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

Latest revision as of 12:19, 21 March 2025

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 $x\in (0.5, 1).$

@@ Line 1: / Line 1: @@
-{{duplicate|[[2024 AMC 12A Problems/Problem 20|2024 AMC 12A #20]] and [[2024 AMC 10A Problems/Problem 23|2024 AMC 10A #23]]}}
+==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>
+<math>\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]</math>
+==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
+<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>. (Note: You can skip most of this by using triangle area ratios ~A_MatheMagician.)
+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>.
-==Problem==
+~Technodoggo
-The figure below shows a dotted grid <math>8</math> cells wide and <math>3</math> cells tall consisting of <math>1^{\prime\prime} \times 1^{\prime\prime}</math> squares. Carl places <math>1</math>-inch toothpicks along some of the sides of the squares to create a closed loop that does not intersect itself. The numbers in the cells indicate the number of sides of that square that are to be covered by toothpicks, and any number of toothpicks are allowed if no number is written. In how many ways can Carl place the toothpicks?
 <asy>
-size(6cm);
+/*Asymptote visual by Technodoggo, 7 November 2024*/
-for (int i=0; i<9; ++i) {
+unitsize(8cm);
-  draw((i,0)--(i,3),dotted);
-}
+draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
-for (int i=0; i<4; ++i){
+label("$0$",(-0.05,-0.05));
-  draw((0,i)--(8,i),dotted);
+label("$1$",(1,-0.05),S);
-}
+label("$1$",(-0.05,1),W);
-for (int i=0; i<8; ++i) {
+draw((-0.05,0)--(1,0)--(1,-0.05));
-  for (int j=0; j<3; ++j) {
+draw((0,-0.05)--(0,1)--(-0.05,1));
-    if (j==1) {
-      label("1",(i+0.5,1.5));
+real f(real x) {return 1/(2*x);}
-}}}
-</asy>
-<math>\textbf{(A)}~130\qquad\textbf{(B)}~144\qquad\textbf{(C)}~146\qquad\textbf{(D)}~162\qquad\textbf{(E)}~196</math>
+path c = graph(f, 0.5,1)--(1,0)--(0,0)--(0,1)--cycle;
-==(Best) Solution 1==
+filldraw(c,blue+white);
-There are two possibilities for the loop if it does not cross the middle row of <math>1</math>'s: a <math>1\times 8</math> rectangle around the top row of cells or a <math>1\times 8</math> rectangle around the bottom row of cells.  Otherwise, notice that wherever the loop crosses this middle row, it proceed straight in both directions after the middle row.  This divides the dotted grid into two components, and both ends of the loop must enter the same connected component.  It follows that the loop must cross one of the leftmost two columns and one of the rightmost two columns for four configurations total.
-<asy>
-unitsize(1.1cm);
-defaultpen(linewidth(0.7)+linetype("1 4"));
-void cross(pair P)
-{
-real r = 0.1;
-draw((P.x-r,P.y-r)--(P.x+r,P.y+r)^^(P.x+r,P.y-r)--(P.x-r,P.y+r),linewidth(1)+solid);
-}
-for(int i=0;i<=3;i=i+1)
-{
-draw((0,i)--(2.5,i));
-}
-for(int j=0;j<=1;j=j+1)
-{
-draw((j,0)--(j,3));
-label("$1$",(j+0.5,1.5),fontsize(15));
-}
-draw((2,0)--(2,3));
-cross((0.5,1)); cross((1,1.5)); cross((0.5,2));
-draw((1,0)--(0,0)--(0,3)--(1,3),linewidth(1.4)+solid);
-real s=4;
+draw((0.5,1)--(0.5,0.5)--(1,0.5),white+dashed+1.1);
-for(int i=0;i<=3;i=i+1)
+draw((0.5,1)--(1,0.5),red+dashed+1.1);
-{
-draw((s,i)--(s+2.5,i));
-}
-for(int j=0;j<=1;j=j+1)
-{
-draw((s+j,0)--(s+j,3));
-label("$1$",(s+j+0.5,1.5),fontsize(15));
-}
-draw((s+2,0)--(s+2,3));
-cross((s+0.5,1)); cross((s,1.5)); cross((s+0.5,2));
-cross((s+1.5,1)); cross((s+2,1.5)); cross((s+1.5,2));
-draw((s+2,0)--(s+1,0)--(s+1,3)--(s+2,3),linewidth(1.4)+solid);
 </asy>
-In each case, there are two ends of the loop that must connect -- one across the top and one across the bottom -- along with some number of <math>1</math>'s in the middle of the grid.  Suppose there are <math>n</math> such <math>1</math>s.  For each <math>1</math>, the loop can cover either the top edge or the bottom edge.  These choices are independent of each other, and so the number of ways to connect both ends of the loop is <math>2^n</math>.  The figure below shows one possibility when <math>n = 6</math>.
-<asy>
-unitsize(1.1cm);
-defaultpen(linewidth(0.7)+linetype("1 4"));
-void cross(pair P)
-{
-real r = 0.1;
-draw((P.x-r,P.y-r)--(P.x+r,P.y+r)^^(P.x+r,P.y-r)--(P.x-r,P.y+r),linewidth(1)+solid);
-}
-for(int i=0;i<=3;i=i+1)
-{
-draw((0.5,i)--(7.5,i));
-}
-for(int j=1;j<=6;j=j+1)
-{
-draw((j,0)--(j,3));
-label("$1$",(j+0.5,1.5),fontsize(15));
-cross((j,1.5));
-}
-draw((7,0)--(7,3)); cross((7,1.5));
-draw((1,2)--(2,2)^^(2,1)--(4,1)^^(4,2)--(5,2)^^(5,1)--(6,1)^^(6,2)--(7,2),linewidth(1.4) + solid + red);
-draw((0.5,3)--(1,3)--(1,2)^^(2,2)--(2,3)--(4,3)--(4,2)^^(5,2)--(5,3)--(6,3)--(6,2)^^(7,2)--(7,3)--(7.5,3),linewidth(1.4)+solid);
-draw((0.5,0)--(2,0)--(2,1)^^(4,1)--(4,0)--(5,0)--(5,1)^^(6,1)--(6,0)--(7.5,0),linewidth(1.4)+solid);
-</asy>
-Returning to the original problem, of the four possibilities, one gives <math>n = 6</math>, two give <math>n = 5</math>, and one gives <math>n = 4</math>.  Remembering to add back the <math>2</math>, it follows that the total number of solutions is <cmath>2^6 + 2\cdot 2^5 + 2^4 + 2 = 64 + 64 + 16 + 2 = \boxed{\textbf{(C)}~146}.</cmath>
-~djmathman
+== Solution 2 ==
+WLOG let <math>AB=AC=1</math>
+<cmath> \frac{AP \cdot AQ \cdot \sin60}{2} < \frac{1 \cdot 1 \cdot \sin60}{4}</cmath>
+<cmath>AP \cdot AQ < \frac{1}{2}</cmath>
+Which we can express as <math>xy < \frac{1}{2}</math>.
+We follow by graphing the equation <math>xy < \frac 1 2</math> on <math>x, y \in [0, 1]</math>, and we follow the same way solution 1 does.
+We see that the probability is slightly less than <math>\frac{7}{8}</math> but definitely greater than <math>\frac{3}{4}</math>
-==Solution 2 (Cheese)==
+Thus answer choice <math>\boxed{\text{(D)} \left(\frac{3}{4},\frac{7}{8} \right]}</math>
-Notice that for any case where the closed loop does not connect from the top side of the ones and bottom side of the ones, there are two of these cases. A cheese solution can be found from this; noting that B and C are the only two options to each other, and, being two apart, with people likely to forget this case, <math>\boxed{\textbf{(C) }146}</math> is likely to be the correct answer.
-Cheese solution done by [[User:Juwushu|juwushu]].
-==Solution 3 (Observation)==
+==Solution 3==
-We have <math>2</math> cases where the loop does not go through the middle.
+The actual probability can be found by using calculus, and taking the integral.
-If the loop goes through the middle, we must have a full column on <math>(0,8),(0,7),(1,8),(1,7).</math> Then we have <math>6,5,5,4</math> empty middle squares. For each one we can have one on top or one on bottom, so <math>2^6+2^5+2^5+2^4=144.</math> Notice that for each case of fixed toothpicks, there is only one way to form the loop. Then we just add <math>2+144=\boxed{\textbf{(C) } 146.}</math>
+<math>P=\int_{0.5}^{1}{\frac{1}{2x}}dx + 0.5= \frac{\ln2+1}{2}\approx 0.84657</math>.
-~nevergonnagiveup
+Thus, since <math>\frac 3 4 < 0.84655 < \frac 7 8</math>, we get <math>\boxed{\text{(D)} \left(\frac{3}{4},\frac{7}{8} \right]}</math>
+~lptoggled
-== Video Solution by Power Solve ==
+Note that if you do not know an approximation for <math>\ln2</math>, you can use the following power series formula: <cmath>\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+...</cmath> Using this formula to four terms gives the apprximation <math>\ln2 \approx \frac{19}{24}</math>, which is good enough to solve the problem.
-https://www.youtube.com/watch?v=FRNbJ5wIGRo
-==Video Solution By SpreadTheMathLove==
+===Another note===
-https://www.youtube.com/watch?v=huMQ9J7rIj0&t=77s
+If you're feeling up to it, you can assume that the curve is actually a <math>90^{\circ}</math> sector of a circle with radius <math>\frac{1}{2}.</math> Doing so yields
+<cmath>P = 1 - \pi\frac{\left(\frac{1}{2}\right)^2}{4} \approx 0.8037,</cmath>
+which is clearly within the interval <math>D</math>. This is also an under-estimate, since the curve of <math>\frac{1}{2x}</math> is not as steep on the interval <math>x\in (0.5, 1).</math>
 ==See also==
 {{AMC12 box|year=2024|ab=A|num-b=19|num-a=21}}
-{{AMC10 box|year=2024|ab=A|num-b=22|num-a=24}}
-[[Category:Intermediate Combinatorics Problems]]
 {{MAA Notice}}

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
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