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(A) Let the coordinates of $A, B, C$ be $(R\cos \alpha, R\sin \alpha), (R\cos \beta, R\sin \beta)$ and $(R\cos \gamma, R\sin \gamma)$ respectively,whe

Vedclass · Accepted Answer

$144$

Vedclass · Answer

(A) Let the coordinates of $A, B, C$ be $(R\cos \alpha, R\sin \alpha), (R\cos \beta, R\sin \beta)$ and $(R\cos \gamma, R\sin \gamma)$ respectively,where $R$ is the circumradius.Since the circumcentre is at the origin $(0,0)$,the distance of $A, B, C$ from the origin is $R$.The orthocentre $H$ of $\Delta ABC$ is given by $(x_H, y_H) = (R(\cos \alpha + \cos \beta + \cos \gamma), R(\sin \alpha + \sin \beta + \sin \gamma))$.Since the orthocentre lies on the $y$-axis,the $x$-coordinate must be zero:$R(\cos \alpha + \cos \beta + \cos \gamma) = 0 \implies \cos \alpha + \cos \beta + \cos \gamma = 0$.Using the identity $\cos^3 	heta = \frac{1}{4}(\cos 3	heta + 3\cos 	heta)$,we have $\cos 3\alpha + \cos 3\beta + \cos 3\gamma = 4(\cos^3 \alpha + \cos^3 \beta + \cos^3 \gamma) - 3(\cos \alpha + \cos \beta + \cos \gamma)$.Since $\cos \alpha + \cos \beta + \cos \gamma = 0$,we use the property that if $a+b+c=0$,then $a^3+b^3+c^3 = 3abc$.Thus,$\cos^3 \alpha + \cos^3 \beta + \cos^3 \gamma = 3 \cos \alpha \cos \beta \cos \gamma$.Substituting this into the expression:$\frac{\cos 3\alpha + \cos 3\beta + \cos 3\gamma}{\cos \alpha \cos \beta \cos \gamma} = \frac{4(3 \cos \alpha \cos \beta \cos \gamma) - 3(0)}{\cos \alpha \cos \beta \cos \gamma} = 12$.Finally,the square of this value is $12^2 = 144$.

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