2
$\begingroup$

$ \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.

$\endgroup$
4
$\begingroup$

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 $\displaystyle\color{red}{2}$.

enter image description here

$\endgroup$
  • $\begingroup$ how can we draw shapes in the site? $\endgroup$ – Sepideh Abadpour Jul 27 '15 at 13:25
  • 2
    $\begingroup$ @sepideh: I draw them using Geogebra, export the figures in the png format and import them here. $\endgroup$ – Jack D'Aurizio Jul 27 '15 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 '15 at 13:43
  • $\begingroup$ By sine rule or some other rule. $\endgroup$ – Vinod Kumar Punia Jul 27 '15 at 13:44
  • 2
    $\begingroup$ @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 '15 at 13:59
1
$\begingroup$

\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}

enter image description here

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 $x=2R=2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.