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?
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.
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?
Many thanks to Junwei of BHASVIC school for the basis of this
solution. A solution was also received from Andrei of School 205,
First draw a circle, centre O and radius of a.
Draw two diameter lines inside the circle, and make sure these
two lines are perpendicular.
From these two lines, we can draw four tangents to the circle
which form a square ABCD touching the circle at HIJK.
Then OH = a = HA.
Thus, OA equals $a\surd 2$.
Using the compasses with focus on O and length OA, make another
circle which has an area two times the area of the original
Extend the original diameter lines to be the diameter lines for
the new circle, and use the same method to form a new square MNOP
which touches the new circle on WXYZ.
Thus,$OZ =a\surd 2. $
Then extend AB to touch the bigger square's side PM on G. Since
$OG^2 = OZ^2 + ZG^2 = 3^2$, then $OG = a \sqrt 3.$ Using the circle
area formula, with radii of the three circles a, $a \sqrt 2$ and
$a\sqrt 3$ respectively, we can easily work out that the ratio of
their areas is 1:2:3.