2025 AMC 10A Problem 25

Set $A = (0,0), B = (1,0);$ $AP$ is the middle length when either $BP \lt AP \lt AB$ or $AB \lt AP \lt BP$

The boundaries are the circle $AP = 1$ (about $A$) and the vertical line $AP = BP;$ use the $60^\circ$ sector where $\triangle ABS$ is equilateral

Place $A = (0,0)$ and $B = (1,0)$ on the unit square. $AP$ is the middle length when $BP \lt AP \lt AB$ or $AB \lt AP \lt BP.$ These regions are bounded by the circle centered at $A$ with radius $1$ (where $AP = AB$) and the line $x = \tfrac12$ (where $AP = BP$). Let $S$ be where the circle meets $x = \tfrac12.$ Then $\triangle ABS$ is equilateral, so $\angle BAS = 60^\circ.$ The larger region has area $\frac{4\pi - 3\sqrt3}{24}$ and the smaller has area $\frac{12 - 2\pi - 3\sqrt3}{24}.$ They add to $\frac{6 + \pi - 3\sqrt3}{12}.$ Thus $a + b + c + d + e$ $= 6 + 1 + 3 + 3 + 12$ $= 25.$ Thus, A is the correct answer.