1. ## trigo

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

2. how should i do it

3. Originally Posted by prasum
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
The area of triangle ABC is $2R^2\sin A\sin B\sin C$, where R is the radius of the circumcircle. Since triangle DEF has the same circumcircle as ABC, its area is $2R^2\sin D\sin E\sin F$. So we want to show that $\sin D\sin E\sin F \geqslant \sin A\sin B\sin C$. But the angles D, E, F are equal to $\frac{B+C}2,\ \frac{C+A}2,\ \frac{A+B}2$ (as you can see by drawing a picture and looking for angles in the same segment). So we want to prove that $\sin \frac{B+C}2\sin \frac{C+A}2\sin\frac{A+B}2 \geqslant \sin A\sin B\sin C$.

Since all these sines are positive, this is equivalent to the inequality $\sin^2 \frac{B+C}2\sin^2 \frac{C+A}2\sin^2\frac{A+B}2 \geqslant \sin^2 A\sin^2 B\sin^2 C$. But we can split that up as the product of three inequalities $\sin^2 \frac{B+C}2\geqslant \sin B\sin C$ (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 $\log\sin x$ is concave for $0\leqslant x\leqslant \pi$. But I expect that there is also a purely trigonometric proof.

4. Originally Posted by Opalg
The area of triangle ABC is $2R^2\sin A\sin B\sin C$, where R is the radius of the circumcircle. Since triangle DEF has the same circumcircle as ABC, its area is $2R^2\sin D\sin E\sin F$. So we want to show that $\sin D\sin E\sin F \geqslant \sin A\sin B\sin C$. But the angles D, E, F are equal to $\frac{B+C}2,\ \frac{C+A}2,\ \frac{A+B}2$ (as you can see by drawing a picture and looking for angles in the same segment). So we want to prove that $\sin \frac{B+C}2\sin \frac{C+A}2\sin\frac{A+B}2 \geqslant \sin A\sin B\sin C$.

Since all these sines are positive, this is equivalent to the inequality $\sin^2 \frac{B+C}2\sin^2 \frac{C+A}2\sin^2\frac{A+B}2 \geqslant \sin^2 A\sin^2 B\sin^2 C$. But we can split that up as the product of three inequalities $\sin^2 \frac{B+C}2\geqslant \sin B\sin C$ (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 $\log\sin x$ is concave for $0\leqslant x\leqslant \pi$. But I expect that there is also a purely trigonometric proof.
thanks

5. Originally Posted by prasum
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

We can obtain the vertex angles for triangle $D$ $E$ $F$ as shown in the attached images.

In Fig.2, using the common chord $BE,\;\;|\angle\ BCE|=|\angle\ BFE|=Z$

Using the common chord $CF,\;\;|\angle\ CBF|=|\angle\ CEF|=Y$

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

The angles of triangle $D$ $E$ $F$ are therefore $Y+Z,\;\;X+Y,\;\;X+Z$

while the angles of triangle $ABC$ are $2X,\;\;2Y,\;\;2Z.$

$R=circumcircle\ radius$

$Area\ \triangle\ ABC=2R^2sin2Xsin2Ysin2Z$

$Area\ \triangle\$ $D$ $E$ $F$ $=2R^2sin(Y+Z)sin(X+Y)sin(X+Z)$

The question becomes

$(sin2X)(sin2Y)(sin2Z)\ \le\ sin(Y+Z)sin(X+Y)sin(X+Z)\;\;?$

$(2sinXcosX)(2sinYcosY)(2sinZcosZ)$

$\le\ (sinYcosZ+sinZcosY)(sinXcosY+sinYcosX)(sinXcosZ+si nZcosX)\;\;?$

$8sinXsinYsinZcosXcosYcosZ$

$\le\ (sinXsinYcosYcosZ+sin^2YcosXcosZ+)(sinXcosZ+sinZco sX)$

$+(sinXsinZcos^2Y+sinYsinZcosXcosY)(sinXcosZ+sinZco sX)\;\;?$

$8sinXsinYsinZcosXcosYcosZ$

$\le\ sin^2XsinYcosYcos^2Z+sinXsinYsinZcosXcosYcosZ$

$+sinXsin^2YcosXcos^2Z+sin^2YsinZcos^2XcosZ$

$+sin^2XsinZcos^2YcosZ+sinXsin^2ZcosXcos^2Y$

$+sinXsinYsinZcosXcosYcosZ+sinYsin^2Zcos^2XcosY\;\; ?$

$6sinXsinYsinZcosXcosYcosZ$

$\le\ sinXcosX\left(sin^2Zcos^2Y+sin^2Ycos^2Z\right)$

$+sinYcosY\left(sin^2Xcos^2Z+sin^2Zcos^2X\right)$

$+sinZcosZ\left(sin^2Xcos^2Y+Sin^2Ycos^2X\right)$

This inequality holds if the following 3 inequalities hold

$(1)\;\;\;2sinYsinZcosYcosZ\ \le\ (sinZcosY)^2+(sinYcosZ)^2$

$(2)\;\;\;2sinXsinZcosXcosZ\ \le\ (sinXcosZ)^2+(sinZcosX)^2$

$(3)\;\;\;2sinXsinYcosXcosY\ \le\ (sinXcosY)^2+(sinYcosX)^2$

$(1)\;\;\;(sinZcosY-sinYcosZ)^2\ \ge\ 0\;\;?\;\;\;Yes$

$(2)\;\;\;(sinXcosZ-sinZcosX)^2\ \ge\ 0\;\;?\;\;\;Yes$

$(3)\;\;\;(sinXcosY-sinYcosX)^2\ \ge\ 0\;\;?\;\;\;Yes$

6. 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 $\sin^2 \frac{B+C}2\geqslant \sin B\sin C$. If we let $\beta = B/2$ and $\gamma = C/2$ then that inequality becomes $(\sin\beta\cos\gamma+\cos\beta\sin\gamma)^2 \geqslant 4\sin\beta\cos\beta\sin\gamma\cos\gamma$. But that is equivalent to $(\sin\beta\cos\gamma-\cos\beta\sin\gamma)^2 \geqslant 0$, which is obviously true.

7. how did yu get those three inequalities

8. Originally Posted by prasum
how did you get those three inequalities ?

$8sinXsinYsinZcosXcosYcosZ\ \le\ 2sinXsinYsinZcosXcosYcosZ+........\;\;?$

we were left with (due to the common terms)

$6sinXsinYsinZcosXcosYcosZ$

$\le\ sinXcosX\left(sin^2Zcos^2Y+sin^2Ycos^2Z\right)$

$+sinYcosY\left(sin^2Xcos^2Z+sin^2Zcos^2X\right)$

$+sinZcosZ\left(sin^2Xcos^2Y+sin^2Ycos^2X\right)\;\ ;?$

Now we split $6sinXsinYsinZcosXcosYcosZ$ into 3 of $2sinXsinYsinZcosXcosYcosZ$

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

For the first of the 3 inequalities

$sinXcosX(2sinYsinZcosYcosZ)\;\le\ sinXcosX\left(sin^2Zcos^2Y+sin^2Ycos^2Z\right)\;\; ?$

Dropping the common factor gives

$2sinYsinZcosYcosZ\ \le\ sin^2Zcos^2Y+sin^2Ycos^2Z\;\;?$

and similarly for the other two.

9. thanks