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

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Orient square <imath>ABCD</imath> such that <imath>A</imath> is the bottom-right corner and <imath>B</imath> is to the left of <imath>A</imath>. Let <imath>E</imath> be the midpoint of <imath>AB</imath> and <imath>F</imath> be the midpoint of <imath>CD</imath>. We will proceed by casework.
Orient square <imath>ABCD</imath> such that <imath>A</imath> is the bottom-right corner and <imath>B</imath> is to the left of <imath>A</imath>. Let <imath>E</imath> be the midpoint of <imath>AB</imath> and <imath>F</imath> be the midpoint of <imath>CD</imath>. We will proceed by casework.


Case <imath>1</imath>: <imath>AB<AP<BP</imath> (note that since we are dealing with geometric probability, it doesn't matter whether one uses <imath><</imath> or <imath>\leq</imath>)
Case <imath>1</imath>: <imath>AB<AP<BP</imath> (note that since we are dealing with geometric probability, it doesn't matter whether one uses <imath><</imath> or <imath>\leq</imath>) \\
Considering only the first part of the inequality (<imath>AB<AP</imath>, so <imath>1<AP</imath>), we have that <imath>P</imath> must be outside the quarter circle with radius <imath>1</imath> going through <imath>B</imath> and <imath>D</imath> centered at <imath>A</imath>. Considering the second part (<imath>AP<BP</imath>), we must have <imath>P</imath> on the right side of <imath>EF</imath> (closer to side <imath>AD</imath>). For all <imath>P</imath> that satisfy the combined inequality must be in the intersection of these two regions. Let this region of points be called <imath>S</imath>.  
Considering only the first part of the inequality (<imath>AB<AP</imath>, so <imath>1<AP</imath>), we have that <imath>P</imath> must be outside the quarter circle with radius <imath>1</imath> going through <imath>B</imath> and <imath>D</imath> centered at <imath>A</imath>. Considering the second part (<imath>AP<BP</imath>), we must have <imath>P</imath> on the right side of <imath>EF</imath> (closer to side <imath>AD</imath>). For all <imath>P</imath> that satisfy the combined inequality must be in the intersection of these two regions. Let this region of points be called <imath>S</imath>.  


Case <imath>2</imath>: <imath>BP<AP<AB</imath>
Case <imath>2</imath>: <imath>BP<AP<AB</imath> \\
Once again, considering the first part of the inequality, we have that <imath>P</imath> must be to the left of <imath>EF</imath> (closer to side <imath>BC</imath>). The second part leaves <imath>P</imath> to be inside the same quarter circle. Let the intersection of the two regions be called <imath>T</imath>.
Once again, considering the first part of the inequality, we have that <imath>P</imath> must be to the left of <imath>EF</imath> (closer to side <imath>BC</imath>). The second part leaves <imath>P</imath> to be inside the same quarter circle. Let the intersection of the two regions be called <imath>T</imath>.


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<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)

Revision as of 14:40, 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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