Difference between revisions of "2024 AMC 12A Problems/Problem 20"
m (→Solution 2: better wording) |
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The actual probability can be found by using calculus, and taking the integral. | The actual probability can be found by using calculus, and taking the integral. | ||
− | <math>P=\int_{0.5}^{1}{\frac{1}{2x}}dx + 0.5= \frac{\ln2+1}{2}\approx 0. | + | <math>P=\int_{0.5}^{1}{\frac{1}{2x}}dx + 0.5= \frac{\ln2+1}{2}\approx 0.84657</math>. |
Thus, since <math>\frac 3 4 < 0.84655 < \frac 7 8</math>, we get <math>\boxed{\text{(D)} \left(\frac{3}{4},\frac{7}{8} \right]}</math> | Thus, since <math>\frac 3 4 < 0.84655 < \frac 7 8</math>, we get <math>\boxed{\text{(D)} \left(\frac{3}{4},\frac{7}{8} \right]}</math> |
Revision as of 16:13, 9 November 2024
Problem
Points and
are chosen uniformly and independently at random on sides
and
respectively, of equilateral triangle
Which of the following intervals contains the probability that the area of
is less than half the area of
Solution 1
Let and
. Applying the sine formula for a triangle's area, we see that
Without loss of generality, we let , and thus
; we therefore require
for
. (Note: You can skip most of this by using triangle area ratios ~A_MatheMagician.)
A quick rough sketch of on the square given by
reveals that the curve intersects the boundaries at
and
, and it is actually quite (very) obvious that the area bounded by the inequality
and the aforementioned unit square is more than
but less than
(cf. the diagram below). Thus, our answer is
.
~Technodoggo
Solution 2
WLOG let
Which we can express as
.
We follow by graphing the equation on
, and we follow the same way solution 1 does.
We see that the probability is slightly less than but definitely greater than
Thus answer choice
Solution 3
The actual probability can be found by using calculus, and taking the integral.
.
Thus, since , we get
~lptoggled
See also
2024 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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.