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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)) .
$$\:\:\:\:\mathrm{Given}\:\mathrm{p},\mathrm{q}\:,\mathrm{r}\:\mathrm{and}\:\mathrm{s}\:\mathrm{real}\:\mathrm{positive}\: \\ $$$$\:\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\:\:\:\:\begin{cases}{\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} =\:\mathrm{r}^{\mathrm{2}} +\mathrm{s}^{\mathrm{2}} }\\{\mathrm{p}^{\mathrm{2}} +\mathrm{s}^{\mathrm{2}} −\mathrm{ps}\:=\:\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} +\mathrm{qr}.}\end{cases} \\ $$$$\:\:\mathrm{Find}\:\:\frac{\mathrm{pq}+\mathrm{rs}}{\mathrm{ps}+\mathrm{qr}}\:. \\ $$
Commented by Frix last updated on 30/May/24
answer is ((√3)/2) p=a q=((a(2b−1)(√3))/3) r=((a(2−b)(√3))/3) s=ab a>0∧(1/2)<b<2 ⇔ p, q, r, s >0
$$\mathrm{answer}\:\mathrm{is}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${p}={a} \\ $$$${q}=\frac{{a}\left(\mathrm{2}{b}−\mathrm{1}\right)\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${r}=\frac{{a}\left(\mathrm{2}−{b}\right)\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${s}={ab} \\ $$$${a}>\mathrm{0}\wedge\frac{\mathrm{1}}{\mathrm{2}}<{b}<\mathrm{2}\:\Leftrightarrow\:{p},\:{q},\:{r},\:{s}\:>\mathrm{0} \\ $$
Commented by efronzo1 last updated on 30/May/24
how to get s=ab , q=((a(2b−1)(√3))/3) r=((a(2−b)(√3))/3)
$$\:\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:{s}={ab}\:,\:\mathrm{q}=\frac{{a}\left(\mathrm{2}{b}−\mathrm{1}\right)\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\:{r}=\frac{{a}\left(\mathrm{2}−{b}\right)\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$
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)
$${p}={a} \\ $$$${q}=\alpha{a} \\ $$$${r}=\beta{a} \\ $$$${s}={ab} \\ $$$$\begin{cases}{\alpha^{\mathrm{2}} +\mathrm{1}={b}^{\mathrm{2}} +\beta^{\mathrm{2}} }\\{{b}^{\mathrm{2}} −{b}+\mathrm{1}=\alpha^{\mathrm{2}} +\alpha\beta+\beta^{\mathrm{2}} }\end{cases} \\ $$$$\mathrm{Subtracting}\:\&\:\mathrm{solving}\:\Rightarrow\:\beta=\frac{\mathrm{2}{b}^{\mathrm{2}} −{b}−\mathrm{2}\alpha^{\mathrm{2}} }{\alpha} \\ $$$$\mathrm{Insert}\:\mathrm{above}\:\mathrm{and}\:\mathrm{factorise} \\ $$$$\left(\alpha^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\left(\alpha^{\mathrm{2}} −\frac{\mathrm{4}{b}^{\mathrm{2}} }{\mathrm{3}}+\frac{\mathrm{4}{b}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{1}.\:\alpha=−{b}\wedge\beta=\mathrm{1} \\ $$$$\:\:\:\:\:{p}={a}\wedge{q}=−{ab}\wedge{r}={a}\wedge{s}={ab} \\ $$$$\mathrm{2}.\:\alpha={b}\wedge\beta=−\mathrm{1} \\ $$$$\:\:\:\:\:{p}={a}\wedge{q}={ab}\wedge{r}=−{a}\wedge{s}={ab} \\ $$$$\mathrm{3}.\:\alpha=\frac{\left(\mathrm{1}−\mathrm{2}{b}\right)\sqrt{\mathrm{3}}}{\mathrm{3}}\wedge\beta=\frac{\left({b}−\mathrm{2}\right)\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\:\:\:\:\:{p}={a}\wedge{q}=\frac{{a}\left(\mathrm{1}−\mathrm{2}{b}\right)\sqrt{\mathrm{3}}}{\mathrm{3}}\wedge{r}=\frac{{a}\left({b}−\mathrm{2}\right)\sqrt{\mathrm{3}}}{\mathrm{3}}\wedge{s}={ab} \\ $$$$\mathrm{4}.\:\alpha=\frac{\left(\mathrm{2}{b}−\mathrm{1}\right)\sqrt{\mathrm{3}}}{\mathrm{3}}\wedge\beta=\frac{\left(\mathrm{2}−{b}\right)\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\:\:\:\:\:{p}={a}\wedge{q}=\frac{{a}\left(\mathrm{2}{b}−\mathrm{1}\right)\sqrt{\mathrm{3}}}{\mathrm{3}}\wedge{r}=\frac{{a}\left(\mathrm{2}−{b}\right)\sqrt{\mathrm{3}}}{\mathrm{3}}\wedge{s}={ab} \\ $$$$\mathrm{But}\:{p},\:{q},\:{r},\:{s}\:>\mathrm{0}\:\Rightarrow\:{a},\:{b}\:>\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{only}\:\mathrm{4}.\:\mathrm{leads}\:\mathrm{to}\:\mathrm{valid}\:\mathrm{solutions} \\ $$$$\Rightarrow\:\frac{{pq}+{rs}}{{ps}+{qr}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$
Answered by mr W last updated on 30/May/24
p^2 +q^2 =r^2 +s^2 =a^2 , say ⇒p=a cos α, q=a sin α ⇒r=a cos β, s=a sin β with 0≤α, β≤(π/2) if only positive numbers 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 2α−cos 2β=sin (α+β) cos^2 2α+cos^2 2β−2 cos 2α cos 2β=sin^2 (α+β) 2−(sin 2α+sin 2β)^2 +2 sin 2α sin 2β−2 cos 2α cos 2β=sin^2 (α+β) 2−(sin 2α+sin 2β)^2 −2 cos 2(α+β)=sin^2 (α+β) 2−(sin 2α+sin 2β)^2 −2+4 sin^2 (α+β)=sin^2 (α+β) (sin 2α+sin 2β)^2 =3 sin^2 (α+β) ⇒sin 2α+sin 2β=±(√3) sin (α+β) ((pq+rs)/(ps+qr)) =((cos α sin α+cos β sin β)/(cos α sin β+sin α cos β)) =((sin 2α+sin 2β)/(2 sin (α+β))) =±((√3)/2) ✓ (only ((√3)/2) if only positive numbers)
$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} ={r}^{\mathrm{2}} +{s}^{\mathrm{2}} ={a}^{\mathrm{2}} ,\:{say} \\ $$$$\Rightarrow{p}={a}\:\mathrm{cos}\:\alpha,\:{q}={a}\:\mathrm{sin}\:\alpha \\ $$$$\Rightarrow{r}={a}\:\mathrm{cos}\:\beta,\:{s}={a}\:\mathrm{sin}\:\beta \\ $$$${with}\:\mathrm{0}\leqslant\alpha,\:\beta\leqslant\frac{\pi}{\mathrm{2}}\:\:{if}\:{only}\:{positive}\:{numbers} \\ $$$${a}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\alpha+{a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\beta−{a}^{\mathrm{2}} \mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta={a}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\alpha+{a}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\beta+{a}^{\mathrm{2}} \mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta \\ $$$$\mathrm{cos}\:\mathrm{2}\alpha−\mathrm{cos}\:\mathrm{2}\beta=\mathrm{sin}\:\left(\alpha+\beta\right) \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\alpha+\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\beta−\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\alpha\:\mathrm{cos}\:\mathrm{2}\beta=\mathrm{sin}^{\mathrm{2}} \:\left(\alpha+\beta\right) \\ $$$$\mathrm{2}−\left(\mathrm{sin}\:\mathrm{2}\alpha+\mathrm{sin}\:\mathrm{2}\beta\right)^{\mathrm{2}} +\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\alpha\:\mathrm{sin}\:\mathrm{2}\beta−\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\alpha\:\mathrm{cos}\:\mathrm{2}\beta=\mathrm{sin}^{\mathrm{2}} \:\left(\alpha+\beta\right) \\ $$$$\mathrm{2}−\left(\mathrm{sin}\:\mathrm{2}\alpha+\mathrm{sin}\:\mathrm{2}\beta\right)^{\mathrm{2}} −\mathrm{2}\:\mathrm{cos}\:\mathrm{2}\left(\alpha+\beta\right)=\mathrm{sin}^{\mathrm{2}} \:\left(\alpha+\beta\right) \\ $$$$\mathrm{2}−\left(\mathrm{sin}\:\mathrm{2}\alpha+\mathrm{sin}\:\mathrm{2}\beta\right)^{\mathrm{2}} −\mathrm{2}+\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\left(\alpha+\beta\right)=\mathrm{sin}^{\mathrm{2}} \:\left(\alpha+\beta\right) \\ $$$$\left(\mathrm{sin}\:\mathrm{2}\alpha+\mathrm{sin}\:\mathrm{2}\beta\right)^{\mathrm{2}} =\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:\left(\alpha+\beta\right) \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{2}\alpha+\mathrm{sin}\:\mathrm{2}\beta=\pm\sqrt{\mathrm{3}}\:\mathrm{sin}\:\left(\alpha+\beta\right) \\ $$$$ \\ $$$$\frac{{pq}+{rs}}{{ps}+{qr}} \\ $$$$=\frac{\mathrm{cos}\:\alpha\:\mathrm{sin}\:\alpha+\mathrm{cos}\:\beta\:\mathrm{sin}\:\beta}{\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta+\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta} \\ $$$$=\frac{\mathrm{sin}\:\mathrm{2}\alpha+\mathrm{sin}\:\mathrm{2}\beta}{\mathrm{2}\:\mathrm{sin}\:\left(\alpha+\beta\right)} \\ $$$$=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\checkmark\:\:\:\left({only}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{if}\:{only}\:{positive}\:{numbers}\right) \\ $$

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