2024 AMC 12A Problems/Problem 20
Contents
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 
![$\textbf{(A) } \left[\frac 38, \frac 12\right] \qquad \textbf{(B) } \left(\frac 12, \frac 23\right] \qquad \textbf{(C) } \left(\frac 23, \frac 34\right] \qquad \textbf{(D) } \left(\frac 34, \frac 78\right] \qquad \textbf{(E) } \left(\frac 78, 1\right]$](http://latex.artofproblemsolving.com/5/8/7/5874157b23f9d385f707b696eee3d0780fd8a45b.png)
Solution 1
Let 
and 
. Applying the sine formula for a triangle's area, we see that
![\[[\Delta APQ]=\dfrac12\cdot x\cdot y\sin(\angle PAQ)=\dfrac{xy}2\sin(60^\circ)=\dfrac{\sqrt3}4xy.\]](http://latex.artofproblemsolving.com/3/c/3/3c3a2afdf31468099408b2729602a6909052e735.png)
Without loss of generality, we let 
, and thus ![$[\Delta ABC]=\dfrac{\sqrt3}4$](http://latex.artofproblemsolving.com/8/6/8/868c7ae6efb0c5f98e6decb90e97ed477048d9a1.png)
; we therefore require 
for 
. A quick rough sketch of 
on the square given by ![$x,y\in[0,1]$](http://latex.artofproblemsolving.com/3/8/6/386ca6d344928391e23697124e9334316f2af0c5.png)
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 ![$\boxed{\textbf{(D) }\left(\dfrac34,\dfrac78\right]}$](http://latex.artofproblemsolving.com/f/3/c/f3c0a390f698502ff5c5df095e8b12a24f51aecf.png)
.
~Technodoggo
![[asy] /*Asymptote visual by Technodoggo, 7 November 2024*/ unitsize(8cm); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); label("$0$",(-0.05,-0.05)); label("$1$",(1,-0.05),S); label("$1$",(-0.05,1),W); draw((-0.05,0)--(1,0)--(1,-0.05)); draw((0,-0.05)--(0,1)--(-0.05,1)); real f(real x) {return 1/(2*x);} path c = graph(f, 0.5,1)--(1,0)--(0,0)--(0,1)--cycle; filldraw(c,blue+white); draw((0.5,1)--(0.5,0.5)--(1,0.5),white+dashed+1.1); draw((0.5,1)--(1,0.5),red+dashed+1.1); [/asy]](http://latex.artofproblemsolving.com/f/6/5/f65f960537043175be10e7f89059569eaedb59a8.png)
Solution 2
WLOG let 
![\[\frac{AP \cdot AQ \cdot sin60}{2} < \frac{1 \cdot 1 \cdot sin60}{4}\]](http://latex.artofproblemsolving.com/a/b/d/abdee800e5fb68d13f90dd8ef6f7520d15a75470.png)
![\[AP \cdot AQ < \frac{1}{2}\]](http://latex.artofproblemsolving.com/3/f/9/3f9e82c940f6901ecc782e084c892771ed1945f1.png)
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 ![$\boxed{(D) \left(\frac{3}{4},\frac{7}{8} \right]}$](http://latex.artofproblemsolving.com/b/4/1/b41e94cfab232273f0d01cb6387398cf25155e64.png)
Note: the actual probability can be found using integration
![\[P=\int_{0.5}^{1}{\frac{1}{2x}}dx+0.5= \frac{ln2+1}{2}\sim0.84655 < \frac{7}{8}\]](http://latex.artofproblemsolving.com/a/1/9/a199d783ffb4b4d21a789a609e97a765aaa4ef19.png)
~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 | |
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