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Re: Heart inscribed in a circle
Posted:
Aug 25, 2010 5:04 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 semicircles 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 = ra/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 = (4sqrt(2))/14 > > In tOTK let <OTK = theta = <AOT. > > Triangles JTK and JEL are similar > .: KT = a/(2*sqrt(2))*r/(ra/2) > and JK = 3a/((2*sqrt(2))r)*r/(ra/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.



