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Pericut

Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts?

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Circumspection

M is any point on the line AB. Squares of side length AM and MB are constructed and their circumcircles intersect at P (and M). Prove that the lines AD and BE produced pass through P.

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Quads

The circumcentres of four triangles are joined to form a quadrilateral. What do you notice about this quadrilateral as the dynamic image changes? Can you prove your conjecture?

Cyclic Quads

Stage: 4 Challenge Level: Challenge Level:1

The following solution is entirely the work of Sarah from The Mount School, York (Yr 11). Nice work Sarah!

I've drawn the triangle and two of the circumcircles and I'll refer to my drawing :

Now the two circles meet in the two points, D and K.

Now angles DAE and EKD add up to 180 degrees, because DAEK is a cyclic quadrilateral.

Similarly, FCD and FKD add up to 180 degrees.

Calling the original angles of the triangle A, B and C, angle EKF is:

(180 - A) + (180 - C) = 360 - (A + C)

So, angle EKF + FBE = 360 - (A + C) + B but A + C = 180 - B, so

EKF + FBE = 360 - (180 - B) + B = 180

Now if I draw the circumcircle through F, B and E it will have to pass through K, as opposite angles of a cyclic quadrilateral add up to 180 degrees.

If K lies within the circle or outside the circle, then this makes the above property of cyclic quadrilaterals a lie ! I noticed, by the way that all three circles pass through the same point.

We've got dynamic geometry software and we were able to experiment with that, once we had it drawn (with a little help from our teacher !)