Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.6 National olympiad Prove it Bulgaria

Points $A$ , $B$ , $Y$ and $C$ lie in this order on circle $k$ with center $O$ , such that $BC = 2$ cm, $\angle BAY = 42^\circ$ and $\angle CAY = 78^\circ$ . It is known that the circle $\omega$ through the points $A$ , $O$ and $B$ is tangent to the line $BY$ .

The circle through the points $A$ and $C$ , tangent to the line $CY$ , intersects $\omega$ for second time at the point $N$ . To be found:

a) the length of the segment $BO$ ;
b) the size of the angle $\angle YAN$ .

Solution

a) Clearly $\angle BAC = \angle BAY + \angle CAY = 120^\circ$ , respectively $\angle BOC = 360^\circ - 2\angle BAC = 120^\circ$ . Thus, if $M$ is the midpoint of $BC$ , then $OM \perp BC$ (because $BO = OC$ ), $\angle BOM = 60^\circ$ and $BM = \frac{BC}{2} = 1$ . Let $BO = x$ and from triangle $BOM$ we have $OM = \frac{x}{2}$ from $\angle OBM = 30^\circ$ and $x^2 = (\frac{x}{2})^2 + 1^2$ from the Pythagorean theorem, respectively $x^2 = \frac{4}{3}$ and $x = \frac{2\sqrt{3}}{3}$ .

b) We have $\angle ANB = \angle AOB = 180^\circ - 2\angle BAO = 180^\circ - 2\angle OBY = 180^\circ - 2(90^\circ - \angle YCB) = 2\angle YCB$ . Hence $\angle ABY = \angle ABO + \angle OBY = 2\angle OAB = 180^\circ - \angle AOB = 180^\circ - 2\angle YCB$ and now from the other circle we calculate $\angle ANC = 180^\circ - \angle ACY = \angle ABY = 180^\circ - 2\angle YCB$ . Therefore, $\angle ANB + \angle ANC = 180^\circ$ , i.e. $N$ lies on $BC$ . It remains to consider that $\angle NAC = \angle BCY = \angle BAY$ from the touching and $\angle YAN = \angle CAY - \angle CAN = \angle CAY - \angle BAY = 36^\circ$ .

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