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

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


Case 2: <imath>AB > AP > BP</imath>
Case 2: <imath>AB > AP > BP</imath>

Revision as of 15:38, 6 November 2025

2025 AMC 10A Problems/Problem 25

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.

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)

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 $\DeltaAMN$ (Error compiling LaTeX. Unknown error_msg) 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 $\frac{1}{2}+\frac{\pi}{12}-\frac{\sqrt{3}}{4} \eq \frac{6+\pi-3\sqrt{3}}{12}$ (Error compiling LaTeX. Unknown error_msg). Formatting this gives us $6+1+3+3+12 \eq \boxed{\textbf{(A)}25}$ (Error compiling LaTeX. Unknown error_msg).

-AVS2010

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