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Difference between revisions of "2020 AMC 12A Problems/Problem 6"

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m (→‎Solution 2 (Analytical): "in additional" -> "in addition")
 
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</asy>
 
</asy>
  
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where the light gray boxes are the ones we have filled. Counting these, we get <math>\boxed{\textbf{(D) } 7}</math> total boxes. ~ciceronii
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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)==
 
==Solution 2 (Analytical)==
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}
 
}
 
}
 
}
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draw((-1,2)--(6,2),dashed+linewidth(2));
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draw((-1,2)--(6,2),linewidth(2)+red);
−
draw((2.5,-1)--(2.5,5),dashed+linewidth(2));
+
draw((2.5,-1)--(2.5,5),linewidth(2)+red);
 
label("$A$",(1.5,3.5));
 
label("$A$",(1.5,3.5));
 
label("$B$",(2.5,1.5));
 
label("$B$",(2.5,1.5));
 
label("$C$",(4.5,0.5));
 
label("$C$",(4.5,0.5));
 
</asy>
 
</asy>
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Since <math>A</math> and <math>C</math> are not on either line of symmetry, they each contribute <math>4</math> shaded unit squares in the resulting figure; since <math>B</math> is on one line of symmetry, it contributes <math>2</math> shaded unit squares in the resulting figure. We need to shade at least <math>4+4+2-3=\boxed{\textbf{(D) } 7}</math> unit squares in additional, as shown below:
+
Note that:
 +
<ol style="margin-left: 1.5em;">
 +
  <li>Since the centers of <math>A</math> and <math>C</math> are on neither line of symmetry, <math>A</math> and <math>C</math> each contribute <math>4</math> shaded unit squares to the resulting figure.</li><p>
 +
  <li>Since the center of <math>B</math> is on one line of symmetry, <math>B</math> contributes <math>2</math> shaded unit squares to the resulting figure.</li><p>
 +
</ol>
 +
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 addition, as shown below:
 
<asy>
 
<asy>
 
import olympiad;
 
import olympiad;
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}
 
}
 
}
 
}
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draw((-1,2)--(6,2),dashed+linewidth(2));
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draw((-1,2)--(6,2),linewidth(2)+red);
−
draw((2.5,-1)--(2.5,5),dashed+linewidth(2));
+
draw((2.5,-1)--(2.5,5),linewidth(2)+red);
 
label("$A$",(1.5,3.5));
 
label("$A$",(1.5,3.5));
 
label("$B$",(2.5,1.5));
 
label("$B$",(2.5,1.5));

Latest revision as of 05:14, 10 September 2021

Problem

In the plane figure shown below, $3$ 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]

$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

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]

where the light gray boxes are the ones we have filled. Counting these, we get $\boxed{\textbf{(D) } 7}$ total boxes.

~ciceronii

Solution 2 (Analytical)

We label the three shaded unit squares $A,B,$ and $C,$ 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),linewidth(2)+red); draw((2.5,-1)--(2.5,5),linewidth(2)+red); label("$A$",(1.5,3.5)); label("$B$",(2.5,1.5)); label("$C$",(4.5,0.5)); [/asy] Note that:

  1. Since the centers of $A$ and $C$ are on neither line of symmetry, $A$ and $C$ each contribute $4$ shaded unit squares to the resulting figure.

  2. Since the center of $B$ is on one line of symmetry, $B$ contributes $2$ shaded unit squares to the resulting figure.

The shaded unit squares contributed by $A,B,$ and $C$ are all distinct, so we need to shade at least $4+4+2-3=\boxed{\textbf{(D) } 7}$ unit squares in addition, 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),linewidth(2)+red); draw((2.5,-1)--(2.5,5),linewidth(2)+red); 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] ~MRENTHUSIASM

Video Solution

https://youtu.be/fzZzGqNqW6U

~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

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