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 will be

> tan(Pi/6)/2;
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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);
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In order to find the minimum value of this function, its derivative should be calculated...

> diff(Area(R),R);
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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 a name.

> DenominEquation := denom(");
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> solve("",R);
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> allvalues("[2]);
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> CriticalPoint1 := "[1];
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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 of circles..

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.


> solve(DenominEquation,R);
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> CriticalPoint2 := "[1];
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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);
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by appealing to Pythagoras' theorem. By similar reasoning, the minimum value occurs when

> CriticalPoint4 := 1/(2*sqrt(3));
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Finally, the absolute minimum must chosen from among the crtitcal points. To this, the function Area must be evaluated at each of these points.

> Area(CriticalPoint1);
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> Area(CriticalPoint2);
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> Area(CriticalPoint3);
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> Area(CriticalPoint4);
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It is difficult to tell by simply looking at these expresions whic is the smallest and which the greatest. However, evalf should help here.

> evalf(Area(CriticalPoint1));
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> evalf(Area(CriticalPoint2));
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> evalf(Area(CriticalPoint3));
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> evalf(Area(CriticalPoint4));
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So the first critical point is the one which yields the minimum total area for the circles covering the unit equilateral triangle.


Instructor

Mike Zabrocki
Email address: zabrocki@yorku.ca
Department of Mathematics and Statistics
York University
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