Just Touching

Stage: 5 Challenge Level: Challenge Level:2 Challenge Level:2

This solution comes from Sue Liu of Madras College


Let the large semi-cirle have diameter $AB$ and centre $X$. Let the two smaller semi-circles have centres $C$ and $D$ and radii $R$ and $r$. Thus $AC = R$, $BD = r$ so that $AX = (2R + 2r)/2 = (R + r)$ and $CX = r$ from which it follows that $XD = R$.

Let the small circle have have centre $O$ and radius $x$. Then $CO = R + x$ and $DO = r + x$. The line $XO$, joining the centre of the large semicircle to the centre of the small circle, cuts the circumference of the large semicircle at $E$ where $XE = XB = R + r$, $OE = x$ and $OX = R + r - x$.

If we now consider the triangle $OCD$ we have
$$\angle OXC + \angle OXD = 180^o$$
So
\begin{eqnarray} \\ \cos \angle OXC + \cos \angle OXD &=& 0 \\ \frac{r^2 + (R + r - x)^2 - (R + x)^2}{2r(R + r - x)} + \frac{R^2 + (R + r - x)^2 - (r + x)^2}{2R(R + r - x)} &=& 0 \\ r^2R + R(R + r - x)^2 - R(R + x)^2 + rR^2 + r(R + r - x)^2 - r(r + x)^2 &=& 0 \\ 4R^2r +4Rr^2 &=& 4xR^2 + 4xr^2 + 4Rrx \\ R^2r + Rr^2 &=& x(R^2 + rR + r^2) \\ x &=& \frac{Rr(R + r)}{R^2 + Rr + r^2} \end{eqnarray}

There is a pleasing symmetry about this formula which gives the radius of the small circle in terms of the radii of the two smaller semicircles. An excellent solution, well done Sue!

Published November 1999.