Consider the following construction in a circle. Choose points $A, B, C$ on the given circle such that $\angle ABC$ is $60^\circ$. Draw another circle that is tangential to the chords $AB$, $BC$ and to the original circle. Do the above construction in the unit circle to obtain a circle $S_1$. Repeat the process in $S_1$ to obtain another circle $S_2$. What is the radius of $S_2$?

In triangle ABC, $\angle \mathrm { A } = \frac { \pi } { 3 }$ and $\overline { \mathrm { AB } } : \overline { \mathrm { AC } } = 3 : 1$. If the circumradius of triangle ABC is 7, let $k$ be the length of segment AC. Find the value of $k ^ { 2 }$. [4 points]

In triangle ABC with $\angle \mathrm { A } = \frac { \pi } { 3 }$ and $\overline { \mathrm { AB } } : \overline { \mathrm { AC } } = 3 : 1$, the radius of the circumcircle of triangle ABC is 7. What is the length of segment AC? [3 points]
(1) $2 \sqrt { 5 }$
(2) $\sqrt { 21 }$
(3) $\sqrt { 22 }$
(4) $\sqrt { 23 }$
(5) $2 \sqrt { 6 }$

As shown in the figure, quadrilateral ABCD is inscribed in a circle and $$\overline { \mathrm { AB } } = 5 , \overline { \mathrm { AC } } = 3 \sqrt { 5 } , \overline { \mathrm { AD } } = 7 , \angle \mathrm { BAC } = \angle \mathrm { CAD }$$ What is the radius of this circle? [4 points]
(1) $\frac { 5 \sqrt { 2 } } { 2 }$
(2) $\frac { 8 \sqrt { 5 } } { 5 }$
(3) $\frac { 5 \sqrt { 5 } } { 3 }$
(4) $\frac { 8 \sqrt { 2 } } { 3 }$
(5) $\frac { 9 \sqrt { 3 } } { 4 }$

In a triangle $ABC$, the circumradius is $r$ and $BC = r/2$. The circumcentre $O$ lies on $AD$ where $D$ is the midpoint of $BC$. Find the ratio $BC : AD$.
(A) $\sqrt{3} : \sqrt{2}$ (B) $\sqrt{2} : \sqrt{3}$ (C) $1 : \sqrt{3}$ (D) $\sqrt{3} : 1$

A triangle has sides of lengths $\sqrt { 5 } , 2 \sqrt { 2 } , \sqrt { 3 }$ units. Then, the radius of its inscribed circle is:
(A) $\frac { \sqrt { 5 } + \sqrt { 3 } + 2 \sqrt { 2 } } { 2 }$
(B) $\frac { \sqrt { 5 } + \sqrt { 3 } + 2 \sqrt { 2 } } { 3 }$
(C) $\sqrt { 5 } + \sqrt { 3 } + 2 \sqrt { 2 }$
(D) $\frac { \sqrt { 5 } + \sqrt { 3 } - 2 \sqrt { 2 } } { 2 }$

Consider a triangle ABC and let $\mathrm { a } , \mathrm { b }$ and c denote the lengths of the sides opposite to vertices $\mathrm { A } , \mathrm { B }$ and C respectively. Suppose $\mathrm { a } = 6 , \mathrm {~b} = 10$ and the area of the triangle is $15 \sqrt { 3 }$. If $\angle \mathrm { ACB }$ is obtuse and if r denotes the radius of the incircle of the triangle, then $r ^ { 2 }$ is equal to

In a triangle $P Q R$, $P$ is the largest angle and $\cos P = \frac { 1 } { 3 }$. Further the incircle of the triangle touches the sides $P Q , Q R$ and $R P$ at $N , L$ and $M$ respectively, such that the lengths of $P N , Q L$ and $R M$ are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)
(A) 16
(B) 18
(C) 24
(D) 22

Consider an obtuse angled triangle $ABC$ in which the difference between the largest and the smallest angle is $\frac { \pi } { 2 }$ and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1.
Then the inradius of the triangle $ABC$ is

If in a triangle $A B C , A B = 5$ units, $\angle B = \cos ^ { - 1 } \left( \frac { 3 } { 5 } \right)$ and radius of circumcircle of $\triangle A B C$ is 5 units, then the area (in sq. units) of $\triangle A B C$ is:
(1) $10 + 6 \sqrt { 2 }$
(2) $8 + 2 \sqrt { 2 }$
(3) $6 + 8 \sqrt { 3 }$
(4) $4 + 2 \sqrt { 3 }$

Let $a , b$ and $c$ be the length of sides of a triangle $ABC$ such that $\frac { a + b } { 7 } = \frac { b + c } { 8 } = \frac { c + a } { 9 }$. If $r$ and $R$ are the radius of incircle and radius of circumcircle of the triangle $ABC$, respectively, then the value of $\frac { R } { r }$ is equal to
(1) 2
(2) $\frac { 3 } { 5 }$
(3) $\frac { 5 } { 2 }$
(4) 1

A triangle ABC satisfies
$$\mathrm{AB} = 7, \quad \mathrm{BC} = 8, \quad \mathrm{CA} = 6.$$
We denote the center and the radius of the circumscribed circle of this triangle ABC by O and $r$, respectively. We draw two straight lines which are tangent to this circumscribed circle at the points B and C, and denote the point of intersection of these straight lines by D.
We see that
$$\cos\angle\mathrm{BAC} = \frac{\mathbf{A}}{\mathbf{B}}, \quad \sin\angle\mathrm{BAC} = \frac{\sqrt{\mathbf{CD}}}{\mathbf{E}},$$
$$r = \frac{\mathbf{FG}\sqrt{\mathbf{HI}}}{\mathbf{JK}}, \quad \mathrm{BD} = \mathbf{LM}.$$
Furthermore, if P is a point on the circumscribed circle, the shortest possible length of the segment DP is $\frac{\mathbf{NO}\sqrt{\mathbf{PQ}}}{\mathbf{R}}$.

Let the lengths of the three sides of the triangle ABC be $\mathrm { AB } = 6 , \mathrm { BC } = 8$ and $\mathrm { CA } = 4$. Let $\mathrm { O } ^ { \prime }$ be the center of the circle which passes through the two points B and C and is tangent to the straight line AB. Let O be the center of the circle circumscribed about triangle ABC. We are to find the length of the line segment $\mathrm { OO } ^ { \prime }$.
(1) First, we have $\cos \angle \mathrm { ABC } = \frac { \mathbf { A } } { \mathbf { B } }$ and $\sin \angle \mathrm { ABC } = \frac { \sqrt { \mathbf { C D } } } { \mathbf { E } }$.
(2) The radius of the circle circumscribed about triangle ABC is $\frac { \mathbf { F G } \sqrt { \mathbf { H I } } } { \mathbf { J K } }$.
(3) When the intersection point of the straight line $\mathrm { OO } ^ { \prime }$ and the side BC is denoted by D, we have
$$\mathrm { OD } = \frac { \mathbf{N} \sqrt { \mathbf { L M } } } { \mathbf { O P } } \text { and } \mathrm { O } ^ { \prime } \mathrm { D } = \frac { \mathbf { Q R } \sqrt { \mathbf { S T } } } { \mathbf { U V } } .$$
Thus we have $\mathrm { OO } ^ { \prime } = \frac { \mathbf { W } \sqrt { \mathbf { X Y } } } { \mathbf { Z } }$.

As shown in the figure, $\triangle A B C$ is an acute triangle, $P$ is a point outside the circumcircle $\Gamma$ of $\triangle A B C$, and both $\overline { P B }$ and $\overline { P C }$ are tangent to circle $\Gamma$. Let $\angle B P C = \theta$. What is the value of $\cos A$?
(1) $\sin 2 \theta$
(2) $\frac { \sin \theta } { 2 }$
(3) $\sin \frac { \theta } { 2 }$
(4) $\frac { \cos \theta } { 2 }$
(5) $\cos \frac { \theta } { 2 }$