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$ \bigtriangleup ABC $is inscribed in a unit circle.If angle bisectors of internal angles at A,B and C meet the circle at D,E and F respectively then value of $\frac{AD \cos\frac{A}{2}+BE \cos\frac{B}{2}+CF \cos\frac{C}{2}}{sin A+sinB+ sinC}$is ?

My attempt:AD,BE,CF are angle bisectors,so they meet at incenter of triangle but i dont know whether this incenter will be the center of unit circle or not.If this incenter is not center of unit circle,then i have no clue to proceed.

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2 Answers 2

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The internal angle bisector of $\widehat{BAC}$ meet the circumcircle of $ABC$ in the midpoint of the $BC$ arc. So, if $O$ is the circumcenter of $ABC$, we have $\widehat{AOD}=\widehat{AOB}+\frac{1}{2}\widehat{BOC}=2\widehat{C}+\widehat{A}$ and:

$$ AD\cos\frac{\widehat{A}}{2} = 2R \sin\left(\frac{1}{2}\widehat{AOD}\right)\cos\frac{\widehat{A}}{2}=\sin(\widehat{C}+\widehat{A})+\sin(\widehat{C})$$ from which $AD\cos\frac{\widehat{A}}{2}=\sin(\widehat{B})+\sin(\widehat{C})$ and the wanted ratio is just $\displaystyle\color{red}{2}$.

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  • $\begingroup$ how can we draw shapes in the site? $\endgroup$
    – Sepideh Abadpour
    Jul 27, 2015 at 13:25
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    $\begingroup$ @sepideh: I draw them using Geogebra, export the figures in the png format and import them here. $\endgroup$
    – Jack D'Aurizio
    Jul 27, 2015 at 13:29
  • $\begingroup$ Sir,how is $AD=2R sin(\frac{1}{2}\angle AOD)$?I could not understand. $\endgroup$
    – Vinod Kumar Punia
    Jul 27, 2015 at 13:43
  • $\begingroup$ By sine rule or some other rule. $\endgroup$
    – Vinod Kumar Punia
    Jul 27, 2015 at 13:44
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    $\begingroup$ @VinodKumarPunia: in a circle with centre $O$ and radius $R$, the length of a chord $AB$ is given by $2R\sin\frac{\widehat{AOB}}{2}$, simple trigonometry. $\endgroup$
    – Jack D'Aurizio
    Jul 27, 2015 at 13:59
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\begin{align} x&=\frac{|AD| \cos\tfrac{\alpha}{2} +|BE|\cos\tfrac{\beta}{2} +|CF|\cos\tfrac{\gamma}{2}}{\sin\alpha+\sin\beta+\sin\gamma} =?\quad (1) \end{align}

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Consider $\triangle ADC$. We know that it shares the circumradius $R=1$ and the side $|AC|$ with $\triangle ABC$, and we also know all the angles: $\angle CAD=\tfrac\alpha2$, $\angle ADC=\beta$ (why?) and $\angle ACD=\pi-\tfrac\alpha2-\beta$. Hence, by the law of sines,

\begin{align} |AD|&= 2R \sin(\pi-\tfrac\alpha2-\beta) = 2R\sin(\tfrac\alpha2+\beta). \end{align} Similarly, from $\triangle BEA$ and $\triangle CFB$,

\begin{align} |BE|&= 2R\sin(\tfrac\beta2+\gamma), \\ |CF|&= 2R\sin(\tfrac\gamma2+\alpha). \end{align}

Numerator of (1) is then \begin{align} &|AD|\cos\tfrac\alpha2 +|BE|\cos\tfrac\beta2 +|CF|\cos\tfrac\gamma2 \\ &=2R\left( \sin(\tfrac\alpha2+\beta)\cos\tfrac\alpha2 + \sin(\tfrac\beta2+\gamma)\cos\tfrac\beta2 + \sin(\tfrac\gamma2+\alpha)\cos\tfrac\gamma2 \right) \\ &= 2R\left( \tfrac12\sin(\alpha+\beta)+\tfrac12\sin(\beta) + \tfrac12\sin(\beta+\gamma)+\tfrac12\sin(\gamma) + \tfrac12\sin(\gamma+\alpha)+\tfrac12\sin(\alpha) \right) \\ &= 2R\left( \tfrac12\sin\gamma+\tfrac12\sin\beta + \tfrac12\sin\alpha+\tfrac12\sin\gamma + \tfrac12\sin\beta+\tfrac12\sin\alpha \right) \\ &= 2R(\sin\alpha+\sin\beta+\sin\gamma). \end{align} Hence, the answer is $x=2R=2$.

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