Keep Your Alphabet Straight Here

Since $a^{2} + b^{2} = c^{2} + d^{2} = 1$ and $ac + bd = 0$ we have that

$ab + cd = ab(c^{2} + d^{2}) + cd(a^{2} + b^{2}) = abc^{2} + abd^{2} + cda^{2} + cdb^{2} =$

$bc(ac + bd) + ad(bd + ac) = (bc + ad)(ac + bd) = (bc + ad) \times 0 = \boxed{0}.$

I considered (a,b) and (c,d) as points on a unit circle. I realized that ac + bd = 0 implies that the vectors (a,b) and (c,d) were perpendicular. This means that (c,d) = (-b,a) [ or (-a,b) ] and when you substitute into the desired expression, you get zero.

Note that the third equation implies $a^2 c^2 = b^2 d^2$ .

Multiply the first equation by $d^2$ to get: $a^2 d^2 + b^2 d^2 = d^2$ $a^2 d^2 + a^2 c^2 = d^2$ By the second equation, this simplifies to $a^2 = d^2$ .

By the first equation, $a$ and $b$ cannot both be zero. Suppose without loss of generality that $a \neq 0$ . Then by the third equation we have $c = \dfrac{-bd}{a}$

$ab+cd= ab+d\left(\dfrac{-bd}{a}\right)$ $= \dfrac{b}{a}\left(a^2-d^2\right) = \boxed{0}$

a = -1 or 1 and b = 0, or
a = 0 and b = -1 or 1, and
c = -1 or 1 and d = 0, or
c = 0 and d = -1 or 1, therefore
ab = 0 and cd = 0
ab + cd = 0

Ojalá esté bien

Without fancy solutions, note that $a$ and $b$ behave symmetrically. Imagine you have boxes for each variable. If you switch the order of the boxes in the equations with the squares and keep that switch in mind for the multiplication equation, the answer is just a rearrangement of the boxes.

Or you could also set a range for each variable and do 500^4 with a random distribution over the ranges. Then calculate the probability of the equation being false after 500^4 positive tests and use that to sleep at night.

My method is more long-winded than the others, but this was the way I thought of:

a^2 + b^2 = 1 (I)

c^2 + d^2 = 1 (II)

ac + bd = 0 (III)

ab + cd = x (IV)

Find x.

Try introducing new variables. WLOG:

let a = cosT, b = sinT, (from (I))

c = cosS, d = sinS. (from (II))

(III):

0 = cosTcosS + sinTsinS

= cos(T - S).

T - S = (2n+1)pi/2 with n being an integer.

Dealing with cos and sin so only need to consider T - S = +- pi/2.

S = T +- pi/2.

(IV):

x = cosTsinT + cosSsinS

= (1/2)[sin2T + sin2S]

= (1/2)[sin2T + sin2(T+-pi/2)]

= (1/2)[sin2T + sin(2T+pi)] or (1/2)[sin2T + sin(2T-pi)]

= 0 or 0

= 0.

You may use trigonometric relations. Clearly, if a=cos x, then b=sin x. Also, if c=cos y, then d=sin y.

Notice, however, that ac+bd=cos x cos y + sin x sin y=cos (x-y). Now it follows that x=y+n pi/2. We'd like to compute

ab+cd=cos x sin x + sin y cos y