2024 AMC 12A Problems/Problem 20
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
. 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
\text{WLOG let } AB=AC=1\
\frac{AP\cdot AQ\cdot sin60}{2}\lt \frac{1\cdot 1\cdot sin60}{4}\
AP\cdot AQ \lt \frac{1}{2}\
\text{Which we can express as for graphing purposes
}\
\text{By graphing it out (someone please insert diagram)}\
\text{We see that the probability is slighty less than
but definitely greater than
}\
\text{Thus answer choice }\fbox{(D)
}\
Note: \text{the actual probability can be found using integration}\
P=\int_{0.5}^{1}{\frac{1}{2x}}dx+0.5=\frac{ln2+1}{2}\sim0.84655 < \frac{7}{8}