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Heart in a circle (Posted on 2024-03-05) Difficulty: 3 of 5
A heart shape is formed by constructing two unit semicircles on consecutive sides of a square of side length 2.

Find the radius of the circumcircle of this shape.

No Solution Yet Submitted by Jer    
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Some Thoughts A start and an answer graphically | Comment 1 of 3
I don't have an exact solution, but here is a graphical answer with the help of Desmos:
https://www.desmos.com/calculator/wcpx1ju2nc

Let the square be positioned at (±1, ±1)
The half circles are:
   (x-1)^2 + y^2 = 1 with x>1  and
   x^2 + (y-1)^2 = 1 with y>1

The circumcircle is centered at (a,a) 
and has radius √2*(1+a)
(x-a)^2 + (y-a)^2 = 2(1+a)^2  <-- Eqn of Circumcircle
    whose radius is  (√2)*(1+a)

Using Desmos to adjust a gives a ~ 0.262
      and radius of circumcircle ~ 1.785

I have not yet found an exact algebraic solution for a or the radius, but I did get the equation for the tangent line in terms of a.

Take the equation for one of the half circles and the equation for the circumcircle.

x^2 + (y-1)^2 = 1
(x-a)^2 + (y-a)^2 = 2(1+a)^2  

x^2 + y^2 - 2y + 1 = 1
x^2 + y^2 = 2y

x^2 - 2ax + a^2 + y^2 - 2ay + a^2 = 2a^2 + 4a + 2
2y -  2ax - 2ay = 4a + 2
2y - 2 = 4a + 2ax + 2ay
y - 1 = 2a + ax + ay
y(1-a) = ax + 2a + 1
y = (a/(1-a))*x + (2a+1)/(1-a)
This line is tangent to the intersection of the half circle and circumcircle.
The slope of the tangent line is a/(1-a)

  Posted by Larry on 2024-03-05 12:02:51
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