Let $A(6,8)$, $B(10\cos\alpha,-10\sin\alpha)$ and $C(-10\sin\alpha,10\cos\alpha)$, be the vertices of a triangle. If $L(a,9)$ and $G(h,k)$ be its orthocenter and centroid respectively, then $(5a-3h+6k+100\sin2\alpha)$ is equal to _______ -.

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $A,B,C$ all lie on the circle $x^2+y^2=100$, so the circumcentre is the origin and the orthocentre is $A+B+C=(6+10\cos\alpha-10\sin\alpha,\ 8+10\cos\alpha-10\sin\alpha)$. Its $y$-coordinate is $9$, so $10(\cos\alpha-\sin\alpha)=1$ and $a=7$; also $\sin2\alpha=1-0.01=0.99$. The centroid is $\frac13$ of the orthocentre: $(h,k)=\left(\frac73,3\right)$. Value $=35-7+18+99=145$. Answer: 145.

Q. Let $A(6,8)$, $B(10\cos\alpha,-10\sin\alpha)$ and $C(-10\sin\alpha,10\cos\alpha)$, be the vertices of a triangle. If $L(a,9)$ and $G(h,k)$ be its orthocenter and centroid respectively, then $(5a-3h+6k+100\sin2\alpha)$ is equal to _______ -.

Explanation

Reviewed by Dr. Abubacker

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