2.5,-1)--(2.5,5),dashed+linewidth(2)); label("$A$",(1.5,3.5)); label("$B$",(2.5,1.5)); label("$C$",(4.5,0.5)); label("$A'$",(3.5,3.5)); label("$A'$",(1.5,0.5)); label("$A'$",(3.5,0.5)); label("$B'$",(2.5,2.5)); label("$C'$",(0.5,0.5)); label("$C'$",(0.5,3.5)); label("$C'$",(4.5,3.5)); [/asy]](http://latex.artofproblemsolving.com/7/4/d/74d5177d76a61b7c75d59e99a92961bb63eec649.png)
~MRENTHUSIASM
Video Solution
~IceMatrix
See Also
| 2020 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
