Difference between revisions of "2020 AMC 12A Problems/Problem 6"
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<math>\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8</math> | <math>\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8</math> | ||
| − | == Solution == | + | == 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: | The two lines of symmetry must be horizontally and vertically through the middle. We can then fill the boxes in like so: | ||
| Line 65: | Line 65: | ||
where the light gray boxes are the ones we have filled. Counting these, we get <math>\boxed{\textbf{(D) } 7}</math> total boxes. ~ciceronii | where the light gray boxes are the ones we have filled. Counting these, we get <math>\boxed{\textbf{(D) } 7}</math> total boxes. ~ciceronii | ||
| + | |||
| + | ==Solution 2 (Analytical)== | ||
| + | We label the three shaded unit squares <math>A,B,</math> and <math>C,</math> 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> | ||
| + | |||
| + | <b>SOLUTION IN PROGRESS, I WILL RETURN</b> | ||
| + | |||
| + | ~MRENTHUSIASM | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 06:38, 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/0/9/8/098f8098c7103b004757d1977d2bd6d459831eb3.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)
SOLUTION IN PROGRESS, I WILL RETURN
~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. 
