Q.125. Geometry/Trigonometry

Question 125.

Rectangle is split, by diagonal, in two right triangles and circles are inscribed in triangles. Distance between the centers of the circles is 1 and width of the rectangle is also 1. Find length of rectangle a=?

Answer 125.

If r = in-radius of the circle, from the traingle,

then, r = area of traingle / semi-perimeter

=> r = a/[1 + a + √(1+a^2)] ... (1)

Constructing a right triangle with a line joining the

centers of the circles as hypotenuse and sides parallel

to sides of the rectangle,

(1 - 2r)^2 + (a - 2r)^2 = 1 ... (2)

Rationalizing the value of r in eqn. (1),

=> r = a[1 + a - √(1+a^2)] / [1 + a + √(1+a^2)]

* [1 + a - √(1+a^2)]

=> r = (1/2) [1 + a - √(1+a^2)]

=> 2r = 1 + a - √(1+a^2)

=> 1 -2r = √(1+a^2) - a

and a - 2r = √(1+a^2) - 1

Plugging in eqn. (2),
[√(1+a^2) - a]^2 + [√(1+a^2) - 1]^2 = 1
=> 2(1 + a^2) + a^2 - 2(a + 1)√(1+a^2) = 0
=> (3a^2 + 2) = 2(a + 1)√(1+a^2)

Squarring,
9a^4 + 12a^2 + 4 = 4(a^2 + 2a + 1)(1 + a^2)
=> 9a^4 + 12a^2 + 4 = 4(a^4 + 2a^3 + 2a^2 + 2a + 1)
=> 5a^4 - 8a^3 + 4a^2 - 8a = 0
=> 5a^3 - 8a^2 + 4a - 8 = 0 ... [because a ≠ 0]

Putting in Wolfram Alpha (Link 1) gives
a ≈ 1.68771 and its exact value as
a = (8/15) + (1/15)*[2492 - 300√(69)]^(1/3)] + (1/15)[623 + 75√(69)]^(1/3)].

Putting the equations (1) and (2) in wolfram alpha (Link 2)
a ≈ 1.68771 and r ≈ 0.362993
For exact values of a and r, click on exact forms in the link.

Wolfram Alpha Link 1
Wolfram Alpha Link 2

LINK to YA!