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  1. #1
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    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
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    how should i do it
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  3. #3
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    Quote Originally Posted by prasum View Post
    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.
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    Quote Originally Posted by Opalg View Post
    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
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  5. #5
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    Quote Originally Posted by prasum View Post
    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
    trigo-circumcircle-1.jpgtrigo-circumcircle-2.jpgtrigo-circumcircle-3.jpgtrigo-circumcircle-4.jpg

    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
    Last edited by Archie Meade; December 28th 2010 at 05:07 PM. Reason: images upload fault corrected
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    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.
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    how did yu get those three inequalities
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  8. #8
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    Quote Originally Posted by prasum View Post
    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.
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    thanks
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