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  • The circum-centre of triangle with vertices <img alt="\mathrm{\mathrm{A}(\mathrm{a}, \mathrm{a} \tan \alpha)\ \mathrm{B}(\mathrm{b}, \mathrm{b} \tan \beta)\ \text { and }\ \mathrm{C}(\mathrm{c

The circum-centre of triangle with vertices \mathrm{\mathrm{A}(\mathrm{a}, \mathrm{a} \tan \alpha)\ \mathrm{B}(\mathrm{b}, \mathrm{b} \tan \beta)\ \text { and }\ \mathrm{C}(\mathrm{c}, \mathrm{c} \tan \gamma)} lies at the origin, where \mathrm{0<\alpha, \beta, \gamma<\pi / 2 \text { and } \alpha+\beta+\gamma=\pi.} Its orthocenter lies on the line \mathrm{k_1\left(\cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos \frac{\gamma}{2}\right) x-k_2\left(\sin \frac{\alpha}{2} \sin \frac{\beta}{2} \sin \frac{\gamma}{2}\right) y=y} , where \mathrm{k_{1}-k_{2}=}

Option: 1

\mathrm{0}


Option: 2

\mathrm{1}


Option: 3

\mathrm{2}


Option: 4

\mathrm{4}


Answers (1)

best_answer

As the circum-centre of the triangle is at the origin \mathrm{O}, we have \mathrm{\mathrm{OA}=\mathrm{OB}=\mathrm{OC}=\mathrm{r},} where \mathrm{{r}} is the radius of the circum-circle.
\mathrm{\therefore \mathrm{OA}^2=\mathrm{r}^2\ \ \Rightarrow \mathrm{a}^2+\mathrm{a}^2 \tan ^2 \alpha=\mathrm{r}^2\ \ \ \ \ \quad \Rightarrow \mathrm{a}=\mathrm{r} \cos \alpha}
Therefore, the coordinates of \mathrm{A} are \mathrm{(r\ \cos\ \alpha,\ r\ \sin\ \alpha).} Similarly, the coordinates of \mathrm{B} are \mathrm{(r\ cos\ \beta\ , r\ sin\ \beta ) } and those of \mathrm{C} are \mathrm{(r\ cos\ \gamma ,r\ sin\ \gamma )}. Thus, the coordinates of the Centeroid \mathrm{G} of \mathrm{\triangle ABC} are
\mathrm{\left({ }^{\frac{1}{3}} \mathrm{r}(\cos \alpha+\cos \beta+\cos \gamma),{ }^{\frac{1}{3}} \mathrm{r}(\sin \alpha+\sin \beta+\sin \gamma)\right)}
Now, if \mathrm{P\left ( h,k \right )} is the orthocenter of \mathrm{\triangle ABC} , then, from geometry, the circum-centre, Centeroid and orthocenter of a triangle lie on a line, and the slope of \mathrm{OG} equals the slope of \mathrm{OP}. 
\mathrm{\therefore \frac{\sin \alpha+\sin \beta+\sin \gamma}{\cos \alpha+\cos \beta+\cos \gamma}=\frac{k}{h} \Rightarrow \frac{4 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos \frac{\gamma}{2}}{1+4 \sin \frac{\alpha}{2} \sin \frac{\beta}{2} \sin \frac{\gamma}{2}}=\frac{k}{h}}
(because \mathrm{\alpha + \beta + \gamma = \pi }). Hence the orthocenter \mathrm{P\left ( h,k \right )} lies on the line 
\mathrm{4\left(\cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos \frac{\gamma}{2}\right) \mathrm{x}-4\left(\sin \frac{\alpha}{2} \sin \frac{\beta}{2} \sin \frac{\gamma}{2}\right) \mathrm{y}=\mathrm{y} }

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