A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle
P is a point on the circumference of a circle radius r which rolls,
without slipping, inside a circle of radius 2r. What is the locus
Three circular medallions fit in a rectangular box. Can you find the radius of the largest one?
This is really an `aha' problem: the sort that
has a very short and sweet solution, if you look at it the right
way. We leave this as a further challenge for you to think about
and send your ideas to `Further Inspirations'. We give the solution
sent by David of Madras College, St Andrews which depends on
splitting the equilateral triangle into four smaller equilateral
Since a straight line touching the circumference of a circle
will unavoidably find itself at right angles to the radius it
``touches'' - in this case line DE - we know angle DEC to be a
right angle. As the whole thing is an equilateral triangle, angle
ACB must be 60 degrees. The ``missing angle'' in triangle ECD
(angle CDE) must therefore be 30 degrees.
Triangle ECD can be reflected along line DE to produce the
equilateral triangle FCD. Triangle FCD is similar to the whole
triangle but with sides half the length, therefore 4 ``FCDs'' can
fit into ABC in the well known pattern:
The height of triangle FCD (or any of the other congruent
triangles) is R and the height of triangle ABC is twice
the height of triangle FCD, that is 2 R .
The upper section of the upper triangle (from the middle of the
small circle upwards - AHJ) can be analysed similarly to the
semi-circle below, so we find the height of triangle AHJ is 2
r . Therefore the height of triangle AFG = the height of
triangle AHJ + r = 3 r . Therefore R = 3
r so the ratio R : r = 3 : 1.