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Difference between revisions of "2024 AMC 12A Problems/Problem 20"

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==Problem==
 
Points <math>P</math> and <math>Q</math> are chosen uniformly and independently at random on sides <math>\overline {AB}</math> and <math>\overline{AC},</math> respectively, of equilateral triangle <math>\triangle ABC.</math> Which of the following intervals contains the probability that the area of <math>\triangle APQ</math> is less than half the area of <math>\triangle ABC?</math>
 
Points <math>P</math> and <math>Q</math> are chosen uniformly and independently at random on sides <math>\overline {AB}</math> and <math>\overline{AC},</math> respectively, of equilateral triangle <math>\triangle ABC.</math> Which of the following intervals contains the probability that the area of <math>\triangle APQ</math> is less than half the area of <math>\triangle ABC?</math>
  
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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>
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==See also==
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{{AMC12 box|year=2024|ab=A|num-b=19|num-a=21}}
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{{MAA Notice}}

Revision as of 18:09, 8 November 2024

Problem

Points $P$ and $Q$ are chosen uniformly and independently at random on sides $\overline {AB}$ and $\overline{AC},$ respectively, of equilateral triangle $\triangle ABC.$ Which of the following intervals contains the probability that the area of $\triangle APQ$ is less than half the area of $\triangle ABC?$

$\textbf{(A) } \left[\frac 38, \frac 12\right] \qquad \textbf{(B) } \left(\frac 12, \frac 23\right] \qquad \textbf{(C) } \left(\frac 23, \frac 34\right] \qquad \textbf{(D) } \left(\frac 34, \frac 78\right] \qquad \textbf{(E) } \left(\frac 78, 1\right]$

Solution 1

Let $\overline{AP}=x$ and $\overline{AQ}=y$. Applying the sine formula for a triangle's area, we see that \[[\Delta APQ]=\dfrac12\cdot x\cdot y\sin(\angle PAQ)=\dfrac{xy}2\sin(60^\circ)=\dfrac{\sqrt3}4xy.\]

Without loss of generality, we let $AB=BC=CA=1$, and thus $[\Delta ABC]=\dfrac{\sqrt3}4$; we therefore require $\dfrac{\sqrt3}4xy\le\dfrac12\cdot\dfrac{\sqrt3}4\implies xy\le\dfrac12$ for $0\le x,y\le1$. A quick rough sketch of $y=\dfrac1{2x}$ on the square given by $x,y\in[0,1]$ reveals that the curve intersects the boundaries at $(0.5,1)$ and $(1,0.5)$, and it is actually quite (very) obvious that the area bounded by the inequality $xy\le0.5$ and the aforementioned unit square is more than $\dfrac34$ but less than $\dfrac78$ (cf. the diagram below). Thus, our answer is $\boxed{\textbf{(D) }\left(\dfrac34,\dfrac78\right]}$.

~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]

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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