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Difference between revisions of "2020 AMC 12A Problems/Problem 6"
(→‎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 ==
 +
 +
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;
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unitsize(25);
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filldraw((1,3)--(1,4)--(2,4)--(2,3)--cycle, gray(0.7));
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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));
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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));
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filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray(0.9));
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for (int i = 0; i < 5; ++i) {
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for (int j = 0; j < 6; ++j) {
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pair A = (j,i);
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}
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}
 +
for (int i = 0; i < 5; ++i) {
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for (int j = 0; j < 6; ++j) {
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if (j != 5) {
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draw((j,i)--(j+1,i));
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}
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if (i != 4) {
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draw((j,i)--(j,i+1));
 +
}
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}
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}
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</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

Revision as of 14:58, 1 February 2020

Problem 6

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

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

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