Baby Circle

255 /www/nrich/html/content/99/02/15plus4/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"> A circle with radius 1 and a circle with radius 2 touch at a point. A third circle fits between these two circles so that all three touch each other and all three have a common tangent. What is the radius of the smallest circle?</mdoxml> <mdoxml version="1.0"> <mdo:image width="276" height="194" src="baby_circ.gif" alt=""></mdo:image> Rosalind of Madras College sent in this solution, well done Rosalind. All three circles touch each other and have a common tangent $DFE$. The radius of the baby circle is $r$ and the radii of the other circles are $BE = 1$ unit and $AD = 2$ units. So $AC = 1$ unit, $AB = 3$ units, $AH = 2 - r$, $AR = 2 +r$, $BR = 1 + r$ and $BG = 1 - r$. Using Pythagoras Theorem: $$\eqalign{ \; RG &=& \sqrt{(1+r)^2 - (1-r)^2} = \sqrt{4r} = 2\sqrt{r} \\ \; HR &=& \sqrt{(2+r)^2 - (2-r)^2} = \sqrt{8r} = 2\sqrt{2r} \\ \; CB &=& 2\sqrt{r} = DE \\ RG + HR = CB &\Rightarrow& 2\sqrt{r} + 2\sqrt{2r} = 2\sqrt{2}.}$$ This gives $\sqrt{r}(1+\sqrt{2}) = \sqrt{2}$ and hence by squaring $$\eqalign{ r &=& \frac{2}{3+2\sqrt{2}} \\ \; &=& \frac{2(3 - 2\sqrt{2})}{(3+2\sqrt{2})(3-2\sqrt{2})} \\ \; &=& \frac{6 - 4\sqrt{2}}{9 - 4x2} \\ \; &=& 6 - 4\sqrt{2}}$$ So the radius of the `baby circle` is $6 - 4\sqrt{2}$. </mdoxml> <mdoxml version="1.0"> The problem <a href="../public/viewer.php?obj_id=620" onclick="mediaSave()">'Surds'</a> published in January 1999 may help with the calculation. </mdoxml> 2 3 0 0 0 0 1 Baby Circle A small circle fits between two touching circles so that all three circles touch each other and have a common tangent? What is the exact radius of the smallest circle? Numbers and the Number System Surds 2D Geometry, Shape and Space Tangents 2D Geometry, Shape and Space Circles Coordinates and Coordinate Geometry Cartesian equations of circles 2D Geometry, Shape and Space Pythagoras' theorem Admin Long problems