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.

Xtra

Stage: 4 and 5 Challenge Level:

Two solutions to this problem have been forthcoming from different students at the same school - Madras College. Thank you to Mike and Euan who used lots of trigonometry as well as to Thom who likewise resorted to double angles and the cosine rule and reduced the problem to solving a quadratic equation. Thom was also able to show the significance of the two roots.

\eqalign{ \beta &=& \frac{\pi}{6} - \frac{\alpha}{2} \\ \cos\beta &=& \frac{\sqrt{3}}{2}\cdot\frac{3\sqrt{3}}{2\sqrt{7}} + \frac{1}{2}\cdot\frac{1}{2\sqrt{7}} \\ \; &=& \frac{10}{4\sqrt{7}} = \frac{5}{2\sqrt{7}}}
Using the cosine rule on $\triangle ABP$ \eqalign{ 4 &=& x^2 + 7 - 2x\sqrt{7}\cos\beta \\ \; &=& x^2 + 7 - 5x} Therefore $x^2 - 5x + 3 = 0$ $$x = \frac{5\pm\sqrt{13}}{2}$$ Both solutions satisfy the triangle inequality for $\triangle ABP$, namely $\sqrt{7} - 2 < x < \sqrt{7} + 2$. The diagram can be redrawn to show the trapezium $BPQC$ flipped down producing the much smaller equilateral triangle of side $x$ units.

A solution which just needs Pythagoras's Theorem was sent in by Ewan from King Edward VII School, Sheffield. See if you can work it out from the diagram below, then reveal the hidden text to check your answer.

The diagram shows the median AU at point A cutting the triangle in half. The length AU is in two parts, $y$ and $z$. Since the triangle is equilateral, $y+z=\frac{\sqrt 3}{2}x$. (You may like to prove this e.g. by trigonometry) To work out $y$ use Pythagoras's Theorem:
\begin{align}
y^2+\left(\frac{1}{2}\right)^2 &=\left(\sqrt 7\right)^2 \\
y^2 &= \frac{27}{4} \\
y &= \frac{3\sqrt 3}{2}
\end{align}
and use $y+z=\frac{\sqrt 3}{2}x$ to give $z = (x-3)\frac{\sqrt 3}{2}$.

Then more Pythagoras's Theorem and our values found above give
\begin{align}
z^2+\left(\frac{x}{2} - \frac{1}{2}\right)^2 &=2^2 \\
\frac{3(x-3)^2}{4} + \frac{(x-1)^2}{4} = 4
\end{align}
which simplifies to get the same equation as Thom and the others: $x^2 -5x + 3 = 0$.

What a simple solution, Well done!