If the line x=y=z intersects the line x A+y B+z C-18=0=x2A+y2B+z2C-9, where A,B,C are the …

Question

If the line $x=y=z$ intersects the line $x\sin A+y\sin B+z\sin C-18=0=x\sin2A+y\sin2B+z\sin2C-9$, where $A,B,C$ are the angles of a triangle $ABC$, then $80\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}$ is equal to:

Answer: 5

Accepted Answer

Since $x=y=z=k$ lies on both planes: $\sin A+\sin B+\sin C=\dfrac{18}{k}$ and $\sin2A+\sin2B+\sin2C=\dfrac9k$.
$$\sin A+\sin B+\sin C=2(\sin2A+\sin2B+\sin2C).$$
Using $\sin A+\sin B+\sin C=4\cos\tfrac{A}2\cos\tfrac{B}2\cos\tfrac{C}2$ and $\sin2A+\sin2B+\sin2C=4\sin A\sin B\sin C$:
$$4\cos\tfrac{A}2\cos\tfrac{B}2\cos\tfrac{C}2=8\sin A\sin B\sin C=64\sin\tfrac{A}2\sin\tfrac{B}2\sin\tfrac{C}2\cos\tfrac{A}2\cos\tfrac{B}2\cos\tfrac{C}2.$$
$$\sin\frac{A}2\sin\frac{B}2\sin\frac{C}2=\frac{1}{16}.$$
$$80\sin\frac{A}2\sin\frac{B}2\sin\frac{C}2=\frac{80}{16}=5.$$

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