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(D) In a triangle $ABC$,$I$ is the incenter and $I_1, I_2, I_3$ are the excenters opposite to vertices $A, B, C$ respectively.It is a known property t

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$16R^2r$

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(D) In a triangle $ABC$,$I$ is the incenter and $I_1, I_2, I_3$ are the excenters opposite to vertices $A, B, C$ respectively.It is a known property that $I I_1 = 4R \sin(\frac{A}{2})$,$I I_2 = 4R \sin(\frac{B}{2})$,and $I I_3 = 4R \sin(\frac{C}{2})$.Therefore,the product is:$(I I_1) \cdot (I I_2) \cdot (I I_3) = (4R \sin \frac{A}{2}) \cdot (4R \sin \frac{B}{2}) \cdot (4R \sin \frac{C}{2})$$= 64R^3 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$Using the identity $r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$,we can write:$64R^3 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = 16R^2 (4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}) = 16R^2r$.

With usual notations in a triangle $ABC$,the product $(I I_1) \cdot (I I_2) \cdot (I I_3)$ has the value equal to

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