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Topic: Heart inscribed in a circle
Replies: 4   Last Post: Aug 25, 2010 5:04 AM

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 Peter Scales Posts: 192 From: Australia Registered: 4/3/05
Re: Heart inscribed in a circle
Posted: Aug 25, 2010 5:00 AM

> Have you or has anyone done the opposite? I.e.,
> circumscribed a given heart shape with a circle?

I found Stuart's problem of inscribing a heart inside an existing circle more difficult than the problem Mary proposed of circumscribing a circle around a given heart. So I tackled them in reverse order.

First I solved the problem analytically:

Consider a square of side 'a' with vertices on the coordinate axes, as A(c,0), B(0,c),C(-c,0),D(0,-c) where c = a/sqrt(2), with semi-circles of diam 'a' on sides AB and BC resp., with E the midpoint of AB and I the midpoint of DA.
Let radius of circumcircle = 'r', and circumcenter J at (0,b), and let T be the common tangent point of circles (E,a/2) and (J,r).

Then r = a/sqrt(2) + b

Let L and K be the projections on Oy of E and T resp.
Then in tJEL we have JE = r-a/2, EL = a/(2*sqrt(2)),
L = a/(2*sqrt(2)) - b = 3*a/(2*sqrt(2)) - r.

Apply Pythagoras to give a/r = (3*sqrt(2)-2)/2 or
r/a = (3*sqrt(2)+2)/7

.: b/a = (4-sqrt(2))/14

In tOTK let <OTK = theta = <AOT.

Triangles JTK and JEL are similar
.: KT = a/(2*sqrt(2))*r/(r-a/2)
and JK = 3a/((2*sqrt(2))-r)*r/(r-a/2)

Then tan(theta) = (b + JK)/KT = 1/sqrt(2)
.: theta = 35.264 deg approx.

Numerically: If a=1, then b=0.1847 and r= 0.8918

Next I looked for a simple graphical construction for the circumcircle of a given heart:

Note the T,E and J are collinear.
With centre D swing arcDI = a/2 to meet Oy at G.
Let HJ with J on Oy be the perpendicular bisector of GE, so that GJ = JE = r - a/2
Then J is the required circumcentre.

To tackle a heart within a given circle I proceeded indirectly as follows:
Inside the given circle draw a square with vertices at (+-r,0) and (0,+-r).[The size of the square is arbitrary, but the inscribed square is convenient.]
Complete the heart on that square and draw the outer circumcircle, radius R. Then complete the inner heart by proportion. One simple way is:
Find point M= (r,R). Join AD and MD.
Draw line y=r to meet MD at N.
Drop a vertical from N to meet AD at P.
Then P is the required square vertex of the inscribed heart.

Regards, Peter Scales.

Date Subject Author
8/16/10 Stuart
8/18/10 Mary Krimmel
8/25/10 Peter Scales
8/25/10 Peter Scales
8/18/10 Avni Pllana