2025 AMC 10A Problem 25

Take the square to have side $1$, with $A = (0,0)$, $B = (1,0)$, $C = (1,1)$, $D = (0,1)$, and let $P = (x,y)$ with $0 < x,y < 1$. The three sides of $\triangle APB$ are $AP$, $BP$ and $AB = 1$. Since $P$ is not on line $AB$, the triangle is genuine, and "neither shortest nor longest" simply means $AP$ is strictly the middle one of the three lengths. So either
$$\textbf{Case 1: } BP < AP < 1, \qquad \text{or} \qquad \textbf{Case 2: } 1 < AP < BP.$$

Two loci do all the work. $AP = BP$ is the perpendicular bisector $x = \tfrac12$, with $BP < AP$ on the right and $AP < BP$ on the left. $AP = 1$ is the quarter of the unit circle centred at $A$ that lies in the square, with $AP < 1$ inside it.

So Case 1 is the part of the quarter disc with $x > \tfrac12$, and Case 2 is the part of the left half of the square lying outside the quarter disc.

Let $L$ be the area of the piece of the quarter disc with $x < \tfrac12$. The quarter disc has area $\tfrac{\pi}{4}$ and the left half of the square has area $\tfrac12$, so
$$[\text{Case 1}] = \frac{\pi}{4} - L, \qquad [\text{Case 2}] = \frac12 - L,$$
and the probability sought is $\dfrac{\pi}{4} + \dfrac12 - 2L$.

Computing $L$. The line $x = \tfrac12$ meets the unit circle at $\left(\tfrac12, \tfrac{\sqrt3}{2}\right)$, a point at angle $60^\circ$ from $A$. Split $L$ along the radius to that point:

• the right triangle with vertices $(0,0)$, $\left(\tfrac12, 0\right)$, $\left(\tfrac12, \tfrac{\sqrt3}{2}\right)$, of area $\tfrac12 \cdot \tfrac12 \cdot \tfrac{\sqrt3}{2} = \tfrac{\sqrt3}{8}$;
• the circular sector from $60^\circ$ to $90^\circ$, of area $\tfrac{30}{360}\pi = \tfrac{\pi}{12}$.

Hence $L = \dfrac{\sqrt3}{8} + \dfrac{\pi}{12}$, and

$$\frac{\pi}{4} + \frac12 - 2\left(\frac{\sqrt3}{8} + \frac{\pi}{12}\right) = \frac{\pi}{4} - \frac{\pi}{6} + \frac12 - \frac{\sqrt3}{4} = \frac{\pi}{12} + \frac12 - \frac{\sqrt3}{4} = \frac{6 + \pi - 3\sqrt3}{12}.$$

The area of the square is $1$, so this is the probability; numerically it is about $0.329$, which is a plausible size for the region. Reading off $a = 6$, $b = 1$, $c = 3$, $d = 3$, $e = 12$ (and $\gcd(6,1,3,12) = 1$, $d=3$ squarefree),
$$a+b+c+d+e = 6+1+3+3+12 = 25.$$

The answer is $\boxed{\textbf{(A)}\ 25}$.

Ordering three lengths is the same as deciding, for each pair, which is larger, and each such comparison is a single curve in the plane: $AP$ versus $BP$ is a perpendicular bisector, $AP$ versus $AB$ is a circle centred at $A$. Two curves cut the square into four regions, and "$AP$ in the middle" is exactly two of them — one on each side of the bisector. Describing the answer as "quarter disc minus $L$" plus "half square minus $L$" avoids computing either region separately and makes the coefficient $-2L$, hence the single $\sqrt3$ and the single $\pi$, appear at once. No calculus is needed: a region bounded by two radii and an arc is a sector, and everything left over is a triangle.

Compute the two regions directly with the sector-plus-triangle decomposition. Case 1 is the quarter disc minus $L$, that is $\tfrac{\pi}{4} - \tfrac{\sqrt3}{8} - \tfrac{\pi}{12} = \tfrac{\pi}{6} - \tfrac{\sqrt3}{8} \approx 0.3071$. Case 2 is the thin region tucked into the corner near $D$: the left half-square minus $L$, i.e. $\tfrac12 - \tfrac{\sqrt3}{8} - \tfrac{\pi}{12} \approx 0.0217$. Their sum is $\approx 0.3288$, agreeing with $\tfrac{6+\pi-3\sqrt3}{12}$. Seeing Case 2 explicitly is worth the extra minute, because its tiny area is what makes it easy to forget.

• Handling only the case BP < AP < AB and missing the sliver near D where AB < AP < BP, which leaves pi/6 - sqrt3/8 and cannot even be written in the required form.
• Getting the perpendicular bisector the wrong way round and pairing "$x < \tfrac12$" with "inside the circle", which computes the complementary regions instead.
• Taking "neither shortest nor longest" to allow ties and worrying about the boundaries $x = \tfrac12$ and $AP = 1$; these are curves of zero area and change nothing.