Difference between revisions of "2020 AMC 10A Problems/Problem 16"
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<math>\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7</math> | <math>\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7</math> | ||
− | == Solution == | + | == Solution 1 == |
− | We consider an individual one by one block. | + | We consider an individual one-by-one block. |
− | If we draw a quarter of a circle from each corner, the area covered by the circles should be <math>0.5</math>. Because of this, and the fact that there are four circles, we write | + | If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius <math>d</math>, the area covered by the circles should be <math>0.5</math>. Because of this, and the fact that there are four circles, we write |
− | <cmath>4 * \frac{1}{4} * \pi | + | <cmath>4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}</cmath> |
− | Solving for <math> | + | Solving for <math>d</math>, we obtain <math>d = \frac{1}{\sqrt{2\pi}}</math>, where with <math>\pi \approx 3</math>, we get <math>d = \frac{1}{\sqrt{6}}</math>, and from here, we simplify and see that <math>d \approx 0.4 \implies \boxed{\textbf{(B) } 0.4.}</math> ~Crypthes |
<math>\textbf{Note:}</math> To be more rigorous, note that <math>d<0.5</math> since if <math>d\geq0.5</math> then clearly the probability is greater than <math>\frac{1}{2}</math>. This would make sure the above solution works, as if <math>d\geq0.5</math> there is overlap with the quartercircles. <math>\textbf{- Emathmaster}</math> | <math>\textbf{Note:}</math> To be more rigorous, note that <math>d<0.5</math> since if <math>d\geq0.5</math> then clearly the probability is greater than <math>\frac{1}{2}</math>. This would make sure the above solution works, as if <math>d\geq0.5</math> there is overlap with the quartercircles. <math>\textbf{- Emathmaster}</math> | ||
+ | |||
+ | == Solution 2 == | ||
+ | As in the previous solution, we obtain the equation <math>4 * \frac{1}{4} * \pi d^2 = \frac{1}{2}</math>, which simplifies to <math>\pi d^2 = \frac{1}{2} = 0.5</math>. Since <math>\pi</math> is slightly more than <math>3</math>, <math>d^2</math> is slightly less than <math>\frac{0.5}{3} = 0.1\bar{6}</math>. We notice that <math>0.1\bar{6}</math> is slightly more than <math>0.4^2 = 0.16</math>, so <math>d</math> is roughly <math>\boxed{\textbf{(B) } 0.4}.</math> ~[[User:emerald_block|emerald_block]] | ||
==Video Solution== | ==Video Solution== |
Revision as of 17:37, 1 February 2020
Problem
A point is chosen at random within the square in the coordinate plane whose vertices are and . The probability that the point is within units of a lattice point is . (A point is a lattice point if and are both integers.) What is to the nearest tenth
Solution 1
We consider an individual one-by-one block.
If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius , the area covered by the circles should be . Because of this, and the fact that there are four circles, we write
Solving for , we obtain , where with , we get , and from here, we simplify and see that ~Crypthes
To be more rigorous, note that since if then clearly the probability is greater than . This would make sure the above solution works, as if there is overlap with the quartercircles.
Solution 2
As in the previous solution, we obtain the equation , which simplifies to . Since is slightly more than , is slightly less than . We notice that is slightly more than , so is roughly ~emerald_block
Video Solution
~IceMatrix
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 10 Problems and Solutions |
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