Given a triangle $ABC$ ($AB> AC$) and a circle circumscribed around it. Construct with a compass and a ruler the midpoint of the arc $BC$ (not containing vertex $A$), with no more than two lines (straight or circles).
Problem
Source: 2018 Oral Moscow Geometry Olympiad grades 8-9 p4
Tags: arc midpoint, construction, geometry
NikoIsLife
25.07.2019 09:43
1. Draw the circle centered at $B$ and passes through $C$.
2. Draw the circle centered at $C$ and passes through $B$.
3. Draw the radical axis of the two circles.
4. The intersection of the radical axis and the circumcircle of $\triangle ABC$ should be the midpoint of arc $BC$.
zuss77
25.07.2019 11:59
@above It would be three lines. The crucial point here is to have solution only with two lines.
NikoIsLife
25.07.2019 12:01
zuss77 wrote: @above It would be three lines. The crucial point here is to have solution only with two lines. I only used one line? As far as I know, lines are different from circles.
MarkBcc168
25.07.2019 12:07
NikoIsLife wrote: zuss77 wrote: @above It would be three lines. The crucial point here is to have solution only with two lines. I only used one line? As far as I know, lines are different from circles. I guess there is a translation issue. The requirement should be not more than two lines and circles in total.
zuss77
25.07.2019 12:15
Yes, in russian word "line" has wider meaning. The correct translation would be "in two steps".
fastlikearabbit
25.07.2019 12:44
You just have to take the intersection of the circle centered at $A$ and radius $AC$ with $\odot(ABC)$ and $AB$, and thus forming a line $\ell$. Then you intersect $\ell$ with the circumcircle.
parmenides51
26.07.2019 00:25
thanks for the clarification, i just added (straight or circles), to make the 'two lines' wording more clear