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Re: Heart inscribed in a circle
Posted:
Aug 25, 2010 5:04 AM
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> > 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.
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