Alice is standing on the circumference of a large circular room of radius $10$. There is a circular pillar in the center of the room of radius $5$ that blocks Alice’s view. The total area in the room Alice can see can be expressed in the form $\frac{m\pi}{n} +p\sqrt{q}$, where $m$ and $n$ are relatively prime positive integers and $p$ and $q$ are integers such that $q$ is square-free. Compute $m + n + p + q$. (Note that the pillar is not included in the total area of the room.)
[img]https://cdn.artofproblemsolving.com/attachments/1/9/a744291a61df286735d63d8eb09e25d4627852.png[/img]
1. **Identify the geometry of the problem:**
   - Alice is standing on the circumference of a large circular room with radius \(10\).
   - There is a circular pillar in the center of the room with radius \(5\).
   - Alice's view is blocked by the pillar, and we need to find the total area in the room that Alice can see.

2. **Determine the lines of sight:**
   - The farthest lines of sight from Alice's perspective are the tangents to the pillar. These tangents form chords \(AX\) and \(AY\) on the circumference of the room.
   - Let \(B\) and \(C\) be the points of tangency on \(AX\) and \(AY\) respectively.

3. **Calculate the length of the chord \(AY\):**
   - Draw the radius \(OC\) to the point of tangency \(C\). Since \(OC\) is perpendicular to \(AY\), it bisects \(AY\).
   - The triangle \(OCA\) is a right triangle with \(OA = 10\) and \(OC = 5\).
   - Using the Pythagorean theorem in \(\triangle OCA\):
     \[
     AC = \sqrt{OA^2 - OC^2} = \sqrt{10^2 - 5^2} = \sqrt{100 - 25} = \sqrt{75} = 5\sqrt{3}
     \]
   - Therefore, \(AY = 2 \times AC = 2 \times 5\sqrt{3} = 10\sqrt{3}\).

4. **Calculate the area of the circular segments:**
   - The angle \(\angle AOY = \angle AOX = 120^\circ\) (since the tangents from a point to a circle are symmetric).
   - The area of the sector \(AOY\) is:
     \[
     \text{Area of sector} = \frac{120^\circ}{360^\circ} \times \pi \times 10^2 = \frac{1}{3} \times 100\pi = \frac{100\pi}{3}
     \]
   - The area of the triangle \(AOY\) is:
     \[
     \text{Area of triangle} = \frac{1}{2} \times 10\sqrt{3} \times 5 = 25\sqrt{3}
     \]
   - The area of the circular segment \(A_1\) is:
     \[
     A_1 = \frac{100\pi}{3} - 25\sqrt{3}
     \]

5. **Calculate the area of the quadrilateral \(OCAB\):**
   - The area of the two right triangles \(OCB\) and \(OCA\) is:
     \[
     \text{Area of } \triangle OCB = \frac{1}{2} \times 5 \times 5\sqrt{3} = \frac{25\sqrt{3}}{2}
     \]
     \[
     \text{Total area of } \triangle OCB \text{ and } \triangle OCA = 2 \times \frac{25\sqrt{3}}{2} = 25\sqrt{3}
     \]
   - The area of the sector \(OCB\) is:
     \[
     \text{Area of sector} = \frac{120^\circ}{360^\circ} \times \pi \times 5^2 = \frac{1}{3} \times 25\pi = \frac{25\pi}{3}
     \]
   - The area of the quadrilateral \(OCAB\) outside the pillar is:
     \[
     A_3 = 25\sqrt{3} - \frac{25\pi}{3}
     \]

6. **Sum the areas:**
   - The total area Alice can see is:
     \[
     A_1 + A_2 + A_3 = 2 \left( \frac{100\pi}{3} - 25\sqrt{3} \right) + 25\sqrt{3} - \frac{25\pi}{3}
     \]
     \[
     = \frac{200\pi}{3} - 50\sqrt{3} + 25\sqrt{3} - \frac{25\pi}{3}
     \]
     \[
     = \frac{175\pi}{3} - 25\sqrt{3}
     \]

7. **Express the area in the required form and compute \(m + n + p + q\):**
   - The area is \(\frac{175\pi}{3} - 25\sqrt{3}\).
   - Here, \(m = 175\), \(n = 3\), \(p = -25\), and \(q = 3\).
   - Therefore, \(m + n + p + q = 175 + 3 - 25 + 3 = 156\).

The final answer is \( \boxed{ 156 } \).