Denote the center of the parallelogram as `O . ` Everything is symmetric with respect to `O .`

Denote `BO = x , ` `CO = y , ` then `d = 2x . ` The angles OBR and OCQ are equal as the angles complementary to the same angle BOC, so `4 / x = 3 / y ` or `y = 3 / 4 x . ` Also, `CQ^2 = y^2 - 9 .`

The area of the triangle BOC is `1 / 2 x CQ = 1 / 2 x sqrt ( 9 / 16 x^2 - 9 ) , ` and also it is a quarter of the area of the entire parallelogram, so we obtain an equation for `x :`

`3 / 2 x sqrt ( 1 / 16 x^2 - 1 ) = 15 / 4 ,`

`x sqrt ( x^2 - 16 ) = 10 ,`

`x^2 ( x^2 - 16 ) = 100 ,`

`x^4 - 16x^2 - 100 = 0 .`

It is a quadratic equation for `x^2 , ` so

`x^2 = 8 + sqrt ( 64 + 100 ) = 8 + 2 sqrt ( 41 )`

(minus before the root is not suitable because `x^2 gt= 0`).

Therefore, `d^2 = 4x^2 = 32 + 8 sqrt ( 41 ) , ` `m = 32 , ` `n = 8 , ` `p = 41 , ` and `m + n + p = 81 .`

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