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

(Created page with "Line <math>\ell</math> in the coordinate plane has the equation <math>3x - 5y + 40 = 0</math>. This line is rotated <math>45^{\circ}</math> counterclockwise about the point <m...")
 
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Line <math>\ell</math> in the coordinate plane has the equation <math>3x - 5y + 40 = 0</math>. This line is rotated <math>45^{\circ}</math> counterclockwise about the point <math>(20, 20)</math> to obtain line <math>k</math>. What is the <math>x</math>-coordinate of the <math>x</math>-intercept of line <math>k?</math>
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==Problem 6==
  
<math>\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30</math>
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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?
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<asy>
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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((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7));
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filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7));
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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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}
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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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}
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</asy>
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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>

Revision as of 15:49, 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]

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