# Tough geometry circle triangle

The triangle ABC with an angle of C 30 degrees at the vertex C is inscribed in a circle with a center O and a radius of 9 cm. If R is the radius of the circle tangent to the segments AO and BO, and the arc AB, then R is:?

This is fairly difficult.

Obviously, the circumcenter is $O$ so the perpendicular bisectors are near.

I used: $x = 9\sin(60)$ as an approximate, but it does not work.

Let $D,E$ be the tangent point of the small circle and $AO,\text{arc$AB$}$ respectively. Also, let $O'$ be the center of the small circle.
Then, considering the right triangle $OO'D$ gives $OO'=2R$. Now, noting that $O'$ is on the line $OE$ gives that $$9=OE=OO'+O'E=2R+R$$$$\Rightarrow R=3.$$
• (+1) I cannot get $OO' = 2R$ – Amad27 Oct 21 '15 at 15:07
• @Amad27: Note that $\angle{AOB}=60^\circ$. – mathlove Oct 21 '15 at 15:08
• Yes, OO'D is a right triangle with the right angle at $D$. So angle, DOO' = 30$. But how does that give OO' still? OE = 9 = O'E + OO', but I still cannot get OO'? – Amad27 Oct 21 '15 at 15:12 • @Amad27: We have$DO':OO':OD=1:2:\sqrt 3$. – mathlove Oct 21 '15 at 15:14 • @Amad27: see en.wikipedia.org/wiki/… – mathlove Oct 21 '15 at 15:16 A sketch can be drawn in which if$a =9 $is radius T is tangent length,$R$is circle radius drawn inside the sector$ABO$touching$AO,BO, arc AB$,$ x $is distance from big circle center to nearest arc point of circle radius$R $, we need to solve 3 equations: $$( T/R = \sqrt 3 , x*a = T^2, ( x - R)/R =2 )$$ which has solutions $$T = a/ \sqrt 3= 3 \sqrt 3, R = x= a/3 = 3.$$ The circle of radius$R\$ and its center trisect the center-line.