Nice ques to solve

1. Aug 18, 2011

Vishalrox

Nice ques to solve !!!

Given a triangle ABC find a point M on its circumcircle such that MA=MB+MC....It is easy for an equilateral triangle... I got it... but I couldn't get how to find for any arbitrary triangle....

2. Aug 18, 2011

gsal

Re: Nice ques to solve !!!

Hhhmmm...in other words, you have 3 points on a circle and you need to find a 4th one that meets such condition.

I would first find out the angle that the radius makes to each point A,B,C; let's call these angles THA, THB, THC (TH for theta).

Then, let's have a point M and its corresponding THM.

Let's call the center of the circle O.

By taking the center of the circle, the point M and one point (A,B,C) at a time, we have 3 isosceles triangles where the base-lines (chords) of those triangles are those lines you want to match the equation MA=MB+MC. Also, it is true that MA/2 = MB/2 + MC/2.

Those base-lines are chords whose lengths are 2Sin(TH/2), where TH is the angle the two sides make at the center of the circle (radius=1).

For triangle AOM, we have one TH, THAOM = THA - THM

and

THBOM = THB - THM
THCOM = THC - THM

And so, we want:

Sin(THAOM/2) = Sin(THBOM/2) + Sin(THCOM/2)

I think it now gets a bit hairy to isolate THM from here; so, you could simply apply brut force and substitute values between 0 and 2Pi and that's it.

3. Aug 18, 2011

Yuqing

Re: Nice ques to solve !!!

The three sides of the triangle as well as the chords MA, MB and MC will form the sides and diagonals of a cyclic quadrilateral. Now if you apply Ptolemy's theorem to the quadrilateral, you should be able to obtain a condition relating the three sides. With that condition, you should be able to derive an expression for the length of either MA, MB or MC and the position of M can be found with that.

Last edited: Aug 18, 2011
4. Aug 18, 2011

DaveC426913

Re: Nice ques to solve !!!

Guys? Help him to the answer. Spoonfeeding is not allowed on PF.

5. Aug 18, 2011

Yuqing

Re: Nice ques to solve !!!

Oh my apologies, since this wasn't posted in the homework forums, I though he just wanted some form of solution. I'll edit my answer to provide hints instead.

6. Aug 19, 2011

superg33k

Re: Nice ques to solve !!!

Now we just do THAOM/2=THA/2-THM/2
Expand out the sines using sin(A-B)=sinAcosB-sinBcosA
Pull out the sin(THM/2) to one side and the cos(THM/2) to the other.
Divide for tan(THM/2)

7. Aug 21, 2011

Vishalrox

Re: Nice ques to solve !!!

i got it mysely...just say whether its r8.....Barycentric coordinates as a function of the side lengths. The circumcenter has trilinear coordinates (cos α, cos β, cos γ) where α, β, γ are the angles of the triangle. The circumcenter M has coordinates:

M = (a^2(-a^2 + b^2 + c^2), b^2(a^2 - b^2 + c^2), c^2(a^2 + b^2 - c^2))

where a, b, c are edge lengths (BC, CA, AB respectively) of the triangle.

and i found the three co-ordinates....now we can find the answer easily....!!!...is it correct.....