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

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


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<cmath>\frac{\pi}{4}+\frac{1}{2}-2 \cdot (\frac{\pi}{12} + \frac{\sqrt{3}}{8}</cmath>
<cmath>\frac{\pi}{4}+\frac{1}{2}-2 \cdot (\frac{\pi}{12} + \frac{\sqrt{3}}{8}</cmath>
<cmath>=\frac{3 \pi}{12} + \frac{6}{12} - \frac{2 \pi}{12} - \frac{3 \sqrt{3}}{12}</cmath>
<cmath>=\frac{3 \pi}{12} + \frac{6}{12} - \frac{2 \pi}{12} - \frac{3 \sqrt{3}}{12}</cmath>
<cmath>=\frac{6 + \pi - 3 \sqrt{3}}{12}</cmath>.
<cmath>=\frac{6 + \pi - 3 \sqrt{3}}{12}.</cmath>


Hence, the answer is <imath>6+1+3+3+12 = \fbox{(A) 25}</imath>.
Hence, the answer is <imath>6+1+3+3+12 = \fbox{(A) 25}</imath>.


~scjh999999 (Thank me later)
~scjh999999 (Thank me later)

Revision as of 14:42, 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)) CURRENTLY UNDERGOING MAINTENANCE

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 $AD$). For 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, we have that $P$ 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 the region bounded by $U$ 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 $6+1+3+3+12 = \fbox{(A) 25}$.

~scjh999999 (Thank me later)

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