The bisectors of triangle ABC meet its circumcircle in points D,E,F respectively show that the area of Triangle ABC cannot be greater then that of triangle DEF

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- December 17th 2010, 01:35 AMprasumtrigo
The bisectors of triangle ABC meet its circumcircle in points D,E,F respectively show that the area of Triangle ABC cannot be greater then that of triangle DEF

- December 24th 2010, 09:40 AMprasum
how should i do it

- December 24th 2010, 11:38 AMOpalg
The area of triangle ABC is , where R is the radius of the circumcircle. Since triangle DEF has the same circumcircle as ABC, its area is . So we want to show that . But the angles D, E, F are equal to (as you can see by drawing a picture and looking for angles in the same segment). So we want to prove that .

Since all these sines are positive, this is equivalent to the inequality . But we can split that up as the product of three inequalities (and two other, similar, inequalities).

I'm sure there are several ways to prove that last inequality. One way would be to take logs and use the fact that the function is concave for . But I expect that there is also a purely trigonometric proof. - December 25th 2010, 03:32 AMprasum
- December 28th 2010, 05:41 AMArchie Meade
Attachment 20270Attachment 20271Attachment 20272Attachment 20273

We can obtain the vertex angles for triangle as shown in the attached images.

In Fig.2, using the common chord

Using the common chord

Similarly for the other angles in Fig.3 and Fig.4.

The angles of triangle are therefore

while the angles of triangle are

The question becomes

This inequality holds if the following 3 inequalities hold

- December 28th 2010, 06:47 AMOpalg
Archie Meade's comment shows how I could have completed the proof of this result. In comment #3 above, I showed how the proof could be reduced to proving inequalities of the form . If we let and then that inequality becomes . But that is equivalent to , which is obviously true.

- December 28th 2010, 08:41 AMprasum
how did yu get those three inequalities

- December 28th 2010, 09:53 AMArchie Meade
When we had

we were left with (due to the common terms)

Now we split into 3 of

and compare each of these to the components on the right-hand side of the inequality.

For the first of the 3 inequalities

Dropping the common factor gives

and similarly for the other two. - December 29th 2010, 03:39 AMprasum
thanks