Covering an equilateral triangle with circle
of minimal area
The problem is to find a value of R such that the
area of the four circles shown in the accompanying diagram is minimal.
The three circles at the corners of the trianble all have then same radius
and this common radius depends on the value of R. The first task is to
calculate the radii of the three corner circles. In order to do this it
will prove to be useful to know the length of the line segment from the
centre of the triangle (which is also the centre of the middle circle)
to the centre of a side. Elementary trigonometry reveals that this length
The total area can now be expressed as a function R. The three corner
circles contribute the first term, while the centre circle contributes
the second term to the following expression.
> Area := unapply(3*Pi*(1/2 - sqrt(R**2 - "**2))**2 + Pi*R**2,R);
In order to find the minimum value of this function, its derivative
should be calculated...
and then this must be set equal to zero in order to find the critical
points. It should be kept in mind that critical points might also occur
where the derivative is not defined. In this case, this might happen when
the denominator is 0. For future reference the denominator will be given
> DenominEquation := denom(");
> CriticalPoint1 := ";
The roots of the equation obtained by setting the derivative equalt
to zero have just been calculated by converting the RooTOf expression using
the allvalues command. Only one point is given a name since the other is
negative and, hence, does not interest us. The solution R=0 will not have
to be considered because it does not correspond to a real configuration
The next step will be to find the places where the derivative is not
defined. The numerator of the first term of the derivative must be set
equal to zero. To this. first use the op command to extract the first term
of the derivative.
> CriticalPoint2 := ";
Once again, only the positive critical point is selected. Finally,
the maximum and minimum possible values of R must be calculated. The maximum
value occurs when the centre circle completely covers the entire triangle.
The value of R in this case is easily seen to be
> CriticalPoint3 := 1/sqrt(3);
by appealing to Pythagoras' theorem. By similar reasoning, the minimum
value occurs when
> CriticalPoint4 := 1/(2*sqrt(3));
Finally, the absolute minimum must chosen from among the crtitcal points.
To this, the function Area must be evaluated at each of these points.
It is difficult to tell by simply looking at these expresions whic
is the smallest and which the greatest. However, evalf should help here.
So the first critical point is the one which yields the minimum total
area for the circles covering the unit equilateral triangle.
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