Difference between revisions of "2021 AMC 10B Problems/Problem 23"
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<b> Near The Center Square </b> | <b> Near The Center Square </b> | ||
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We can have the center of the coin land within <math>\frac{1}{2}</math> of the center square, or inside of the center square. We have that the center lands either outside of the square, or inside. So, we have a region with <math>\frac{1}{2}</math> emanating from every point on the exterior of the square, forming <math>4</math> quarter circles and <math>4</math> rectangles. <img> https://latex.artofproblemsolving.com/3/6/d/36ddfcdad3ad425c1a9c62d2380136463179dff1.png </img> The <math>4</math> quarter circles combine to make a full circle, with radius of <math>\frac{1}{2}</math>, so that has an area of <math>\frac{\pi}{4}</math>. The area of a rectangle is <math>2 \sqrt 2 \cdot \frac{1}{2} = \sqrt 2</math>, so <math>4</math> of them combine to an area of <math>4 \sqrt 2</math>. The area of the black square is simply <math>(2\sqrt 2)^2 = 8</math>. So, for this case, we have a combined total of <math>8 + 4\sqrt 2 + \frac{\pi}{4}</math>. Onto the second (and last) case. | We can have the center of the coin land within <math>\frac{1}{2}</math> of the center square, or inside of the center square. We have that the center lands either outside of the square, or inside. So, we have a region with <math>\frac{1}{2}</math> emanating from every point on the exterior of the square, forming <math>4</math> quarter circles and <math>4</math> rectangles. <img> https://latex.artofproblemsolving.com/3/6/d/36ddfcdad3ad425c1a9c62d2380136463179dff1.png </img> The <math>4</math> quarter circles combine to make a full circle, with radius of <math>\frac{1}{2}</math>, so that has an area of <math>\frac{\pi}{4}</math>. The area of a rectangle is <math>2 \sqrt 2 \cdot \frac{1}{2} = \sqrt 2</math>, so <math>4</math> of them combine to an area of <math>4 \sqrt 2</math>. The area of the black square is simply <math>(2\sqrt 2)^2 = 8</math>. So, for this case, we have a combined total of <math>8 + 4\sqrt 2 + \frac{\pi}{4}</math>. Onto the second (and last) case. | ||
<b> Near A Triangle </b> | <b> Near A Triangle </b> | ||
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We can also have the coin land within <math>\frac{1}{2}</math> of one of the triangles. By symmetry, we can just find the successful region for one of them, then multiply by <math>4</math>. Consider this diagram. <img> https://latex.artofproblemsolving.com/e/7/1/e7194733f1299176a89a56fb28ba3cf247e736c5.png </img> We can draw in an altitude from the bottom corner of the square to hit the hypotenuse of the blue triangle. The length of this when passing through the black region is <math>\sqrt 2</math>, and when passing through the white region (while being contained in the blue triangle) is <math>\frac{1}{2}</math>. However, we have to subtract off when it doesn't pass through the red square. Then, it's the hypotenuse of a small isosceles right triangle with side lengths of <math>\frac{1}{2}</math>, or <math>\frac{\sqrt 2}{2}</math>. So, our altitude of the blue triangle is <math>\sqrt 2 + \frac{1}{2} - \frac{\sqrt 2}{2} = \frac{\sqrt 2 + 1}{2}</math>. Then, recall, the area of an isosceles right triangle is <math>h^2</math>, where <math>h</math> is the altitude from the right angle. So, squaring this, we get <math>\frac{3 + 2\sqrt 2}{4}</math>. Now, we have to multiply this by <math>4</math> to account for all of the black triangles, to get <math>3 + 2\sqrt 2</math> as the final area for this case. | We can also have the coin land within <math>\frac{1}{2}</math> of one of the triangles. By symmetry, we can just find the successful region for one of them, then multiply by <math>4</math>. Consider this diagram. <img> https://latex.artofproblemsolving.com/e/7/1/e7194733f1299176a89a56fb28ba3cf247e736c5.png </img> We can draw in an altitude from the bottom corner of the square to hit the hypotenuse of the blue triangle. The length of this when passing through the black region is <math>\sqrt 2</math>, and when passing through the white region (while being contained in the blue triangle) is <math>\frac{1}{2}</math>. However, we have to subtract off when it doesn't pass through the red square. Then, it's the hypotenuse of a small isosceles right triangle with side lengths of <math>\frac{1}{2}</math>, or <math>\frac{\sqrt 2}{2}</math>. So, our altitude of the blue triangle is <math>\sqrt 2 + \frac{1}{2} - \frac{\sqrt 2}{2} = \frac{\sqrt 2 + 1}{2}</math>. Then, recall, the area of an isosceles right triangle is <math>h^2</math>, where <math>h</math> is the altitude from the right angle. So, squaring this, we get <math>\frac{3 + 2\sqrt 2}{4}</math>. Now, we have to multiply this by <math>4</math> to account for all of the black triangles, to get <math>3 + 2\sqrt 2</math> as the final area for this case. | ||
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<b> Finishing </b> | <b> Finishing </b> | ||
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Then, to have the coin touching a black region, we add up the area of our successful regions, or <math>8 + 4\sqrt 2 + \frac{\pi}{4} + 3 + 2\sqrt 2 = 11 + 6\sqrt 2 + \frac{\pi}{4} = \frac{44 + 24\sqrt 2 + \pi}{4}</math>. The total region is <math>49</math>, so our probability is <math>\frac{\frac{44 + 24\sqrt 2 + \pi}{4}}{49} = \frac{44 + 24\sqrt 2 + \pi}{196}</math>, which implies <math>a+b = 44+24 = 68</math>. This corresponds to answer choice <math>\boxed{\textbf{(A)} ~68}</math>. ~rocketsri (diagrams from the MathJam) | Then, to have the coin touching a black region, we add up the area of our successful regions, or <math>8 + 4\sqrt 2 + \frac{\pi}{4} + 3 + 2\sqrt 2 = 11 + 6\sqrt 2 + \frac{\pi}{4} = \frac{44 + 24\sqrt 2 + \pi}{4}</math>. The total region is <math>49</math>, so our probability is <math>\frac{\frac{44 + 24\sqrt 2 + \pi}{4}}{49} = \frac{44 + 24\sqrt 2 + \pi}{196}</math>, which implies <math>a+b = 44+24 = 68</math>. This corresponds to answer choice <math>\boxed{\textbf{(A)} ~68}</math>. ~rocketsri (diagrams from the MathJam) | ||
Revision as of 14:06, 13 February 2021
Problem
A square with side length is colored white except for black isosceles right triangular regions with legs of length in each corner of the square and a black diamond with side length in the center of the square, as shown in the diagram. A circular coin with diameter is dropped onto the square and lands in a random location where the coin is completely contained within the square. The probability that the coin will cover part of the black region of the square can be written as , where and are positive integers. What is ?
Solution
To find the probability, we look at the . For the coin to be completely contained within the square, we must have the distance from the center of the coin to a side of the square to be at least , as it's the radius of the coin. This implies the is a square with side length , with an area of . Now, we consider cases on where needs to land to partially cover a black region.
Near The Center Square
We can have the center of the coin land within of the center square, or inside of the center square. We have that the center lands either outside of the square, or inside. So, we have a region with emanating from every point on the exterior of the square, forming quarter circles and rectangles. <img> https://latex.artofproblemsolving.com/3/6/d/36ddfcdad3ad425c1a9c62d2380136463179dff1.png </img> The quarter circles combine to make a full circle, with radius of , so that has an area of . The area of a rectangle is , so of them combine to an area of . The area of the black square is simply . So, for this case, we have a combined total of . Onto the second (and last) case.
Near A Triangle
We can also have the coin land within of one of the triangles. By symmetry, we can just find the successful region for one of them, then multiply by . Consider this diagram. <img> https://latex.artofproblemsolving.com/e/7/1/e7194733f1299176a89a56fb28ba3cf247e736c5.png </img> We can draw in an altitude from the bottom corner of the square to hit the hypotenuse of the blue triangle. The length of this when passing through the black region is , and when passing through the white region (while being contained in the blue triangle) is . However, we have to subtract off when it doesn't pass through the red square. Then, it's the hypotenuse of a small isosceles right triangle with side lengths of , or . So, our altitude of the blue triangle is . Then, recall, the area of an isosceles right triangle is , where is the altitude from the right angle. So, squaring this, we get . Now, we have to multiply this by to account for all of the black triangles, to get as the final area for this case.
Finishing
Then, to have the coin touching a black region, we add up the area of our successful regions, or . The total region is , so our probability is , which implies . This corresponds to answer choice . ~rocketsri (diagrams from the MathJam)
Video Solution by OmegaLearn (Similar Triangles and Area Calculations)
~ pi_is_3.14
2021 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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All AMC 10 Problems and Solutions |