2025 AMC 10A Problems/Problem 25

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 (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

Solution 2

$[asy] size(200); import olympiad; pair A = (0,1), B = (1,1), C = (1,0), D = (0,0); pair M = (0.5,1), Q = (0.5,0); draw((0.5,0)--(0.5,1)); draw(A--B--C--D--cycle); draw(arc(A,1,270,360)); pair N = intersectionpoint(arc(A,1,270,360), (0.5,0)--(0.5,1)); dot(N); label("$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 1, it is easy to see that the regions are 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 all regions is just the area of the square and also, $2b+a=\frac{\pi}{4}$ , just the area of a quarter-circle.

Next, $\triangle ABN$ 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, $2k+[ABN]=2k+\frac{\sqrt3}{4}=2b$ . Subsequently, we see that sector $ADN$ has an angle of $90-60=30$ and is the sum of $a+k$ . Therefore, $a+k=\frac{\pi}{12}$ .

Multiply this equation by 2: $2a+2k=\frac{\pi}{6}$ and combining it with $2b-2k=\frac{\sqrt{3}}{4}$ yields $2a+2b=\frac{\pi}{6}+\frac{\sqrt{3}}{4}$ . Since we also have that $2b+a=\frac{\pi}{4}$ , subtracting 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 $b+c=\frac{1}{2}-a=\frac{1}{2} - (\frac{\sqrt{3}}{4} - \frac{\pi}{12}) = \frac{1}{2} - \frac{\sqrt{3}}{4} + \frac{\pi}{12} = \frac{6+\pi-3\sqrt{3}}{12}$ .

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

~hxve

Solution 3 (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)

Video Solution (In 2 Mins)

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

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

Video Solution by OmegaLearn

https://youtu.be/2_znFsACFZI

2025 AMC 10A (Problems • Answer Key • Resources)
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