2024 AMC 12A Problems/Problem 20

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 $0\le x,y\le1$ . (Note: You can skip most of this by using triangle area ratios ~A_MatheMagician.)

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

WLOG let $AB=AC=1$ $\frac{AP \cdot AQ \cdot \sin60}{2} < \frac{1 \cdot 1 \cdot \sin60}{4}$ $AP \cdot AQ < \frac{1}{2}$ Which we can express as $xy < \frac{1}{2}$ .

We follow by graphing the equation $xy < \frac 1 2$ on $x, y \in [0, 1]$ , and we follow the same way solution 1 does.

We see that the probability is slightly less than $\frac{7}{8}$ but definitely greater than $\frac{3}{4}$

Thus answer choice $\boxed{\text{(D)} \left(\frac{3}{4},\frac{7}{8} \right]}$

Solution 3

The actual probability can be found by using calculus, and taking the integral.

$P=\int_{0.5}^{1}{\frac{1}{2x}}dx + 0.5= \frac{\ln2+1}{2}\approx 0.84657$ .

Thus, since $\frac 3 4 < 0.84655 < \frac 7 8$ , we get $\boxed{\text{(D)} \left(\frac{3}{4},\frac{7}{8} \right]}$ ~lptoggled

Note that if you do not know an approximation for $\ln2$ , you can use the following power series formula: $\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+...$ Using this formula to four terms gives the apprximation $\ln2 \approx \frac{19}{24}$ , which is good enough to solve the problem.

Another note

If you're feeling up to it, you can assume that the curve is actually a $90^{\circ}$ sector of a circle with radius $\frac{1}{2}.$ Doing so yields $P = 1 - \pi\frac{\left(\frac{1}{2}\right)^2}{4} \approx 0.8037,$ which is clearly within the interval $D$ . This is also an under-estimate, since the curve of $\frac{1}{2x}$ is not as steep on the interval $x\in (0.5, 1).$

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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