<B=? AP+CM=AC, AM,CP angle bisectors (2020 All-Siberian Qualifying 10.4)

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<B=? AP+CM=AC, AM,CP angle bisectors (2020 All-Siberian Qualifying 10.4)

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#1 h May 18, 2021, 9:13 PM

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In the triangle $ABC$ , the segments $AM$ and $CP$ are the bisectors of the angles $A$ and $C$ respectively, with $AP+CM=AC$ . Find the angle $B$ .

This post has been edited 1 time. Last edited by parmenides51, May 18, 2021, 9:13 PM

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#2 h May 19, 2021, 4:09 PM • 3 Y

Y by Mango247, Mango247, Mango247

Rule of sines in $\triangle ACP\ :\ \frac{AP}{\sin \frac{\gamma}{2}}=\frac{AC}{\sin(\alpha+\frac{\gamma}{2})}$ .
Rule of sines in $\triangle ACM\ :\ \frac{CM}{\sin \frac{\alpha}{2}}=\frac{AC}{\sin(\frac{\alpha}{2}+\gamma)}$ .

We add: $AP=\frac{\sin \frac{\gamma}{2} \cdot AC}{\sin(\alpha+\frac{\gamma}{2})}$ and $CM=\frac{\sin \frac{\alpha}{2} \cdot AC}{\sin(\frac{\alpha}{2}+\gamma)}$ .
$AP+CM=\frac{\sin \frac{\gamma}{2} \cdot AC}{\sin(\alpha+\frac{\gamma}{2})} + \frac{\sin \frac{\alpha}{2} \cdot AC}{\sin(\frac{\alpha}{2}+\gamma)}=AC$ ,
$\frac{\sin \frac{\gamma}{2}}{\sin(\alpha+\frac{\gamma}{2})} + \frac{\sin \frac{\alpha}{2}}{\sin(\frac{\alpha}{2}+\gamma)}=1$ ,
$\sin \frac{\gamma}{2}\sin(\frac{\alpha}{2}+\gamma)+\sin \frac{\alpha}{2}\sin(\alpha+\frac{\gamma}{2})=\sin(\alpha+\frac{\gamma}{2})\sin(\frac{\alpha}{2}+\gamma)$ .

$\cos \frac{\alpha+\gamma}{2}-\cos \frac{\alpha+3\gamma}{2}+\cos \frac{\alpha+\gamma}{2}-\cos \frac{3\alpha+\gamma}{2}=\cos \frac{\alpha-\gamma}{2}-\cos \frac{3\alpha+3\gamma}{2}$ .
$\cos \frac{\alpha+\gamma}{2}+\cos \frac{3\alpha+3\gamma}{2}-[\cos \frac{3\alpha+3\gamma}{2}+\cos \frac{3\alpha+\gamma}{2}]+\cos \frac{\alpha+\gamma}{2}-\cos \frac{\alpha-\gamma}{2}=0$ .

$2\cos(\alpha+\gamma)\cos \frac{\alpha+\gamma}{2}-2\cos(\alpha+\gamma)\cos \frac{\gamma-\alpha}{2}+\cos \frac{\alpha+\gamma}{2}-\cos \frac{\alpha-\gamma}{2}=0$ .
$2\cos(\alpha+\gamma)[\cos \frac{\alpha+\gamma}{2}-\cos \frac{\gamma-\alpha}{2}]+\cos \frac{\alpha+\gamma}{2}-\cos \frac{\alpha-\gamma}{2}=0$ .
$2\cos(\alpha+\gamma)+1=0$ .
$\alpha+\gamma=120^{\circ}$ and then $\beta=60^{\circ}$ .

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