Area of the locus

Geometry Level 5

Let C 1 C_1 and C 2 C_2 be circles with centres 10 10 units apart, with radii of length 1 1 and 3 3 respectively. Firstly find the locus of all points M M for which there exist points X X on C 1 C_1 and Y Y on C 2 C_2 such that M M is the midpoint of the line segment X Y XY . Then find the area of the closed region inscribed by the locus.


The answer is 9.42.

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2 solutions

William Allen
Jul 22, 2019

Let X = a + e i α Y = b + 3 e i β X=a+e^{i\alpha} \quad Y=b+3e^{i\beta}

M = X + Y 2 = a + b 2 + e i α + 3 e i β 2 M=\frac{X+Y}{2}=\frac{a+b}{2}+\frac{e^{i\alpha}+3e^{i\beta}}{2}

Now let α \alpha and β \beta vary for π α , β π -\pi \leq \alpha , \beta \leq \pi\quad then e i α + 3 e i β 2 \frac{e^{i\alpha}+3e^{i\beta}}{2} is simply a circle of radius 1 2 \frac{1}{2} with its centre on the circumference of a circle with radius 3 2 \frac{3}{2} . So the locus of points is an annulus with inner radius 1 1 and outer radius 2 2 .

Area of the annulus is π ( 2 2 1 2 ) = 3 π 9.42 \pi(2^2-1^2)=3\pi \approx \boxed{9.42}

Steven Chase
Jul 22, 2019

Discretizing the two angular parameters and plotting results in a cool looking pattern. In the first plot, the angle step is 2 π 50 \frac{2 \pi}{50} , and the plotting is done with two nested "for" loops; one for each parameter. In the second plot, the angle step is 2 π 100 \frac{2 \pi}{100} , and the curves fill the space between the inner and outer circles more thoroughly.

Of course, this "filling" is illusory, given that the locus itself is one-dimensional. Maybe we could define the "filling" (for a particular angle step size) by asking "What is the largest circle within the region bounded by the two greater circles, such that the locus does not pass through it?"

That's very pretty. I suppose you could use this to make a trammel to draw this annulus (although, there are easier ways!)

Chris Lewis - 1 year, 10 months ago

That would be interesting to try. The trammel would force some extra constraints, namely that the distance between the two points would have to have a constant length.

Steven Chase - 1 year, 10 months ago

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