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Source: 2019 VU MIF Olympiad XI, XI p3 - Vilnius University, Lithuania

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#1 h Oct 4, 2020, 10:54 AM

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The point $C \ne A, B$ belongs to a circle of diameter $AB$ . Point $D$ divides the shorter arc $BC$ in half. Lines $AD$ and $BC$ intersects at point $E$ . Find the length of the segment $AB$ if $CE = 3$ , $BD = 2\sqrt5$ .

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#2 h Nov 27, 2020, 5:02 PM

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Oh, d*mn, I wonder is there an easier solution. I actually hoped that you do not have to use trig, so I looked like dumb trying angle-bisector and Ptolemy and stuff.

$\sin{\angle DAB}=\frac{DB}{2R}$
$\sin{(90^\circ-2\angle DAB)}=\frac{AC}{2R}$
$\tan{(90^\circ-\angle DAB)}=\frac{AC}{CE}$
$\sin{(90^\circ-2\angle DAB)}=\frac{\tan{(90^\circ-\angle DAB)}CE}{2R}$
$\sin{(90^\circ-2\angle DAB)}=\frac{\sin{(90^\circ-\angle DAB)}CE}{\cos{(90^\circ-\angle DAB)}2R}=\frac{\sin{(90^\circ-\angle DAB)}CE}{\sin{\angle DAB}\cdot 2R}$ $\cos{2\angle DAB}=\frac{\cos{\angle DAB} \cdot CE}{DB}$ $2\sqrt5(2\cos^2{\angle DAB}-1)=3\cos{\angle DAB}$
$4\sqrt5 \cos^2{\angle DAB}-3\cos{\angle DAB}-2\sqrt5=0$
$\cos{\angle DAB}=\frac{3\pm\sqrt{9-4\cdot (-2\sqrt5)4\sqrt5 }}{2\cdot 4\sqrt5}=\frac{3\pm13}{2\cdot 4\sqrt5}=\frac{3\pm13}{2\cdot 4\sqrt5}$ .
Negative value is false, since then $\angle DAB>90^\circ$ , which means that $\angle CAB>180^\circ$ , which is false since max $\angle CAB$ value can be $\to 90^\circ$ (approaching to it) .
Hence, $\cos{\angle DAB}=\frac{2\sqrt5}{5}$
Let $F=BC\cap OD$ , where $O$ is the centre of that circle.
Thus, $\frac{BF}{BD}=\cos{\angle DAB}=\frac{2\sqrt5}{5}\implies BF=4$ .
Thus, $BC=8$ .
And we have that $\cos{\angle CAB}=\cos{2\angle DAB}=2\cos^2{\angle DAB}-1=\frac{8\cdot 5}{25}-1=\frac{15}{25}=\frac{3}{5}$ .
$\frac{AC}{AB}=\cos{\angle CAB}=\frac{3}{5}\implies AC=\frac{3\cdot AB}{5}$
By Pythagorean theorem, $AC^2+BC^2=AB^2$ , we get that $\frac{9\cdot AB^2}{25}+64=AB^2\implies \frac{16\cdot AB^2}{25}=64\implies AB^2=100\implies \boxed{AB=10}.$

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