Question

Let A(6,8), B(10 cos $\alpha$, −10 sin $\alpha$) and C(−10 sin $\alpha$, 10cos $\alpha$), 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 sin 2$\alpha$) is equal to _____.

Answer 1

145

Answer 2

The origin O(0,0) is the circumcenter because OA=OB=OC=10. For a circumcenter at the origin, the orthocenter H satisfies $\vec{OH} = \vec{OA} + \vec{OB} + \vec{OC}$. Equating y-coordinates of L(a,9): $9 = 8 - 10\sin\alpha + 10\cos\alpha \Rightarrow 1 = 10(\cos\alpha - \sin\alpha)$. Squaring gives $1 - \sin(2\alpha) = 1/100 \Rightarrow 100\sin(2\alpha) = 99$. Equating x-coordinates: $a = 6 + 10(\cos\alpha - \sin\alpha) = 6+1=7$. Centroid G(h,k) is the average of vertices: $h = (6+1)/3 = 7/3$ and $k = (8+1)/3 = 3$. The expression $5a - 3h + 6k + 100 \sin 2\alpha = 5(7) - 3(7/3) + 6(3) + 99 = 35 - 7 + 18 + 99 = 46 + 99 = 145$.