computational with tangent of circles (Lomonosov Tournament 2019 p3)

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computational with tangent of circles (Lomonosov Tournament 2019 p3)

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#1 h Apr 21, 2020, 10:19 AM • 3 Y

Y by Mango247, Mango247, Mango247

Two non intersecting circles of radius $1$ and $3$ are inscribed in the angle $\angle POQ$ , where $P$ is the point of contact of the side of the corner with the first circle, and $Q$ is the point of contact of the other side of the angle with the second circle. The common internal tangent of circles $AB$ intersects the ray $OP$ at point $A$ , and the segment $OQ$ at point $B$ . Find $AB$ if $OP = 3$ .

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#2 h May 12, 2020, 6:51 PM

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$\triangle OO_{1}P \sim \triangle OO_{2}Q' \Rightarrow OQ'=OQ=9$ .
$O_{1}(3,1)$ and $O_{2}(9,3)$ , then $O_{1}O_{2}=2\sqrt{10}$ .
$\triangle CO_{1}S \sim \triangle DO_{2}S \Rightarrow SO_{1}=\frac{\sqrt{10}}{2},SO_{2}=\frac{3\sqrt{10}}{2},SC=\frac{\sqrt{6}}{2},SD=\frac{3\sqrt{6}}{2}$
$CD=2\sqrt{6}$ .

$\triangle OO_{1}P\ :\ \sin \angle OO_{2}P=\sin \beta=\frac{3}{\sqrt{10}}$ .
$\triangle CO_{1}S\ :\ \sin \angle CO_{1}S=\sin \gamma=\frac{\sqrt{6}}{\sqrt{10}}$ .
$\sin \angle CO_{1}P=\sin(2\alpha)=\sin(\beta+\gamma)=\frac{6+\sqrt{6}}{10}$ .

$\sin(2\alpha)=\frac{2\tan \alpha}{1+\tan^{2}\alpha}=\frac{6+\sqrt{6}}{10} \Rightarrow \tan \alpha=3-\sqrt{6}$ .
$\triangle ACO_{1}\ :\ AC=\tan \alpha=3-\sqrt{6}$ .
Idem $BD=3-\sqrt{6}$ .
$AB=6$ .

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