Just Touching

Age 16 to 18

Challenge Level Yellow star

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

$\angle OXC + \angle OXD = 180^o$

\begin{eqnarray}cos \angle OXC + cos \angle OXD = 0\end{eqnarray}
\begin{eqnarray}\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\end{eqnarray}

\begin{eqnarray}r^2R + R(R + r - x)^2 - R(R + x)^2 + rR^2 + r(R + r - x)^2 - r(r + x)^2 = 0 \end{eqnarray}

Simplifying gives:

\begin{eqnarray}4R^2r +4Rr^2 &=& 4xR^2 + 4xr^2 + 4Rrx\end{eqnarray}

\begin{eqnarray}R^2r + Rr^2 &=& x(R^2 + rR + r^2)\end{eqnarray}

\begin{eqnarray}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!

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