Q.208. Trapezium

Question 208.
Trapezoid (trapezium) has sides tanx, sinx, cotx, cosx and diagonals d1 and d2, as shown in the figure. Find the sum of the squares of its diagonals, d1^2 + d2^2.

This question was asked by Dragan K whose questions cannot be taken lightly. I get thrilled whenever I am able to do the problem given by him.


I have not posted here my original answer given in YA! as
Mr. Duke, who is well versed in geometry drew my attention to the known properties of trapezoid and I have modified my answer here on those lines. He pointed out the following known formulae of trapezoid that were not known to me till now and all that I had done in answering the question was just proving these formulae.
 
d'^2 + d"^2 = m^2 + n^2 + 2ab
is valid in every trapezoid with bases a, b, lateral sides m, n, diagonals d', d"
which => d'^2 + d"^2 = (sin x)^2 + (cos x)^2 + 2 * tan x * cot x = 3.

As  special cases,
( i ) for isosceles trapeoid,
      m = n and d' = d" = d => d^2 = m^2 + ab and
( ii ) for parallelogram,
       m = n, a = b => d'^2 + d"^2 = 2 (a^2 + m^2).
 
Answer 208.
Applying cosine rule to one of the base angles of the trapezoid and its supplementory angle, 
(m^2 + a^2 - d'^2) / 2am = - (m^2 + b^2 - d"^2) / 2bm
=> bm^2  + ba^2  - bd'^2  = -  (am^2 + ab^2 - ad"^2)
=> ad"^2 + bd'^2 = (a + b) (ab + m^2)   ...   ( 1 )
 Similarly, applying cosine rule to the other base angle of the trapezoid and its supplementary angle,
=> bd"^2 + ad'^2 = (a + b) (ab + n^2)   ...   ( 2 )
 Adding ( 1 ) and ( 2 ) and simplifying,
d'^2 + d"^2 = m^2 + n^2 + 2ab
 Plugging m = cosx, n = sinx, a = tanx and b = cotx in the above derivation,
=> d'^2 + d"^2
= cos^2 x + sin^2 x + 2 tanx cotx
= 3.

gianlino  made some interesting observations about the trapezium of the question with lengths of sides in terms of trigo functions. He found the range of the values of x for which the unique trapezium can be constructed with the exception of x = π/4 for which trapezium is either a rectangle or one of an infinite number of parallelograms. He found the values of the range of x as under.

The minimum value of x is given by
sinx + cosx + tanx = cotx
=> sinx + cosx = cotx - tanx
=> sinx + cosx = (cos^2 x - sin^2 x) / sinx cosx
=> 1 = (cosx - sinx) / sinxcosx
=> sinx cosx = cosx - sinx
=> sin^2 x cos^2 x = 1 - 2sinx cosx
=> (sin2x)^2 = 4 - 4sin2x
=> (sin2x)^2 + 4sin2x - 4 = 0
=> sin2x = (1/2) [- 4 + √(16 + 16)] = - 2 + 2√2
=> x ≈ 27.96° and the maximum value will
be x ≈ 90° - 27.97° ≈ 62.03° (rounded off to 2nd place)
=> the range of values of x = (27.96°, 62.03°).

Link to YA!