The Circle-Parabola Problem

Adam Hrankowski

No Calculus. No Algebra. Just Plane Geometry.

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Actual image of Euclid solving a problem he found on InstagramWikipedia Commons

In June 2021, Dr. Peyam (of YouTube fame) published a video of a geometry problem he found on Instagram. The solution, he says, “uses a lot of beautiful geometry and even more beautiful calculus.” (It does indeed. Check it out.)

I am going to attack the same problem using methods that predate calculus — and even algebra. Here goes!

The Problem

A parabola formed by the graph of y=x² cradles a circle of radius 1. The circle touches the parabola at exactly two points. What are the coordinates of the centre of the circle?

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Illustration by Author

My Approach

We’ll solve this by constructing the diagram, using only a compass and straight-edge. (And Geogebra.)

Begin with a vertical line. Find a point on it, C. Draw a circle around that point. We’ll define the radius of the circle as our unit.

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Illustration by Author

Mark as F the lower point of intersection between the circle and the vertical line. Construct a circle through C with F as the centre. The two circles share the radius CF. Therefore they both have a radius of 1.

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Illustration by Author

Mark the points where the two circles intersect: P and P’. Construct a third circle through C and F with a centre at P. This circle also has a unit radius.

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Illustration by Author

Find the intersection (other than C) between circles P and F. Label this point D and connect points P and D. Segment PD will have a unit length.

Construct a line through D tangent to circle P and perpendicular to the vertical line. (This is not a primitive construction, like drawing a circle or a straight line. You can find the detailed construction in Euclid 1.11.)

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Illustration by Author

Mark 3 more points: The second intersection between F and the vertical line, C’; The intersection between the two straight lines, F’; and the midpoint between F and F’, V (Euclid 1.10).

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Illustration by Author

Since FDC’ are the vertices of an equilateral triangle (Euclid 1.1), F’ is the midpoint of FC’. Therefore Segment CV has a length of 5/4.

Finally, consider a parabola with V as the vertex and F as its focus. A parabola is that set of points that are equidistant from a focus, F, and a given straight line, the directrix, F’D. Since PF and PD are both of unit length, point P must be on the parabola. By a similar argument, so is point P’.

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Illustration by Author

It remains to place the problem in its original context, a parabola and circle on the Cartesian plane.

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Illustration by Author

With the vertex at the origin, the centre of the circle, C is at (0, 5/4). Points P and P’ are vertices of equilateral triangles with unit sides. Thus, they each have a distance of √3/2 from the y-axis. Each has a y-component 1/2 less than that of C. That sets their coordinates at (√3/2, 3/4) and (-√3/2, 3/4), respectively.

These points establish the parabola as y=x². The circle is x²+(y-5/4)²=1

Which approach do you prefer?

[Originally published June 7, 2021 on Medium]

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