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

(→‎Solution 1)
(→‎Solution 1)
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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==
 +
\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 <math>xy<\frac{1}{2}</math> for graphing purposes <math>(x,y<1)</math>}\\
 +
\text{By graphing it out (someone please insert diagram)}\\
 +
\text{We see that the probability is slighty less than <math>\frac{7}{8}</math> but definitely greater than <math>\frac{3}{4}</math>}\\
 +
\text{Thus answer choice }\fbox{(D) <math>\left(\frac{3}{4},\frac{7}{8} \right]</math>}\\
 +
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}

Revision as of 17:49, 8 November 2024

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]

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 $xy<\frac{1}{2}$ for graphing purposes $(x,y<1)$}\\ \text{By graphing it out (someone please insert diagram)}\\ \text{We see that the probability is slighty less than $\frac{7}{8}$ but definitely greater than $\frac{3}{4}$}\\ \text{Thus answer choice }\fbox{(D) $\left(\frac{3}{4},\frac{7}{8} \right]$}\\ 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}

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