Aditya Gupta
Last Activity: 5 Years ago
there is a slight mistake in ques, but anyways we can correct it.
we see that a ray y= tan(q) will intersect the circle x^2+y^2= a^2 at (acosq, asinq) which can be obtained by solving the two eqns together.
so, A(acosq1, asinq1); B(acosq2, asinq2); C(acosq3, asinq3)
now, we know that the circumcentre of ABC is obviously origin (0, 0). its centroid is [a(cosq1 + cosq2 + cosq3)/3, a(sinq1 + sinq2 + sinq3)/3].
now, we use the fact that eulers line passes through orthocentre, centroid and circumcentre dividing them in ratio 2:1.
let coordinates of orthocentre be (c, d). then by section formula
a(cosq1 + cosq2 + cosq3)/3 = (2*0+1*c)/(2+1)= c/3
or a(cosq1 + cosq2 + cosq3)= c......(1)
similarly, a(sinq1 + sinq2 + sinq3)/3= (2*0+1*d)/(2+1)= d/3
or a(sinq1 + sinq2 + sinq3)= d.......(2)
it is pretty obvious to see that (c, d) always lies on the line y/x = (sinq1 + sinq2 + sinq3)/(cosq1 + cosq2 + cosq3)
or y= (∑sinqi)x/∑cosqi
kindly approve :=)