Difference between revisions of "2020 AMC 12A Problems/Problem 6"
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− | ==Problem | + | ==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== | ||
+ | |||
+ | {{AMC12 box|year=2020|ab=A|num-b=5|num-a=7}} | ||
+ | {{MAA Notice}} |
Latest revision as of 06:14, 10 September 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?
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:
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:
Note that:
- Since the centers of
and
are on neither line of symmetry,
and
each contribute
shaded unit squares to the resulting figure.
- Since the center of
is on one line of symmetry,
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 addition, as shown below:
~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.