Just Touching

Three semi-circles have a common diameter, each touches the other two and two lie inside the biggest one. What is the radius of the circle that touches all three semi-circles?

Four rods are hinged at their ends to form a convex quadrilateral. Investigate the different shapes that the quadrilateral can take. Be patient this problem may be slow to load.

Biggest Bendy

Four rods are hinged at their ends to form a quadrilateral with fixed side lengths. Show that the quadrilateral has a maximum area when it is cyclic.

Hexy-metry

Stage: 4 and 5 Challenge Level:

This very good solution came from The Maths Club at Wilson's School :

The interior angles of the hexagon are all equal to 120 degrees by symmetry.

Let O be the centre of the circle, with radius $r$. Considering triangle BCD, by the cosine rule,
$$BD^2 = a^2 + b^2 - 2ab \cos 120.$$
Considering triangle BOD, by the cosine rule,
$$BD^2 = 2r^2 - 2r^2 \cos 120.$$

As $\cos 120 = -\frac {1}{2}$, we have $a^2 + b^2 + ab = 3r^2$ so
$$r = \sqrt {{a^2 + b^2 + ab\over 3}}.$$

 Ruth from the Manchester High School for Girls proves the figure has rotational symmetry of order 3 and the angles are 120 degrees as follows, and then goes on to find the solution using the Cosine Rule as above: Let O be the centre of the circle and $R$ be the radius. Construct the radii OA, OB, OC, OD, OE and OF. This creates 6 isoceles triangles, 3 with sides $R, R$ and $a$ and 3 with sides $R, R$ and $b$, therefore by SSS congruence, angles FAO=AFO=CBO=BCO=DEO=EDO and BAO=ABO=CDO=DCO=FEO=EFO therefore the angles of the hexagon are all 120 degrees. ACE and FBD are therfore equilateral triangles.
 Matt, who did not reveal his school, used a different method not involving the Cosine Rule. This hexagon is formed by drawing an equilateral triangle over a circle where the centre of the circle coincides with the centre of the triangle and joining the points at which the triangle crosses the circle. The shorter edge of the hexagon has length $FB= a$ and the longer has length $b$. Therefore line AB has length $\frac {a}{2}$ and line BD has length $b$. Constructing the equilateral triangle BCD defines line BC as having length $b$. So line AC has length $\frac{a}{2}+b$.
Let line AE have length $x$. The radius has length $r$. From triangle ABE and using Pythagoras:
$$r^2 = x^2 +\left({\frac{a}{2}}\right)^2.$$
From triangle ACE: $$\tan \theta = \frac{x}{\frac{a}{2} +b}$$ but, $\theta = 30^o$ (since it is half the internal angle of an equilateral triangle) so, $$x = \frac{1}{\sqrt 3}\left( \frac{a}{2} +b \right).$$ Therefore, by eliminating $x$ between these expressions and simplifying, we get:
$$r = \sqrt {{a^2 + b^2 + ab\over 3}}.$$