Given-p-q-r-and-s-real-positive-numbers-such-that-p-2-q-2-r-2-s-2-p-2-s-2-ps-q-2-r-2-qr-Find-pq-rs-ps-qr-

Question Number 207920 by efronzo1 last updated on 30/May/24

Given p, q, r and s real positive

numbers such that

{\begin{cases} p^{2} + q^{2} = r^{2} + s^{2} \\ p^{2} + s^{2} - ps = q^{2} + r^{2} + qr . \end{cases}

Find \frac{pq + rs}{ps + qr} .

Commented by Frix last updated on 30/May/24

answer is \frac{\sqrt{3}}{2}

p = a

q = \frac{a (2 b - 1) \sqrt{3}}{3}

r = \frac{a (2 - b) \sqrt{3}}{3}

s = a b

a > 0 \land \frac{1}{2} < b < 2 \Leftrightarrow p, q, r, s > 0

Commented by efronzo1 last updated on 30/May/24

how to get s = a b, q = \frac{a (2 b - 1) \sqrt{3}}{3}

r = \frac{a (2 - b) \sqrt{3}}{3}

Answered by Frix last updated on 30/May/24

p = a

q = α a

r = β a

s = a b

{\begin{cases} α^{2} + 1 = b^{2} + β^{2} \\ b^{2} - b + 1 = α^{2} + α β + β^{2} \end{cases}

Subtracting & solving \Rightarrow β = \frac{2 b^{2} - b - 2 α^{2}}{α}

Insert above and factorise

(α^{2} - b^{2}) (α^{2} - \frac{4 b^{2}}{3} + \frac{4 b}{3} - \frac{1}{3}) = 0

\Rightarrow

1 . α = - b \land β = 1

p = a \land q = - a b \land r = a \land s = a b

2 . α = b \land β = - 1

p = a \land q = a b \land r = - a \land s = a b

3 . α = \frac{(1 - 2 b) \sqrt{3}}{3} \land β = \frac{(b - 2) \sqrt{3}}{3}

p = a \land q = \frac{a (1 - 2 b) \sqrt{3}}{3} \land r = \frac{a (b - 2) \sqrt{3}}{3} \land s = a b

4 . α = \frac{(2 b - 1) \sqrt{3}}{3} \land β = \frac{(2 - b) \sqrt{3}}{3}

p = a \land q = \frac{a (2 b - 1) \sqrt{3}}{3} \land r = \frac{a (2 - b) \sqrt{3}}{3} \land s = a b

But p, q, r, s > 0 \Rightarrow a, b > 0

and only 4 . leads to valid solutions

\Rightarrow \frac{p q + r s}{p s + q r} = \frac{\sqrt{3}}{2}

Answered by mr W last updated on 30/May/24

p^{2} + q^{2} = r^{2} + s^{2} = a^{2}, s a y

\Rightarrow p = a \cos α, q = a \sin α

\Rightarrow r = a \cos β, s = a \sin β

w i t h 0 ⩽ α, β ⩽ \frac{π}{2} i f o n l y p o s i t i v e n u m b e r s

a^{2} \cos^{2} α + a^{2} \sin^{2} β - a^{2} \cos α \sin β = a^{2} \sin^{2} α + a^{2} \cos^{2} β + a^{2} \sin α \cos β

\cos 2 α - \cos 2 β = \sin (α + β)

\cos^{2} 2 α + \cos^{2} 2 β - 2 \cos 2 α \cos 2 β = \sin^{2} (α + β)

2 - {(\sin 2 α + \sin 2 β)}^{2} + 2 \sin 2 α \sin 2 β - 2 \cos 2 α \cos 2 β = \sin^{2} (α + β)

2 - {(\sin 2 α + \sin 2 β)}^{2} - 2 \cos 2 (α + β) = \sin^{2} (α + β)

2 - {(\sin 2 α + \sin 2 β)}^{2} - 2 + 4 \sin^{2} (α + β) = \sin^{2} (α + β)

{(\sin 2 α + \sin 2 β)}^{2} = 3 \sin^{2} (α + β)

\Rightarrow \sin 2 α + \sin 2 β = \pm \sqrt{3} \sin (α + β)

\frac{p q + r s}{p s + q r}

= \frac{\cos α \sin α + \cos β \sin β}{\cos α \sin β + \sin α \cos β}

= \frac{\sin 2 α + \sin 2 β}{2 \sin (α + β)}

= \pm \frac{\sqrt{3}}{2} ✓ (o n l y \frac{\sqrt{3}}{2} i f o n l y p o s i t i v e n u m b e r s)

Facebook Tweet Pin