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Source: SRMO 2005

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#1 h Apr 10, 2005, 8:46 AM • 2 Y

Y by Adventure10, Mango247

Assume $A,B,C$ are three collinear points that $B \in [AC]$ . Suppose $AA'$ and $BB'$
are to parrallel lines that $A'$ , $B'$ and $C$ are not collinear. Suppose $O_1$ is circumcenter of circle passing through $A$ , $A'$ and $C$ . Also $O_2$ is circumcenter of circle passing through $B$ , $B'$ and $C$ . If area of $A'CB'$ is equal to area of $O_1CO_2$ , then find all possible values for $\angle CAA'$

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#2 h May 9, 2005, 7:36 AM • 2 Y

Y by Adventure10, Mango247

Let $C'$ be the intersection of the circles $(O_1), (O_2)$ other than the point $C$ . The radical axis $CC'$ of these 2 circles is perpendicular to their center line $O_1O_2$ and the center line cuts the segment $CC'$ at its midpoint $D$ . Since the lines $AA' \parallel BB'$ are parallel, the angles $\angle CAA' = \angle CBB'$ are equal. Thus the angles spanning the major arcs $CA', CB'$ of the circles $(O_1), (O_2)$ are equal. In particular, the angles $\angle CC'A' = \angle CC'B'$ are equal, which means that the points $A', B', C'$ are collinear. The triangles $\triangle CC'O_1, \triangle CC'O_2$ are both isosceles. Since $O_1O_2 \perp CC'$ , the center line $O_1O_2$ bisects the angles $\angle CO_1C', \angle CO_2C'$ , i.e.,

$\angle CO_1O_2 = \frac{\angle CO_1C'}{2},\ \ \angle CO_2D = \frac{\angle CO_2C'}{2}$

The angles $\angle CA'C', \angle CB'C'$ span the major arcs $CC'$ of the circles $(O_1), (O_2)$ with the central angles $\angle CO_1C', \angle CO_2C'$ . Hence, we also have

$\angle CA'B' \equiv CA'C' = \frac{\angle CO_1C'}{2},\ \ \angle CB'C' = \frac{\angle CO_2C'}{2}$

Thus the internal angles $\angle CA'B' = \angle CO_1O_2$ and the corresponding external angles $\angle CB'C' = \angle CO_2D$ of the triangles $\triangle A'CB', \triangle O_1CO_2$ are equal, which means that these 2 triangles are always similar, regardless of the angle $\angle CAA' = \angle CBB'$ .

The similar triangles $\triangle A'CB' \sim \triangle O_1CO_2$ are congruent, iff their corresponding sides $CO_1 = CA'$ are equal. The segment $CO_1$ is the circumradius of the triangle $\triangle CAA'$ and the segment $CA'$ its side against the angle $\angle CAA'$ . Using the extended sine theorem for this triangle,

$\sin{\widehat{CAA'}} = \frac{CA'}{2 CO_1}$

Hence, $CO_1 = CA'$ iff $\sin{\widehat{CAA'}} = \frac 1 2$ . i.e., iff $\angle CAA' = 30^o$ or $150^o$ .

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