This is hard geometry!!

Two quarter-circles are drawn inside a unit square, as shown. A circle is drawn that is tangent to the quarter-circles, and a side of the square. Find the radius of the circle.

wiseowl Aug 24, 2023

#3**+1 **

Two methods.

Using CPhill's diagram,

E, H and D are collinear, (the line EHD is at rt-angles to the common tangent at E).

Let the radius of the small circle be c, then from the triangle GHD, c^2 + (1/2)^2 = HD^2,

so HD = sqrt(c^2 + 1/4),

so ED = HD + c = sqrt(c^2 + 1/4) + c = 1, (the radius of the big circle).

c^2 + 1/4 = (1 - c)^2 = 1 -2c + c^2,

2c = 3/4,

c = 3/8.

Let the radius of the small circle be c, then its Cartesian equation is

x^2 + (y - c)^2 = c^2

x^2 + y^2 -2yc = 0.................(1)

The equation of the big circle centre D is

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

x^2 - x + y^2 -3/4 = 0..................(2)

Solving simultaneously, (subtract (1) - (2)),

x - 2yc + 3/4 = 0.

Substitute for x in (1),

(2yc - 3/4)^2 + y^2 - 2yc = 0,

4y^2c^2 - 3yc + 9/16 + y^2 - 2yc = 0,

y^2(4c^2 + 1) - 5yc + 9/16 = 0.

For tangency, the equation needs to have a single root, so

25c^2 - 4(4c^2 +1)(9/16) = 0,

16c^2 - 9/4 = 0,

c^2 = 9/64,

c = 3/8.

Tiggsy Aug 25, 2023

#2**+1 **

Let the intersection of the circle with the bottom edge of the square = ( 0 , 0) = G

Through this point draw lines y = -sqrt (3) x and y = sqrt (3) x

We can show that this will lead to the (approximate) radius of the circle

The equation of one circle is (x-.5)^2 + y^2 =1

And the intersection of this cirlce with the line y =sqrt (3) x can be found as

(x -.5)^2 + (sqrt (3) * x)^2 =1

Solving this for (x,y) produces ≈ (.3256 , .5641) = F

And using symmetry the intersection of the other circle with the line y = -sqrt (3) x is

(-.3256 , .5641) = E

The distance between these points is 2 (.3256) ≈ .651 = EF

And the distance between one of these intersection points and the intersection of the circle with the bottom of the square = sqrt [ (.3256)^2 + (.5641)^2 ] ≈ .651 = EG = FG

So we have an iinscribed equilateral triangle in the circle with sides ≈ .651

Angle EGF = 60° so angle EHF =120°

Using the Law of Cosines, we can find the (approximate) radius of the circle as follows

.651^2 = 2* r^2 - 2*(r^2) cos (120°)

.651^2 = 2r^2 - 2r^2 (-1/2)

.651^2 = 3r^2

r^2 = .651^2 / 3

r = sqrt [ .651^2 / 3 ] ≈ .3759

I'm sure that someone else may be able to give a more precise answer !!!!

CPhill Aug 25, 2023

#3**+1 **

Best Answer

Two methods.

Using CPhill's diagram,

E, H and D are collinear, (the line EHD is at rt-angles to the common tangent at E).

Let the radius of the small circle be c, then from the triangle GHD, c^2 + (1/2)^2 = HD^2,

so HD = sqrt(c^2 + 1/4),

so ED = HD + c = sqrt(c^2 + 1/4) + c = 1, (the radius of the big circle).

c^2 + 1/4 = (1 - c)^2 = 1 -2c + c^2,

2c = 3/4,

c = 3/8.

Let the radius of the small circle be c, then its Cartesian equation is

x^2 + (y - c)^2 = c^2

x^2 + y^2 -2yc = 0.................(1)

The equation of the big circle centre D is

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

x^2 - x + y^2 -3/4 = 0..................(2)

Solving simultaneously, (subtract (1) - (2)),

x - 2yc + 3/4 = 0.

Substitute for x in (1),

(2yc - 3/4)^2 + y^2 - 2yc = 0,

4y^2c^2 - 3yc + 9/16 + y^2 - 2yc = 0,

y^2(4c^2 + 1) - 5yc + 9/16 = 0.

For tangency, the equation needs to have a single root, so

25c^2 - 4(4c^2 +1)(9/16) = 0,

16c^2 - 9/4 = 0,

c^2 = 9/64,

c = 3/8.

Tiggsy Aug 25, 2023