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2011-02-01T00:00:01
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<div>Two semi-circles are drawn on one side of a line segment.
Another semi-circle touches them externally as shown in the
diagram.</div>
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<div>What is the radius of the circle that touches all three
semi-circles?</div>
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<p>This solution comes from Sue Liu of Madras College</p>
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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 $XD = R$.<br></br>
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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 $OX = R + r - x$.<br></br>
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If we now consider the triangle $OCD$ we have <br></br>
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<td>$$\angle OXC + \angle OXD = 180^o$$</td>
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<div>So
<div class="math">\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}</div>
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<div>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!</div>
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<p>Find lengths in terms of the radii of the two red semicircles.</p></mdoxml>
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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?
Algebra
Creating expressions/formulae
2D Geometry, Shape and Space
Circles
Trigonometry
Cosine rule
Admin
Long problems
2D Geometry, Shape and Space
Radius (radii) & diameters