Library / Ukrainian National Mathematical Olympiad - Third Round (Second Tour) /7 of 10

Ukrainian National Mathematical Olympiad - Third Round (Second Tour)

Geometry Difficulty 6.7 National Olympiad Prove it Ukraine

Circles $w_1$ and $w_2$ with centers $O_1$ and $O_2$ respectively intersect in points $A$ and $B$ . A straight line $O_1O_2$ intersects $w_1$ in a point $Q$ , that is not inside $w_2$ , and $w_2$ in a point $X$ , that is inside $w_1$ . Around the triangle $O_1AX$ a circle $w_3$ is circumscribed and intersects $w_1$ for the second time in a point $T$ . A straight line $QT$ intersects $w_3$ in a point $K$ , and a straight line $QB$ intersects $w_2$ a second time in a point $H$ . Prove that

Fig. 24

a) points $T$ , $X$ , $B$ are collinear;
b) points $K$ , $X$ , $H$ are collinear.

Solution

a) Let $\angle AO_1X = \alpha$ , then $\angle ATX = \alpha$ , because they are subtended by the same arc of the circle $w_3$ (Fig. 24), moreover $\angle AO_1X = \angle ATB$ , so $\angle ATX = \angle ATB$ , therefore $T$ , $X$ , $B$ are collinear.

b) Let $K_1 = XH \cap TQ$ , $\angle O_2XH = \alpha$ , $\angle HXB = \beta$ , $\angle XHB = \varphi$ , then:

$\angle XQB = \alpha - \varphi, \quad \angle QTB = 90^\circ - \alpha + \varphi,$
$\angle TQX = 180^\circ - (\beta + \alpha) - (90^\circ - \alpha + \varphi) = 90^\circ - \beta - \varphi = \alpha,$
in other words $\angle TQX = \angle QTO_1$ , since $O_1T = QO_1$ , then $\angle K_1XQ = \alpha$ , hence $\angle K_1XQ = \angle QTO_1$ , that is $K_1 \in w_3$ , so $K_1 = K$ and points $K$ , $X$ , $H$ are collinear.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.