Difference between revisions of "2024 AMC 12A Problems/Problem 20"
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draw((0.5,1)--(1,0.5),red+dashed+1.1); | draw((0.5,1)--(1,0.5),red+dashed+1.1); | ||
</asy> | </asy> | ||
− | + | == Solution 2 == | |
+ | WLOG let <math>AB=AC=1</math> | ||
+ | <cmath>\frac{AP\cdot AQ\cdot sin60}{2}\lt \frac{1\cdot 1\cdot sin60}{4}</cmath> | ||
+ | <cmath>AP\cdot AQ \lt \frac{1}{2}</cmath> | ||
+ | Which we can express as <math>xy<\frac{1}{2}</math> for graphing purposes <math>(x,y<1)</math>} | ||
+ | By graphing it out (someone please insert diagram)} | ||
+ | We see that the probability is slighty less than <math>\frac{7}{8}</math> but definitely greater than <math>\frac{3}{4}</math> | ||
+ | Thus answer choice <math>\fbox{(D) \left(\frac{3}{4},\frac{7}{8} \right]</math> | ||
+ | Note: the actual probability can be found using integration | ||
+ | <cmath>P=\int_{0.5}^{1}{\frac{1}{2x}}dx+0.5=\frac{ln2+1}{2}\sim0.84655 < \frac{7}{8}</cmath> | ||
==See also== | ==See also== | ||
{{AMC12 box|year=2024|ab=A|num-b=19|num-a=21}} | {{AMC12 box|year=2024|ab=A|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:21, 8 November 2024
Contents
[hide]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 . 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
\[\frac{AP\cdot AQ\cdot sin60}{2}\lt \frac{1\cdot 1\cdot sin60}{4}\] (Error compiling LaTeX. Unknown error_msg)
\[AP\cdot AQ \lt \frac{1}{2}\] (Error compiling LaTeX. Unknown error_msg)
Which we can express as for graphing purposes } By graphing it out (someone please insert diagram)} We see that the probability is slighty less than but definitely greater than Thus answer choice $\fbox{(D) \left(\frac{3}{4},\frac{7}{8} \right]$ (Error compiling LaTeX. Unknown error_msg) Note: the actual probability can be found using integration
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.