Difference between revisions of "2024 AMC 12A Problems/Problem 20"
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==Solution 1== | ==Solution 1== | ||
| − | Let <math>\overline{AP}=x</math> and <math>\overline{AQ}=y</math>. Applying the sine formula for a triangle's area, we see that < | + | Let <math>\overline{AP}=x</math> and <math>\overline{AQ}=y</math>. Applying the sine formula for a triangle's area, we see that |
| + | <cmath>[\Delta APQ]=\dfrac12\cdot x\cdot y\sin(\angle PAQ)=\dfrac{xy}2\sin(60^\circ)=\dfrac{\sqrt3}4xy.</cmath> | ||
| + | |||
| + | Without loss of generality, we let <math>AB=BC=CA=1</math>, and thus <math>[\Delta ABC]=\dfrac{\sqrt3}4</math>; we therefore require <math>\dfrac{\sqrt3}4xy\le\dfrac12\cdot\dfrac{\sqrt3}4\implies xy\le\dfrac12</math> for <math>0\le x,y\le1</math>. A quick rough sketch of <math>y=\dfrac1{2x}</math> on the square given by <math>x,y\in[0,1]</math> reveals that the curve intersects the boundaries at <math>(0.5,1)</math> and <math>(1,0.5)</math>, and it is actually quite (very) obvious that the area bounded by the inequality <math>xy\le0.5</math> and the aforementioned unit square is more than <math>\dfrac34</math> but less than <math>\dfrac78</math> (cf. the diagram below). Thus, our answer is <math>\boxed{\textbf{(D) }\left(\dfrac34,\dfrac78\right]}</math>. | ||
~Technodoggo | ~Technodoggo | ||
Revision as of 17:48, 8 November 2024
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)
