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Difference between revisions of "2020 AMC 12A Problems/Problem 6"
m (→‎Solution 2 (Analytical): "in additional" -> "in addition")
 
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==Problem 6==
+
==Problem==
  
 
In the plane figure shown below, <math>3</math> 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?
 
In the plane figure shown below, <math>3</math> 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?
Line 12: Line 12:
 
for (int j = 0; j < 6; ++j) {
 
for (int j = 0; j < 6; ++j) {
 
pair A = (j,i);
 
pair A = (j,i);
−
 
 
}
 
}
 
}
 
}
Line 29: Line 28:
 
<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 64: Line 63:
 
</asy>
 
</asy>
  
−
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),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:
 +
<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>
 +
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==
 
==See Also==
  
−
{{AMC10 box|year=2020|ab=A|num-b=9|num-a=11}}
+
{{AMC12 box|year=2020|ab=A|num-b=5|num-a=7}}
−
{{AMC12 box|year=2020|ab=A|num-b=6|num-a=8}}
 
 
{{MAA Notice}}
 
{{MAA Notice}}

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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