Properties and Incircle of Triangle XYZ

Let $\frac{s - x}{4} = \frac{s - y}{3} = \frac{s - z}{2} = k$ for some $k > 0$.

From these equations, we can express the terms in terms of $k$: $s - x = 4k$ $s - y = 3k$ $s - z = 2k$

Summing these three equations yields: $(s - x) + (s - y) + (s - z) = 4k + 3k + 2k$ $3s - (x + y + z) = 9k$

Since $2s = x + y + z$, we substitute this into the equation: $3s - 2s = 9k \implies s = 9k$

Now, we can find the side lengths $x, y,$ and $z$ in terms of $k$: $x = s - 4k = 9k - 4k = 5k$ $y = s - 3k = 9k - 3k = 6k$ $z = s - 2k = 9k - 2k = 7k$

1. Area of Triangle $XYZ$ ($\Delta$) and Inradius ($r$)

Using Heron's formula, the area of triangle $XYZ$ is: $\Delta = \sqrt{s(s - x)(s - y)(s - z)}$ $\Delta = \sqrt{9k \cdot 4k \cdot 3k \cdot 2k} = \sqrt{216 k^4} = 6\sqrt{6}k^2$

The inradius $r$ of the triangle is given by: $r = \frac{\Delta}{s} = \frac{6\sqrt{6}k^2}{9k} = \frac{2\sqrt{6}}{3}k$

We are given that the area of the incircle is $\frac{8\pi}{3}$: $\pi r^2 = \frac{8\pi}{3}$ $\pi \left(\frac{2\sqrt{6}}{3}k\right)^2 = \frac{8\pi}{3}$ $\frac{24}{9}k^2 = \frac{8}{3} \implies \frac{8}{3}k^2 = \frac{8}{3} \implies k = 1$

Since $k = 1$, we have:

• $s = 9$
• $x = 5, y = 6, z = 7$
• $\Delta = 6\sqrt{6}$
• $r = \frac{2\sqrt{6}}{3}$

Thus, the area of the triangle $XYZ$ is $6\sqrt{6}$. Statement (A) is correct.

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2. Radius of Circumcircle ($R$)

The circumradius $R$ is given by the formula: $R = \frac{xyz}{4\Delta} = \frac{5 \cdot 6 \cdot 7}{4 \cdot 6\sqrt{6}} = \frac{35}{4\sqrt{6}} = \frac{35\sqrt{6}}{24}$

Since $\frac{35\sqrt{6}}{24} \neq \frac{35}{6}\sqrt{6}$, statement (B) is incorrect.

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3. Evaluating $\sin\frac{X}{2}\sin\frac{Y}{2}\sin\frac{Z}{2}$

Using the identity $r = 4R \sin\frac{X}{2}\sin\frac{Y}{2}\sin\frac{Z}{2}$, we get: $\sin\frac{X}{2}\sin\frac{Y}{2}\sin\frac{Z}{2} = \frac{r}{4R}$

Substitute the values of $r$ and $R$: $\sin\frac{X}{2}\sin\frac{Y}{2}\sin\frac{Z}{2} = \frac{\frac{2\sqrt{6}}{3}}{4 \cdot \frac{35\sqrt{6}}{24}} = \frac{\frac{2\sqrt{6}}{3}}{\frac{35\sqrt{6}}{6}} = \frac{2\sqrt{6}}{3} \times \frac{6}{35\sqrt{6}} = \frac{4}{35}$

Statement (C) is correct.

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4. Evaluating $\sin^2\left(\frac{X + Y}{2}\right)$

Since $X + Y + Z = \pi$, we have $\frac{X + Y}{2} = \frac{\pi}{2} - \frac{Z}{2}$. Thus: $\sin\left(\frac{X + Y}{2}\right) = \sin\left(\frac{\pi}{2} - \frac{Z}{2}\right) = \cos\left(\frac{Z}{2}\right)$

Squaring both sides and using the half-angle trigonometric formula for cosine: $\sin^2\left(\frac{X + Y}{2}\right) = \cos^2\left(\frac{Z}{2}\right) = \frac{s(s - z)}{xy}$

Substituting the known values: $\sin^2\left(\frac{X + Y}{2}\right) = \frac{9 \cdot 2}{5 \cdot 6} = \frac{18}{30} = \frac{3}{5}$

Statement (D) is correct.

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Conclusion

The correct choices are (A), (C), and (D).