Question:

Step 1: Let $\dfrac{s-a}{4} = \dfrac{s-b}{3} = \dfrac{s-c}{2} = k$. From this, we have: $s - a = 4k$ $s - b = 3k$ $s - c = 2k$ Adding these three equations yields: $(s-a) + (s-b) + (s-c) = 4k + 3k + 2k$ $3s - (a+b+c) = 9k$ Since $2s = a+b+c$, we substitute $a+b+c = 2s$: $3s - 2s = 9k$ $s = 9k$ Step 2: Using $s=9k$ and the expressions from Step 1, we find the side lengths: $a = s - 4k = 9k - 4k = 5k$ $b = s - 3k = 9k - 3k = 6k$ $c = s - 2k = 9k - 2k = 7k$ Step 3: The area of the inscribed circle is given as $\dfrac{8\pi}{3}$. Let $r$ be the inradius. $\pi r^2 = \dfrac{8\pi}{3}$ $r^2 = \dfrac{8}{3}$ $r = \sqrt{\dfrac{8}{3}} = \dfrac{2\sqrt{2}}{\sqrt{3}}$ Step 4: The area of the triangle, $\Delta$, is related to the inradius $r$ and semi-perimeter $s$ by the formula $\Delta = rs$. $\Delta = \left(\dfrac{2\sqrt{2}}{\sqrt{3}}\right) (9k) = \dfrac{18\sqrt{2}k}{\sqrt{3}}$ Step 5: The area of the triangle can also be calculated using Heron's formula: $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$. Substituting the expressions in terms of $k$: $\Delta = \sqrt{(9k)(4k)(3k)(2k)}$ $\Delta = \sqrt{216k^4}$ $\Delta = 6\sqrt{6}\,k^2$ Step 6: Equating the two expressions for $\Delta$ from Step 4 and Step 5: $6\sqrt{6}\,k^2 = \dfrac{18\sqrt{2}k}{\sqrt{3}}$ $6\sqrt{6}\,k^2 = 18k\sqrt{\dfrac{2}{3}}$ $6\sqrt{6}\,k^2 = 18k\dfrac{\sqrt{2}\sqrt{3}}{3}$ $6\sqrt{6}\,k^2 = 6k\sqrt{6}$ Since $k$ must be positive (as it relates to side lengths), we can divide by $6\sqrt{6}\,k$: $k = 1$ Step 7: With $k=1$, the side lengths are $a=5, b=6, c=7$. The semi-perimeter is $s=9$. The area of $\triangle ABC$ is $\Delta = 6\sqrt{6}(1)^2 = 6\sqrt{6}$. Step 8: The circumradius $R$ of $\triangle ABC$ is given by the formula $R = \dfrac{abc}{4\Delta}$. $R = \dfrac{5 \cdot 6 \cdot 7}{4 \cdot 6\sqrt{6}} = \dfrac{210}{24\sqrt{6}} = \dfrac{35}{4\sqrt{6}}$ Step 9: To find angle $A$, we use the Law of Cosines: $\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}$. $\cos A = \dfrac{6^2 + 7^2 - 5^2}{2 \cdot 6 \cdot 7} = \dfrac{36 + 49 - 25}{84} = \dfrac{60}{84} = \dfrac{5}{7}$ Thus, $A = \cos^{-1}\left(\dfrac{5}{7}\right)$. Step 10: We evaluate the expression $\dfrac{8\sin^2\left(\frac{A+B}{2}\right)}{21\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)}$. Since $A+B+C = \pi$, we have $\dfrac{A+B}{2} = \dfrac{\pi - C}{2}$. Therefore, $\sin\left(\dfrac{A+B}{2}\right) = \sin\left(\dfrac{\pi}{2} - \dfrac{C}{2}\right) = \cos\left(\dfrac{C}{2}\right)$. The expression becomes $\dfrac{8\cos^2\left(\frac{C}{2}\right)}{21\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)}$. We know that $r = 4R\sin\left(\dfrac{A}{2}\right)\sin\left(\dfrac{B}{2}\right)\sin\left(\dfrac{C}{2}\right)$, so $\sin\left(\dfrac{A}{2}\right)\sin\left(\dfrac{B}{2}\right)\sin\left(\dfrac{C}{2}\right) = \dfrac{r}{4R}$. Substituting the values of $r = \dfrac{2\sqrt{2}}{\sqrt{3}}$ and $R = \dfrac{35}{4\sqrt{6}}$: $\sin\left(\dfrac{A}{2}\right)\sin\left(\dfrac{B}{2}\right)\sin\left(\dfrac{C}{2}\right) = \dfrac{\frac{2\sqrt{2}}{\sqrt{3}}}{4 \cdot \frac{35}{4\sqrt{6}}} = \dfrac{\frac{2\sqrt{2}}{\sqrt{3}}}{\frac{35}{\sqrt{6}}} = \dfrac{2\sqrt{2}}{\sqrt{3}} \cdot \dfrac{\sqrt{6}}{35} = \dfrac{2\sqrt{12}}{35\sqrt{3}} = \dfrac{2 \cdot 2\sqrt{3}}{35\sqrt{3}} = \dfrac{4}{35}$. Next, we find $\cos C$ using the Law of Cosines: $\cos C = \dfrac{a^2+b^2-c^2}{2ab}$. $\cos C = \dfrac{5^2+6^2-7^2}{2 \cdot 5 \cdot 6} = \dfrac{25+36-49}{60} = \dfrac{12}{60} = \dfrac{1}{5}$. Now, we find $\cos^2\left(\dfrac{C}{2}\right)$ using the half-angle identity $\cos^2\left(\dfrac{C}{2}\right) = \dfrac{1+\cos C}{2}$. $\cos^2\left(\dfrac{C}{2}\right) = \dfrac{1+\frac{1}{5}}{2} = \dfrac{\frac{6}{5}}{2} = \dfrac{3}{5}$. Finally, substitute these values into the expression: $\dfrac{8\cos^2\left(\frac{C}{2}\right)}{21\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)} = \dfrac{8 \cdot \frac{3}{5}}{21 \cdot \frac{4}{35}} = \dfrac{\frac{24}{5}}{\frac{3 \cdot 7 \cdot 4}{5 \cdot 7}} = \dfrac{\frac{24}{5}}{\frac{12}{5}} = 2$.