Incentre without Incircle

Geometry Level 3

In triangle $ABC$ , we have $BC = 10$ and $\angle BAC = 120^\circ$ . Let $I$ be the incenter of triangle $ABC$ . What is the radius of the circumcircle of triangle $BIC$ ?

The answer is 10.

2 solutions

Himanshu Arora
Jun 2, 2014

Sum of the other two angles is $60 ^\circ$ . Now, when we draw their angle bisectors and form a triangle with the In-center the sum of the angles would be $30 ^\circ$ . Hence the third angle of this triangle is $150 ^\circ$ .

Since, chord BC subtends angle $150 ^\circ$ on the circumference, it must subtend angle $300 ^\circ$ on the center. i.e. $60 ^\circ$ . Hence the triangle formed by B,C and the center is an equilateral triangle. Hence the answer $\boxed{10}$

FYI, The latex command for degrees is given by ^\circ to produce $^\circ$ , (as opposed to \degree). I've updated your solution accordingly.

There also isn't a need to state "\usepackage{gensymb}", because we do not Latex compile the text.

Calvin Lin Staff - 7 years ago

Thanks! :)

Himanshu Arora - 7 years ago

Saurabh Nenawati
Jun 4, 2014

The radius of circumcircle of the triangle is given by R=a/2sinA here a=10 and A=120. Now if a circle passes two vertices and incenter and has diameter equal to distance between incenter and ex-center corresponding to that side than its center lies at circumcenter. The radius of such circle is given by r= 2 R sin(A/2). So r= 2 10 sin60/2sin120 i.e. r=10

Incentre without Incircle

The answer is 10.

2 solutions

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