A point $O$ is choosen inside a triangle $ABC$ so that the length of segments $OA$, $OB$ and $OC$ are equal to $15$,$12$ and $20$, respectively. It is known that the feet of the perpendiculars from $O$ to the sides of the triangle $ABC$ are the vertices of an equilateral triangle. Find the value of the angle $BAC$.
Problem
Source: 2018 Belarusian National Olympiad 9.7
Tags: geometry
Pluto1708
29.03.2019 20:02
Since $O$ is the first isodynamic point hence $15:12:20=OA:OB:OC=AB.AC:BA.BC:CA.CB \implies BC:CA:AB=4:5:3$.So $\angle BAC=53^{\circ}$
Wiz-of-Oz
30.03.2019 12:11
Let the feet of perpendiculars be $X,Y,Z$ on $BC,CA,AB$ $OB$ is diameter of circumcircle of cyclic quadrilateral $OZBX$ By sine rule, $XZ = 2R \sin \angle ZOX = OB \sin B$ Since $XZ = YZ = XY$ We have $15 \sin A = 12 \sin B = 20 \sin C$ So, $\angle BAC = \sin ^{-1} \frac {4}{5} $
Wiz-of-Oz
30.03.2019 12:14
Pluto1708 wrote: Since $O$ is the first isodynamic point hence $15:12:20=OA:OB:OC=AB.AC:BA.BC:CA.CB \implies BC:CA:AB=4:5:3$.So $\angle BAC=53^{\circ}$ It's not $53 ^{\circ}$ but a little greater ($53.13...^{\circ}$)
RC.
30.03.2019 13:33
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