Difference between revisions of "2020 AMC 10A Problems/Problem 16"
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-Solution by Joeya | -Solution by Joeya | ||
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+ | === Solution 6 (Estimating but differently again, again)=== | ||
+ | As above, we have the equation <math>\pi d^2 = \frac{1}{2}, and we want to find the most accurate value of </math>d<math>. We resort to the answer choices and can plug those values of</math>d<math>in and see which value of </math>d<math> will lead to the most accurate value of </math>pi<math>. | ||
+ | |||
+ | Starting off in the middle, we try option C with </math>d=0.5<math>. Plugging this in, we get </math>\pi (\frac{1}{2})^2 = \frac{1}{2},<math> and after simplifying we get </math>\pi = \frac{1}{2} \cdot 4 = 2.<math> That's not very good. We know </math>\pi \approx 3.14.<math> Let's see if we can do better.Trying option A with</math>d = 0.3,<math> we get </math>\pi = \frac{1}{2} \cdot {100/9} = \frac{50}{9} = 5 \frac{5}{9}.<math> Hm, let's try option B with </math>d = 0.4.<math> We get </math>\pi = \frac{1}{2} \cdot \frac{25}{4} = \frac{25}{8} = 3 \frac{1}{8}<math>. This is very close to </math>\pi<math> and is the best value of the 5 options. Therefore, the answer is </math>\boxed{\textbf{(B) } 0.4}.$ ~ epiconan | ||
== Solution 6 (Sol. 1, but rigorous (and excessive)) == | == Solution 6 (Sol. 1, but rigorous (and excessive)) == |
Revision as of 01:05, 11 November 2023
- The following problem is from both the 2020 AMC 12A #16 and 2020 AMC 10A #16, so both problems redirect to this page.
Contents
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
Solutions
Diagram
The diagram represents each unit square of the given square.
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 see that
~Crypthes
~ Minor Edits by BakedPotato66
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
Solution 3 (Estimating)
As above, we find that we need to estimate .
Note that we can approximate and so
.
And so our answer is .
~Silverdragon
Solution 4 (Estimating but a bit different)
We only need to figure out the probability for a unit square, as it will scale up to the square. Since we want to find the probability that a point inside a unit square that is
units away from a lattice point (a corner of the square) is
, we can find which answer will come the closest to covering
of the area.
Since the closest is which turns out to be
which is about
, we find that the answer rounded to the nearest tenth is
or
.
~RuiyangWu
Solution 5 (Estimating but differently again)
As per the above diagram, realize that , so
.
.
is between
and
and
, so we can say
.
So . This is slightly above
, since
.
-Solution by Joeya
Solution 6 (Estimating but differently again, again)
As above, we have the equation d
d
d
pi$.
Starting off in the middle, we try option C with$ (Error compiling LaTeX. Unknown error_msg)d=0.5\pi (\frac{1}{2})^2 = \frac{1}{2},
\pi = \frac{1}{2} \cdot 4 = 2.
\pi \approx 3.14.
d = 0.3,
\pi = \frac{1}{2} \cdot {100/9} = \frac{50}{9} = 5 \frac{5}{9}.
d = 0.4.
\pi = \frac{1}{2} \cdot \frac{25}{4} = \frac{25}{8} = 3 \frac{1}{8}
\pi
\boxed{\textbf{(B) } 0.4}.$ ~ epiconan
Solution 6 (Sol. 1, but rigorous (and excessive))
Let be the side length of a square. When
, the shaded areas represent half of the total area:
When :
For :
We can calculate the total number of shaded circles given some . There are
full circles on the inside,
semicircles on the sides, and
quarter circles for the corners.
Full circles are, of course, worth one circle. Semicircles are worth half a circle each, and quarter circles are worth of a circle. Thus, weighing our sum gives
Thus, there is
worth of the shaded area for any
, and since the area of each circle is
if
is the radius of each.
We want the ratio of this shaded area to the entire to be . The area of the entire square is
, so dividing, we see that
.
The rest is the same as solution .
Video Solutions
Video Solution 1
Education, The Study of Everything
Video Solution 2
~IceMatrix
Video Solution 3
https://youtu.be/R220vbM_my8?t=238
~ amritvignesh0719062.0
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 |
2020 AMC 12A (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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.