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December 17th 2010, 01:35 AM
prasum

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
December 24th 2010, 09:40 AM
prasum

how should i do it
December 24th 2010, 11:38 AM
Opalg

Quote:

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.
December 25th 2010, 03:32 AM
prasum

Quote:

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
December 28th 2010, 05:41 AM
Archie Meade

4 Attachment(s)

Quote:

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

Attachment 20270 Attachment 20271 Attachment 20272 Attachment 20273

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

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

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

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

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

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

$(3)\;\;\;(sinXcosY-sinYcosX)^2\ \ge\ 0\;\;?\;\;\;Yes$
December 28th 2010, 06:47 AM
Opalg

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.
December 28th 2010, 08:41 AM
prasum

how did yu get those three inequalities
December 28th 2010, 09:53 AM
Archie Meade

Quote:

Originally Posted by prasum

how did you get those three inequalities ?

When we had

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

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.
December 29th 2010, 03:39 AM
prasum

thanks