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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.

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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 $$9=OE=OO'+O'E=2R+R$$$$\Rightarrow R=3.$$

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  • $\begingroup$ (+1) I cannot get $OO' = 2R$ $\endgroup$ – Amad27 Oct 21 '15 at 15:07
  • $\begingroup$ @Amad27: Note that $\angle{AOB}=60^\circ$. $\endgroup$ – mathlove Oct 21 '15 at 15:08
  • $\begingroup$ 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'? $\endgroup$ – Amad27 Oct 21 '15 at 15:12
  • $\begingroup$ @Amad27: We have $DO':OO':OD=1:2:\sqrt 3$. $\endgroup$ – mathlove Oct 21 '15 at 15:14
  • $\begingroup$ @Amad27: see en.wikipedia.org/wiki/… $\endgroup$ – mathlove Oct 21 '15 at 15:16
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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.

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