2025 AMC 10A Problems/Problem 25: Difference between revisions

Revision as of 00:04, 7 November 2025

Problem

A point $P$ is chosen at random inside square $ABCD$ . The probability that $\overline{AP}$ is neither the shortest nor the longest side of $\triangle APB$ can be written as $\frac{a + b \pi - c \sqrt{d}}{e}$ , where $a, b, c, d,$ and $e$ are positive integers, $\text{gcd}(a, b, c, e) = 1$ , and $d$ is not divisible by the square of a prime. What is $a+b+c+d+e$ ?

$\textbf{(A) }25 \qquad \textbf{(B) }26 \qquad \textbf{(C) }27 \qquad \textbf{(D) }28 \qquad \textbf{(E)} 29 \qquad$

Solution 1 (Calculus (the actual way I used))

Note: this solution is only recommended for those who have integrated $\text{cos}^2(x)$ too many times.

WLOG, assume that the square has side length $1$ . Orient square $ABCD$ such that $A$ is the bottom-right corner and $B$ is to the left of $A$ . Let $E$ be the midpoint of $AB$ and $F$ be the midpoint of $CD$ . We will proceed by casework.

Case $1$ : $AB<AP<BP$ (note that since we are dealing with geometric probability, it doesn't matter whether one uses " $<$ " or " $\leq$ ")

Considering only the first part of the inequality ( $AB<AP$ , so $1<AP$ ), we have that $P$ must be outside the quarter circle with radius $1$ going through $B$ and $D$ centered at $A$ . Considering the second part ( $AP<BP$ ), we must have $P$ on the right side of $EF$ (closer to side $AD$ ). All $P$ that satisfy the combined inequality must be in the intersection of these two regions. Let this region of points be called $S$ .

Case $2$ : $BP<AP<AB$

Once again, considering the first part of the inequality, $P$ must be to the left of $EF$ (closer to side $BC$ ). The second part leaves $P$ to be inside the same quarter circle. Let the intersection of the two regions be called $T$ .

We wish to find the area of $S \cup T$ , and notice that there is no overlap between the two regions. To do this, we first see that the area of the quarter circle minus the area of $T$ is equal to the area of rectangle $ADFE$ minus the area of $S$ . Let this equal area be $U$ . We can rewrite the area we wish to find (where the lowercase version of each set of points represent the area of the region and " $[x]$ " represents the area of $x$ ) as $s+t=[\text{quarter circle}]-u+[\text{ADFE}]-u=\frac{\pi}{4}+\frac{1}{2}-2u$ .

We will now find the area of $U$ using calculus. Let $A$ be point $(0, 0)$ (then $B$ would be $(-1, 0)$ ). The graph of the quarter circle is given by $y=\sqrt{1-x^2}$ . Thus, the area is $\int_{-\frac{1}{2}}^{0} \sqrt{1-x^2} \; dx.$ Because the quarter circle is symmetric, we can rewrite the bounds as from $0$ to $\frac{1}{2}$ . We then proceed by trigonometric substitution where $x=\text{sin}(u)$ and $dx=\text{cos}(u) \; du$ as follows. In addition, we will find the indefinite integral first before considering the bounds.

$\int \sqrt{1-\text{sin}^2(u)} \cdot \text{cos}(u) \; du$ $=\int \text{cos}^2(u) \; du$ $=\int \frac{1+\text{cos}(2u)}{2} \; du$ $=\int \frac{1}{2} \; du + \int \frac{\text{cos}(2u)}{2} \; du$ $=\frac{u}{2} + \frac{\frac{1}{2} \; \text{sin}(2u)}{2}$ $=\frac{\text{arcsin}(x)}{2} + \frac{\text{sin}(2 \; \text{arcsin}(x))}{4}$ $=\frac{\text{arcsin}(x)}{2} + \frac{2 \; \text{sin}(\text{arcsin}(x)) \; \text{cos}(\text{arcsin}(x))}{4}$ $=\frac{\text{arcsin}(x)}{2} + \frac{x \sqrt{1-\text{sin}^2(\text{arcsin}(x))}}{2}$ $=\frac{\text{arcsin}(x)}{2} + \frac{x \sqrt{1-x^2}}{2}$

Substituting in the bounds ( $0$ to $\frac{1}{2}$ ) (at $x=0$ , the expression is $0$ ), we have $\frac{\text{arcsin}(\frac{1}{2})}{2} + \frac{\frac{1}{2} \cdot \sqrt{\frac{3}{4}}}{2}$ $=\frac{\pi}{12} + \frac{\sqrt{3}}{8}.$

Combining this with the rest of the areas, we have $\frac{\pi}{4}+\frac{1}{2}-2 \cdot (\frac{\pi}{12} + \frac{\sqrt{3}}{8})$ $=\frac{3 \pi}{12} + \frac{6}{12} - \frac{2 \pi}{12} - \frac{3 \sqrt{3}}{12}$ $=\frac{6 + \pi - 3 \sqrt{3}}{12}.$

Hence, the answer is $6+1+3+3+12 = \boxed{\textbf{(A) } 25}$ .

~scjh999999 (Thank me later)

Solution 2 (Geometry)

$[asy] size(200); pair A = (0,1), B = (1,1), C = (1,0), D = (0,0); pair M = (0.5,1), Q = (0.5,0); draw((0.5,0)--(0.5,1)); draw(A--B--C--D--cycle); draw(arc(A,1,270,360)); pair N = intersectionpoint(arc(A,1,270,360), (0.5,0)--(0.5,1)); dot(N); label("$N$", N, NE); dot(M); label("$M$", M, N); dot(Q); label("$Q$", Q, S); label("$A$", A, NW); label("$B$", B, NE); label("$C$", C, SE); label("$D$", D, SW); [/asy]$

Say WLOG that $AB$ is the top side of the square, and the square is of side length 1. Let us say that the midpoint of $AB$ is $M$ , while the midpoint of $CD$ is $Q$ . Drawing a vertical line to split the square in half, we notice that if $P$ is to the left of the line, $AP < BP$ , and if P is to the right of the line, $AP > BP$ . Also, drawing a quarter circle of radius 1 from point $A$ , we can split the area into points P for which $AP < AB$ and $AP > AB$ . Because of our constraints, there are 2 cases:

Case 1: $AB < AP < BP$ In this case, $P$ will be to the right of the vertical line and inside of the quarter circle. Let us say that the intersection of the vertical line and quarter circle is $N$ . The distance from $N$ to $AD$ is 1/2, and we can say that $\angle BAN$ is $60^\circ$ . Sector $BAN$ of circle $A$ would therefore have an area of $\frac{\pi}{6}$ . Because $\triangle AMN$ is a 30-60-90 triangle, the area of $AMN$ is $\frac{\sqrt{3}}{8}$ . The probability of case 1 happening should then be $\frac{\pi}{6}-\frac{\sqrt{3}}{8}$ .

Case 2: $AB > AP > BP$ In this case, $P$ will be to the left of the vertical line and outside of the quarter circle. Knowing that the quarter circle's area is $\frac{\pi}{4}$ , we can subtract the probability of Case 1 happening to get the chance that $P$ is on the left of the vertical line and in circle $A$ . Doing this would give $\frac{\pi}{12}+\frac{\sqrt{3}}{8}$ . To get the probability of Case 2 happening, we can subtract this from the area of rectangle $AMQD$ . This would give us $\frac{1}{2}-\frac{\pi}{12}-\frac{\sqrt{3}}{8}$ .

Adding both Cases, we get the total probability as $\frac{1}{2}+\frac{\pi}{12}-\frac{\sqrt{3}}{4} = \frac{6+\pi-3\sqrt{3}}{12}$ . Formatting this gives us $6+1+3+3+12 = \boxed{\textbf{(A) } 25}$ .

-AVS2010

Video Solution (In 2 Mins)

https://youtu.be/wpGW0PDsbdw?si=HvgB4-h7UPAhdWGX ~ Pi Academy

Solution 3

$[asy] size(200); import olympiad; pair A = (0,1), B = (1,1), C = (1,0), D = (0,0); pair M = (0.5,1), Q = (0.5,0); draw((0.5,0)--(0.5,1)); draw(A--B--C--D--cycle); draw(arc(A,1,270,360)); pair N = intersectionpoint(arc(A,1,270,360), (0.5,0)--(0.5,1)); dot(N); label("$N$", N, NE); draw(arc(B,1,180,270)); dot(M); label("$M$", M, N); dot(Q); label("$Q$", Q, S); draw(A--N--cycle); label("$A$", A, NW); label("$B$", B, NE); label("$C$", C, SE); label("$D$", D, SW); label("$a$", (0.15,0.2), W); label("$a$", (0.85,0.2), N); label("$b$", (0.65,0.6), E); label("$c$", (0.35,0.01), M); label("$c$", (0.65,0.03), Q); label("$k$", (0.2,0.45), B); [/asy]$

Assume the sides of this square is 1, hence we only need to find the area of the desired regions. From Solution 2 (took your diagram and also ty), it is easy to see that the region is the bottom left region c and the top right region b, hence we must compute $b+c$ . Also, define $k$ to be the circular segment. We have two equations right off the bat:

$2a+2b+2c=1 \implies a+b+c=\frac{1}{2}$ since the sum of them all is just the area of the square and also $2b+a=\frac{\pi}{4}$ , just the area of a quarter-circle.

Next $\triangle ABF$ has a area of $\frac{\sqrt{3}}{4}$ since it is just an equilateral triangle with length 1 (each side is a radius of a circle with radius of 1). From the diagram, $2k+[ABF]=2k+\frac{\sqrt{3}}{4}=2b$ . Next, we see that sector $ADN$ has an angle of $90-60=30$ and is the sum of $a+k$ . Therefore $a+k=\frac{\pi}{12}$ .

Multiplying this equation by 2: $2a+2k=\frac{\pi}{6}$ and adding it with $2b-2k=\frac{\sqrt{3}}{4}$ yields $2a+2b=\frac{\pi}{6}+\frac{\sqrt{3}}{4}$ . Since we also have that $2b+a=\frac{\pi}{4}$ , subtract this from the equation yields $a=(\frac{\pi}{6}+\frac{\sqrt{3}}{4}) - (\frac{\pi}{4}) = \frac{-\pi}{12}+\frac{\sqrt{3}}{4}$ . We are to find $b+c=\frac{1}{2}-a=\frac{1}{2} - (\frac{\sqrt{3}}{4} - \frac{\pi}{12}) = \frac{1}{2} - \frac{\sqrt{3}}{4} + \frac{\pi}{4} = \frac{6+\pi-3\sqrt{3}}{12}$ .

At last, $a+b+c+d+e=6+1+3+3+12=\boxed{\textbf{(A) } 25}$ .

~hxve

@@ Line 130: / Line 130: @@
 Next <imath>\triangle ABF</imath> has a area of <imath>\frac{\sqrt{3}}{4}</imath> since it is just an equilateral triangle with length 1 (each side is a radius of a circle with radius of 1). From the diagram, <imath>2k+[ABF]=2k+\frac{\sqrt{3}}{4}=2b</imath>. Next, we see that sector <imath>ADN</imath> has an angle of <imath>90-60=30</imath> and is the sum of <imath>a+k</imath>. Therefore <imath>a+k=\frac{\pi}{12}</imath>.
-Multiplying this equation by 2: <imath>2a+2k=\frac{\pi}{6}</imath> and adding it with <imath>2b-2k=\frac{\sqrt{3}}{4}</imath> yields <imath>2a+2b=\frac{\pi}{6}+\frac{\sqrt{3}}{4}</imath>. Since we also have that <imath>2b+a=\frac{\pi}{4}</imath>, subtract this from the equation yields <imath>a=(\frac{\pi}{6}+\frac{\sqrt{3}}{4}) - (\frac{\pi}{4}) = \frac{-\pi}{12}+\frac{\sqrt{3}}{4}</imath>. We are to find <imath>b+c=\frac{1}{2}-a=\frac{1}{2} - (\frac{\sqrt{3}}{4} - \frac{\pi}{4}) = \frac{1}{2} - \frac{\sqrt{3}}{4} + \frac{\pi}{4} = \frac{6+\pi-3\sqrt{3}}{12}</imath>.
+Multiplying this equation by 2: <imath>2a+2k=\frac{\pi}{6}</imath> and adding it with <imath>2b-2k=\frac{\sqrt{3}}{4}</imath> yields <imath>2a+2b=\frac{\pi}{6}+\frac{\sqrt{3}}{4}</imath>. Since we also have that <imath>2b+a=\frac{\pi}{4}</imath>, subtract this from the equation yields <imath>a=(\frac{\pi}{6}+\frac{\sqrt{3}}{4}) - (\frac{\pi}{4}) = \frac{-\pi}{12}+\frac{\sqrt{3}}{4}</imath>. We are to find <imath>b+c=\frac{1}{2}-a=\frac{1}{2} - (\frac{\sqrt{3}}{4} - \frac{\pi}{12}) = \frac{1}{2} - \frac{\sqrt{3}}{4} + \frac{\pi}{4} = \frac{6+\pi-3\sqrt{3}}{12}</imath>.
 At last, <imath>a+b+c+d+e=6+1+3+3+12=\boxed{\textbf{(A) } 25}</imath>.

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