Q.455. Geometry challenge

Question 455.
A circle (radius = r), and an equilateral triangle (side = 2r), fit perfectly in a square, as shown in the diagram.
What is (length CD) divided by (height of triangle)?

Answer 455.

Trigonometric Proof:

Refer to the figure:

Let O be the center of the circle.
With r = 1, (The required ratio is independent of the value of r.)
vertical side of the square
= 1 + ODcos60° + CDcos30°
= 1 + 1/2 + (√3/2) CD
= 3 + (√3/2) CD ... ( 1 )

Horizontal side of the square
= 1 + ODcos30° + (BC - CDcos60°)
= 1 + √3/2 + 2 - CD/2
= 3 + √3/2 - (1/2) CD ... ( 2 )
Equatting ( 1 ) and ( 2 ),
3/2 + (√3/2) CD = 3 + √3/2 - (1/2) CD
=> (√3 - 1)/2 CD = (√3 - 3)/2
=> CD = √3
and Height of the triangle = 2 cos30° = √3
=> CD/Height of the triangle
= √3 / √3
= 1.
=======================================…
Proof using Co-ordinate Geometry:
Refer to the figure:

Consider the given drawn inverted as above.

The required ratio, being independent of the radius, let r = 1
=> The eqn. of the circle with its center as origin is
x^2 + y^2 = 1 ... ( 1 )

Let the length of side of the square = a
=> B = (a-1, a-1), C = (a-3, a-1) and A = (a-2, a-1-√3)
Slope of AC = - √3
Let the eqn. of the tangent AC be y = - √3x + c
=> c = r √(1 + m^2) = 2
=> eqn. of AC is y = - √3x + 2 ... ( 2 )

Solving eqn. ( 1 ) and ( 2 ) gives
x^2 + (-√3x + 2)^2 = 1
=> 4x^2 - 4√3x + 3 = 0
=> (2x - √3) = 0
=> x-coordinate of D is √3/2
Plugging in eqn. ( 2 ),
y-coordinate is 1/2
=> D = (- √3/2, 1/2)

Plugging coordinates of C in eqn. ( 2 ),
a -1 = - √3 (a - 3) + 2
=> a = (3√3 + 3) / (√3 + 1) = 3
=> C = (0, 2)

CD^2 = (0 + √3/2)^2 + (3/2)^2 = 3
=> CD = √3
and Height of the triangle = 2 cos30° = √3
=> CD/Height of the triangle
= √3 / √3
= 1.

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