Question

Triangle $ABC$ is inscribed in a circle with center $O'.$ A circle with center $O$ is inscribed in triangle $ABC.$ $AO$ is drawn, and extended to intersect the larger circle in $D.$ Then we must have:

$CD=BD=O'D$

$AO=CO=OD$

$CD=CO=BD$

$CD=OD=BD$

$O'B=O'C=OD$

Answer: D

Concepts:angle chasing circumcircle, circumcenter, and circumradius incircle, incenter, and inradius

Difficulty rating: 1880

Small Hint:

Line $AD$ bisects $\angle A,$ so $D$ is the midpoint of arc $BC$

Big Hint:

Compare angles in triangle $COD$ to show $CD=OD$

Solution:

Because $AO$ is an angle bisector, the inscribed angles $\angle BAD$ and $\angle CAD$ are equal. Hence arcs $BD$ and $CD$ , and therefore chords $BD$ and $CD,$ are equal.

Let $\angle BAD=\alpha$ and $\angle BCO=\beta.$ Then $\angle OCD=\alpha+\beta.$ In triangle $AOC,$ the exterior angle $\angle COD$ also equals $\alpha+\beta.$ Thus $CD=OD,$ so $CD=OD=BD.$

Therefore, the correct answer is D.

1966 AMC 12 Problem 31

Answer: D

Difficulty rating: 1880

Small Hint:

Big Hint:

Solution:

Problem 31 in Other Years