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# help!!

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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.

Aug 24, 2023

#3
+397
+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.

Aug 25, 2023

#2
+128732
+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  !!!!

Aug 25, 2023
#3
+397
+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
#4
+128732
+1

THX, Tiggsy  !!!!

(For me....the second  method seems  more straightforward....)

CPhill  Aug 25, 2023
edited by CPhill  Aug 25, 2023