Problem 8
Let S and S’ be two circles whose centers are O and O’ respectively. The circles intersect at A and B. The point A is rotated clockwise about O by an angle θ to get C. The point A is also rotated clockwise about O’ by the same angle to get D. Then, prove that:
B, C, and D are collinear
Here’s a simulation that demonstrates the problem.
You can drag the point C anywhere on S. The point D will also move on S’ such that ∠COA = ∠DO’A. Observe the points B, C, and D. Are they always collinear?
Solution
To solve this problem, you’ll need to apply the following property of circles.
Angles Subtended by a Chord
The angle subtended by a chord of a circle at its center is twice the angle subtended by it at the circumference.
Let’s observe this in the simulation below.
You can drag the points P, Q, and R, and observe the angles POQ and PRQ. Is ∠POQ always twice of ∠PRQ?
To prove the three points as collinear, you can either prove
∠CBA = ∠DBA, or
∠CDA + ∠BDA = 180°
That’s all that you need to know here. I suggest you proceed on your own from this point. Good luck!
Were you able to solve it using the above hints? Or, did you solve using any other method? Or, do you need some more help? You can reach out at the Instagram handle or the Telegram group.
Source: Math Club
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