2025 AMC 10A Problem 25

The probability is an area. Split the square into regions where $AP$ is the middle length, and add up their areas.

1. Set up. Let the square have side $1$, with $A = (0, 0)$, $B = (1, 0)$, $C = (1, 1)$, $D = (0, 1)$, and $P = (x, y)$ inside it. The triangle $APB$ has sides $AP$, $BP$ and $AB = 1$.

2. What "neither shortest nor longest" means. $AP$ must be the middle length. That can happen in two ways:
- Case 1: $BP < AP < 1$ (the side $BP$ is shortest, $AB$ is longest);
- Case 2: $1 < AP < BP$ (the side $AB$ is shortest, $BP$ is longest).

3. Draw the two dividing curves. Each comparison splits the square along a curve, so draw both.
- $AP$ versus $BP$: points that are equally far from $A$ and $B$ lie on the vertical line $x = \frac12$. To the right of it, $P$ is closer to $B$, so $BP < AP$. To the left, $AP < BP$.
- $AP$ versus $1$: points at distance $1$ from $A$ lie on the quarter circle with center $A$ and radius $1$, running from $B$ to $D$. Inside it, $AP < 1$; outside it (the corner near $C$), $AP > 1$.

4. Read off the two cases.
- Case 1 is the part of the quarter circle to the right of the line $x = \frac12$.
- Case 2 is the part of the left half of the square that is outside the quarter circle. This is a thin sliver up near corner $D$.

5. One shared piece. Let $L$ be the area of the part of the quarter circle to the left of the line $x = \frac12$. The whole quarter circle has area $\frac{\pi}{4}$ and the left half of the square has area $\frac12$, so
$$\text{Case 1} = \frac{\pi}{4} - L, \qquad \text{Case 2} = \frac12 - L.$$
Together they give $\dfrac{\pi}{4} + \dfrac12 - 2L$.

6. Find $L$. The line $x = \frac12$ meets the quarter circle at $\left(\frac12, \frac{\sqrt3}{2}\right)$ (check: $\frac14 + \frac34 = 1$). The segment from $A$ to this point has length $1$ and its horizontal part is $\frac12$, so it makes a $60^\circ$ angle with $\overline{AB}$ (a $30$-$60$-$90$ triangle). Cut $L$ along this segment into two pieces:
- a right triangle with corners $(0, 0)$, $\left(\frac12, 0\right)$, $\left(\frac12, \frac{\sqrt3}{2}\right)$, with area $\frac12 \cdot \frac12 \cdot \frac{\sqrt3}{2} = \frac{\sqrt3}{8}$;
- a pie slice of the circle from $60^\circ$ to $90^\circ$. That is $\frac{30}{360} = \frac{1}{12}$ of a full circle of area $\pi$, so its area is $\frac{\pi}{12}$.

So $L = \dfrac{\sqrt3}{8} + \dfrac{\pi}{12}$.

7. Put it together.
$$\frac{\pi}{4} + \frac12 - 2\left(\frac{\sqrt3}{8} + \frac{\pi}{12}\right) = \frac{\pi}{4} - \frac{\pi}{6} + \frac12 - \frac{\sqrt3}{4} = \frac{\pi}{12} + \frac12 - \frac{\sqrt3}{4} = \frac{6 + \pi - 3\sqrt3}{12}.$$
The square has area $1$, so this is the probability (about $0.33$).

8. Read off the numbers. $a = 6$, $b = 1$, $c = 3$, $d = 3$, $e = 12$. The only number dividing all of $6, 1, 3, 12$ is $1$, and $3$ has no square factor. So
$$a + b + c + d + e = 6 + 1 + 3 + 3 + 12 = 25.$$

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

Putting three lengths in order means comparing them two at a time, and each comparison is a curve in the square: "closer to $A$ or to $B$" is the middle line, and "closer or farther than $1$ from $A$" is a circle. Two curves cut the square into four regions, and "$AP$ in the middle" is two of them, one on each side of the line. Every region here is built from triangles and pie slices, so no advanced tools are needed.

Work out each case on its own. This takes a minute longer, but it shows how small Case 2 is, which is why it is so easy to forget.

• Case 1: $\dfrac{\pi}{4} - \dfrac{\sqrt3}{8} - \dfrac{\pi}{12} = \dfrac{\pi}{6} - \dfrac{\sqrt3}{8} \approx 0.307$.
• Case 2: $\dfrac12 - \dfrac{\sqrt3}{8} - \dfrac{\pi}{12} \approx 0.022$.

Their sum is about $0.329$, which matches $\dfrac{6 + \pi - 3\sqrt3}{12}$.

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 $\boxed{\textbf{(A)}\ 25}$.