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)) .


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Question Number 207920 by efronzo1 last updated on 30/May/24
$Given p,q ,r and s real positive numbers such that { ((p^2 +q^2 = r^2 +s^2 )),((p^2 +s^2 −ps = q^2 +r^2 +qr.)) :} Find ((pq+rs)/(ps+qr)) .$
$Given p, q, r and s real positive$ $numbers such that$ ${\begin{cases} p^{2} + q^{2} = r^{2} + s^{2} \\ p^{2} + s^{2} - ps = q^{2} + r^{2} + qr . \end{cases}$ $Find \frac{pq + rs}{ps + qr} .$
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
	$p=a q=αa r=βa s=ab { ((α^2 +1=b^2 +β^2 )),((b^2 −b+1=α^2 +αβ+β^2 )) :} Subtracting & solving ⇒ β=((2b^2 −b−2α^2 )/α) Insert above and factorise (α^2 −b^2 )(α^2 −((4b^2 )/3)+((4b)/3)−(1/3))=0 ⇒ 1. α=−b∧β=1 p=a∧q=−ab∧r=a∧s=ab 2. α=b∧β=−1 p=a∧q=ab∧r=−a∧s=ab 3. α=(((1−2b)(√3))/3)∧β=(((b−2)(√3))/3) p=a∧q=((a(1−2b)(√3))/3)∧r=((a(b−2)(√3))/3)∧s=ab 4. α=(((2b−1)(√3))/3)∧β=(((2−b)(√3))/3) p=a∧q=((a(2b−1)(√3))/3)∧r=((a(2−b)(√3))/3)∧s=ab But p, q, r, s >0 ⇒ a, b >0 and only 4. leads to valid solutions ⇒ ((pq+rs)/(ps+qr))=((√3)/2)$
	$p = a$ $q = α a$ $r = β a$ $s = a b$ ${\begin{cases} α^{2} + 1 = b^{2} + β^{2} \\ b^{2} - b + 1 = α^{2} + α β + β^{2} \end{cases}$ $Subtracting & solving \Rightarrow β = \frac{2 b^{2} - b - 2 α^{2}}{α}$ $Insert above and factorise$ $(α^{2} - b^{2}) (α^{2} - \frac{4 b^{2}}{3} + \frac{4 b}{3} - \frac{1}{3}) = 0$ $\Rightarrow$ $1 . α = - b \land β = 1$ $p = a \land q = - a b \land r = a \land s = a b$ $2 . α = b \land β = - 1$ $p = a \land q = a b \land r = - a \land s = a b$ $3 . α = \frac{(1 - 2 b) \sqrt{3}}{3} \land β = \frac{(b - 2) \sqrt{3}}{3}$ $p = a \land q = \frac{a (1 - 2 b) \sqrt{3}}{3} \land r = \frac{a (b - 2) \sqrt{3}}{3} \land s = a b$ $4 . α = \frac{(2 b - 1) \sqrt{3}}{3} \land β = \frac{(2 - b) \sqrt{3}}{3}$ $p = a \land q = \frac{a (2 b - 1) \sqrt{3}}{3} \land r = \frac{a (2 - b) \sqrt{3}}{3} \land s = a b$ $But p, q, r, s > 0 \Rightarrow a, b > 0$ $and only 4 . leads to valid solutions$ $\Rightarrow \frac{p q + r s}{p s + q r} = \frac{\sqrt{3}}{2}$
	Answered by mr W last updated on 30/May/24

	$p^{2} + q^{2} = r^{2} + s^{2} = a^{2}, s a y$ $\Rightarrow p = a \cos α, q = a \sin α$ $\Rightarrow r = a \cos β, s = a \sin β$ $w i t h 0 ⩽ α, β ⩽ \frac{π}{2} i f o n l y p o s i t i v e n u m b e r s$ $a^{2} \cos^{2} α + a^{2} \sin^{2} β - a^{2} \cos α \sin β = a^{2} \sin^{2} α + a^{2} \cos^{2} β + a^{2} \sin α \cos β$ $\cos 2 α - \cos 2 β = \sin (α + β)$ $\cos^{2} 2 α + \cos^{2} 2 β - 2 \cos 2 α \cos 2 β = \sin^{2} (α + β)$ $2 - {(\sin 2 α + \sin 2 β)}^{2} + 2 \sin 2 α \sin 2 β - 2 \cos 2 α \cos 2 β = \sin^{2} (α + β)$ $2 - {(\sin 2 α + \sin 2 β)}^{2} - 2 \cos 2 (α + β) = \sin^{2} (α + β)$ $2 - {(\sin 2 α + \sin 2 β)}^{2} - 2 + 4 \sin^{2} (α + β) = \sin^{2} (α + β)$ ${(\sin 2 α + \sin 2 β)}^{2} = 3 \sin^{2} (α + β)$ $\Rightarrow \sin 2 α + \sin 2 β = \pm \sqrt{3} \sin (α + β)$ $\frac{p q + r s}{p s + q r}$ $= \frac{\cos α \sin α + \cos β \sin β}{\cos α \sin β + \sin α \cos β}$ $= \frac{\sin 2 α + \sin 2 β}{2 \sin (α + β)}$ $= \pm \frac{\sqrt{3}}{2} ✓ (o n l y \frac{\sqrt{3}}{2} i f o n l y p o s i t i v e n u m b e r s)$
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