Problem Analysis #11

To solve the problem, we need to determine the values of $s_1$ and $s_2$ for the given isosceles triangle $ABC$ with $AB = AC = 1$ and $\angle BAC = \pi/3$ and $\angle BAC = 2\pi/3$, respectively. The goal is to compute $(2s_1) + (4s_2^2)$.

Step-by-Step Solution:

1. Understanding the Problem:

   • We have an isosceles triangle $ABC$ with $AB = AC = 1$ and $\angle BAC = \pi/3$ and $\angle BAC = 2\pi/3$.
   • We need to find the smallest radius $s$ such that disks centered at $A$, $B$, and $C$ cover all three sides.
   • Given $s_1 = 1/2$ for $\angle BAC = \pi/3$ and $s_2 = \sqrt{3}/2$ for $\angle BAC = 2\pi/3$.

2. Calculating $s_1$ and $s_2$:

   • For $\angle BAC = \pi/3$, $s_1 = 1/2$.
   • For $\angle BAC = 2\pi/3$, $s_2 = \sqrt{3}/2$.

3. Compute the Required Expression:

   • We need to evaluate $(2s_1) + (4s_2^2)$.
   • Substitute $s_1 = 1/2$ and $s_2 = \sqrt{3}/2$: $$2s_1 = 2 \times \frac{1}{2} = 1$$ $$4s_2^2 = 4 \left(\frac{\sqrt{3}}{2}\right)^2 = 4 \times \frac{3}{4} = 3$$
   • Adding these results: $$(2s_1) + (4s_2^2) = 1 + 3 = 4$$

Final Answer: