Consider the set $$\mathcal{T}$$ of all triangles whose sides are distinct prime numbers which are also in arithmetic progression. Let $$\triangle\in\mathcal{T}$$ be the triangle with the least perimeter. If $$a^\circ$$ is the largest angle of $$\triangle$$ and if $$L$$ is its perimeter, determine the value of $$\frac{a}{L}$$.

Let the three distinct prime sides in increasing order be $$p-d$$, $$p$$ and $$p+d$$, with common difference $$d \gt 0$$. To be members of $$\mathcal{T}$$ these three numbers must

(i) be prime, and
(ii) satisfy the triangle inequality $$\bigl(p-d\bigr)+p \gt p+d \; \Rightarrow \; 2p-d \gt p+d \;\Rightarrow\; p \gt 2d$$.

Testing consecutive sets of three primes in arithmetic progression:

• $$2,\,3,\,5$$ has perimeter $$10$$, but $$2+3=5$$ so it violates the strict triangle inequality.
• $$3,\,5,\,7$$ has perimeter $$15$$ and satisfies $$3+5=8 \gt 7$$. This is the first (hence least-perimeter) valid triple because any smaller perimeter must include the prime $$2$$, which already failed.

Therefore the triangle with the least perimeter in $$\mathcal{T}$$ has sides $$3,\,5,\,7$$.

Perimeter: $$L = 3+5+7 = 15.$

The largest angle is opposite the longest side $$7$$. Using the Law of Cosines:

$$$$\cos$$ a = $$\frac{3^{2}$$+5^{2}-7^{2}}{2$$\cdot$$3$$\cdot$$5} = $$\frac{9+25-49}{30} = \frac{-15}{30}$$ = -\tfrac12.$$

Hence $$a = 120^\circ.$$

Finally, $$$$\frac{a}{L} = \frac{120}{15}$$ = 8.$$

Answer: $$8$$