Let triangle ABC be an acute angle triangle. The circumcircle of triangle ABC is drawn.

Let X, Y, Z be the midpoints of the minor arcs of AB, BC, CA respectively. Prove that the area of hexagon AXBYCZ is at least twice the area of triangle ABC.

Is this also true when X, Y, Z are any points on the minor arcs, without being the midpoints?

I thought this was going to be easy.  It isn't turning out that way.

We can make the circumcircle have radius 1.  Call the angles of the triangle A, B, 180-A-B.  The circumcenter, O, then has 3 radii to the corners of the triangle which is split into 3 smaller isosceles triangles.
The total area of triangle ABC is then
.5(sin(2A)+sin(2B)-sin(360-2A-2B)
which after many identities simplifies to
2sin(A)sin(B)sin(A+B)

The hexagon is similarly split into 6 isosceles triangles (3 congruent pairs.)  The total area of the hexagon simplifies to
sin(A) + sin(B) + sin(A+B)

What we are to prove then is
4sin(A)sin(B)sin(A+B) < sin(A) + sin(B) + sin(A+B)
for A<90, B<90, A+B>90

Graphical evidence seems to indicate this is true.  Proving it is eluding me.

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The second question is a simple no.  By choosing X, Y and Z very close to the corners of the triangle we can make the hexagon have an area only very slightly bigger than the triangle.

Posted by Jer on 2013-04-09 15:29:03