Flying saucer II

If X is the midpoint of the segment connecting the centers of the upper and lower arcs, and JK is perpendicular to XH, then show that H is the midpoint of JK.

Scroll down for a solution to this problem.


We have the PoP condition IH*HJ = KH*HL with us at all times.

Let a = IN = NJ,  b = KM = ML, c = NH = HM.

Then IH*HJ = (a+c)(a-c) = a² -c²,

KH* HL = (b-c)*(b+c) = b² – c².

Thus a = b, and KH = b-c = a-c = HJ.

🤞 Don’t miss these puzzles!

Subscribe to the weekly geometry puzzle e-mail.

Leave a Reply

Your email address will not be published. Required fields are marked *

Optionally add an image (JPEG only)