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Question-169815

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Question Number 169815 by mr W last updated on 09/May/22
Commented by mr W last updated on 10/May/22
can you please give a explanation?
$${can}\:{you}\:{please}\:{give}\:{a}\:{explanation}? \\ $$
Commented by mr W last updated on 09/May/22
Given a triangle ΔABC with sides a,b,c. Find the maximum area of the inscribed triangle whose sides are perpendicular to the sides of ΔABC as shown.
$${Given}\:{a}\:{triangle}\:\Delta{ABC}\:{with}\:{sides} \\ $$$${a},{b},{c}. \\ $$$${Find}\:{the}\:{maximum}\:{area}\:{of}\:{the}\: \\ $$$${inscribed}\:{triangle}\:{whose}\:{sides}\:{are} \\ $$$${perpendicular}\:{to}\:{the}\:{sides}\:{of}\:\Delta{ABC} \\ $$$${as}\:{shown}. \\ $$
Commented by Rasheed.Sindhi last updated on 10/May/22
I think at most only one such traingle is possible.
$${I}\:{think}\:{at}\:{most}\:{only}\:\:{one}\:{such}\:{traingle} \\ $$$$\:{is}\:{possible}. \\ $$
Commented by mr W last updated on 10/May/22
you are right sir.
$${you}\:{are}\:{right}\:{sir}. \\ $$
Commented by Rasheed.Sindhi last updated on 10/May/22
sir,I can say only intuitively.All such triangles should be parallel to each other.I mean their sides must be parallel.I saw only one such triangle (intuitively),maybe I′m wrong!
$$\boldsymbol{{sir}},{I}\:{can}\:{say}\:{only}\:{intuitively}.{All} \\ $$$$\:{such}\:{triangles}\:{should}\:{be} \\ $$$${parallel}\:{to}\:{each}\:{other}.{I}\:{mean}\:{their} \\ $$$${sides}\:{must}\:{be}\:{parallel}.{I}\:{saw}\:{only} \\ $$$${one}\:{such}\:{triangle}\:\left({intuitively}\right),{maybe} \\ $$$${I}'{m}\:{wrong}! \\ $$
Commented by Rasheed.Sindhi last updated on 10/May/22
THanX Sir!
$$\mathcal{TH}{an}\mathcal{X}\:\mathcal{S}{ir}! \\ $$
Commented by mr W last updated on 10/May/22
Commented by mr W last updated on 10/May/22
if such a triangle exists, say the red one, then it exists only one time. we can get two sides (blue) parallel to the sides of red triangle, but the third side (green) is no longer parallel to the corresponding red triangle side. i.e. the red triangle is unique.
$${if}\:{such}\:{a}\:{triangle}\:{exists},\:{say}\:{the}\:{red} \\ $$$${one},\:{then}\:{it}\:{exists}\:{only}\:{one}\:{time}. \\ $$$${we}\:{can}\:{get}\:{two}\:{sides}\:\left({blue}\right)\:{parallel} \\ $$$${to}\:{the}\:{sides}\:{of}\:{red}\:{triangle},\:{but}\:{the} \\ $$$${third}\:{side}\:\left({green}\right)\:{is}\:{no}\:{longer}\: \\ $$$${parallel}\:{to}\:{the}\:{corresponding}\:{red} \\ $$$${triangle}\:{side}.\:{i}.{e}.\:{the}\:{red}\:{triangle} \\ $$$${is}\:{unique}. \\ $$
Answered by mr W last updated on 10/May/22
Commented by mr W last updated on 10/May/22
we can see that ΔQRP∼ΔABC ⇒(p/c)=(q/a)=(r/b)=k, say a=(q/(tan B))+(r/(sin C)) a=((ka)/(tan B))+((kb)/(sin C)) a((1/k)−(1/(tan B)))=(b/(sin C)) ⇒((a(sin B−k cos B))/(k sin B))=(b/(sin C)) ...(i) similarly ⇒((b(sin C−k cos C))/(k sin C))=(c/(sin A)) ...(ii) ⇒((c(sin A−k cos A))/(k sin A))=(a/(sin B)) ...(iii) (i)×(ii)×(iii): (((sin A−k cos A)(sin B−k cos B)(sin C−k cos C))/k^3 )=1 (1+cos A cos B cos C)k^3 −(sin A sin B sin C)k^2 +(1+cos A cos B cos C)k−sin A sin B sin C=0 (k−((sin A sin B sin C)/(1+cos A cos B cos C)))(k^2 +1)=0 ⇒k=((sin A sin B sin C)/(1+cos A cos B cos C)) ⇒k=((2 sin A sin B sin C)/(sin^2 A+sin^2 B+sin^2 C)) sin A=(a/(2R)), sin B=(b/(2R)), sin C=(c/(2R)) ⇒sin A sin B sin C=((abc)/(8R^3 )) ⇒sin^2 A+sin^2 B+sin^2 C=((a^2 +b^2 +c^2 )/(4R^2 )) k=((abc)/(8R^3 ))×((8R^2 )/(a^2 +b^2 +c^2 )) k=((abc)/((a^2 +b^2 +c^2 )R)) ⇒k=((4Δ_(ABC) )/(a^2 +b^2 +c^2 )) Δ_(PQR) =k^2 Δ_(ABC) =((16Δ_(ABC) ^3 )/((a^2 +b^2 +c^2 )^2 )) ✓
$${we}\:{can}\:{see}\:{that} \\ $$$$\Delta{QRP}\sim\Delta{ABC} \\ $$$$\Rightarrow\frac{{p}}{{c}}=\frac{{q}}{{a}}=\frac{{r}}{{b}}={k},\:{say} \\ $$$${a}=\frac{{q}}{\mathrm{tan}\:{B}}+\frac{{r}}{\mathrm{sin}\:{C}} \\ $$$${a}=\frac{{ka}}{\mathrm{tan}\:{B}}+\frac{{kb}}{\mathrm{sin}\:{C}} \\ $$$${a}\left(\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{\mathrm{tan}\:{B}}\right)=\frac{{b}}{\mathrm{sin}\:{C}} \\ $$$$\Rightarrow\frac{{a}\left(\mathrm{sin}\:{B}−{k}\:\mathrm{cos}\:{B}\right)}{{k}\:\mathrm{sin}\:{B}}=\frac{{b}}{\mathrm{sin}\:{C}}\:\:\:…\left({i}\right) \\ $$$${similarly} \\ $$$$\Rightarrow\frac{{b}\left(\mathrm{sin}\:{C}−{k}\:\mathrm{cos}\:{C}\right)}{{k}\:\mathrm{sin}\:{C}}=\frac{{c}}{\mathrm{sin}\:{A}}\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow\frac{{c}\left(\mathrm{sin}\:{A}−{k}\:\mathrm{cos}\:{A}\right)}{{k}\:\mathrm{sin}\:{A}}=\frac{{a}}{\mathrm{sin}\:{B}}\:\:\:…\left({iii}\right) \\ $$$$\left({i}\right)×\left({ii}\right)×\left({iii}\right): \\ $$$$\frac{\left(\mathrm{sin}\:{A}−{k}\:\mathrm{cos}\:{A}\right)\left(\mathrm{sin}\:{B}−{k}\:\mathrm{cos}\:{B}\right)\left(\mathrm{sin}\:{C}−{k}\:\mathrm{cos}\:{C}\right)}{{k}^{\mathrm{3}} }=\mathrm{1} \\ $$$$\left(\mathrm{1}+\mathrm{cos}\:{A}\:\mathrm{cos}\:{B}\:\mathrm{cos}\:{C}\right){k}^{\mathrm{3}} −\left(\mathrm{sin}\:{A}\:\mathrm{sin}\:{B}\:\mathrm{sin}\:{C}\right){k}^{\mathrm{2}} +\left(\mathrm{1}+\mathrm{cos}\:{A}\:\mathrm{cos}\:{B}\:\mathrm{cos}\:{C}\right){k}−\mathrm{sin}\:{A}\:\mathrm{sin}\:{B}\:\mathrm{sin}\:{C}=\mathrm{0} \\ $$$$\left({k}−\frac{\mathrm{sin}\:{A}\:\mathrm{sin}\:{B}\:\mathrm{sin}\:{C}}{\mathrm{1}+\mathrm{cos}\:{A}\:\mathrm{cos}\:{B}\:\mathrm{cos}\:{C}}\right)\left({k}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{k}=\frac{\mathrm{sin}\:{A}\:\mathrm{sin}\:{B}\:\mathrm{sin}\:{C}}{\mathrm{1}+\mathrm{cos}\:{A}\:\mathrm{cos}\:{B}\:\mathrm{cos}\:{C}} \\ $$$$\Rightarrow{k}=\frac{\mathrm{2}\:\mathrm{sin}\:{A}\:\mathrm{sin}\:{B}\:\mathrm{sin}\:{C}}{\mathrm{sin}^{\mathrm{2}} \:{A}+\mathrm{sin}^{\mathrm{2}} \:{B}+\mathrm{sin}^{\mathrm{2}} \:{C}} \\ $$$$\mathrm{sin}\:{A}=\frac{{a}}{\mathrm{2}{R}},\:\mathrm{sin}\:{B}=\frac{{b}}{\mathrm{2}{R}},\:\mathrm{sin}\:{C}=\frac{{c}}{\mathrm{2}{R}} \\ $$$$\Rightarrow\mathrm{sin}\:{A}\:\mathrm{sin}\:{B}\:\mathrm{sin}\:{C}=\frac{{abc}}{\mathrm{8}{R}^{\mathrm{3}} } \\ $$$$\Rightarrow\mathrm{sin}^{\mathrm{2}} \:{A}+\mathrm{sin}^{\mathrm{2}} \:{B}+\mathrm{sin}^{\mathrm{2}} \:{C}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{4}{R}^{\mathrm{2}} } \\ $$$${k}=\frac{{abc}}{\mathrm{8}{R}^{\mathrm{3}} }×\frac{\mathrm{8}{R}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$${k}=\frac{{abc}}{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right){R}} \\ $$$$\Rightarrow{k}=\frac{\mathrm{4}\Delta_{{ABC}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$\Delta_{{PQR}} ={k}^{\mathrm{2}} \Delta_{{ABC}} =\frac{\mathrm{16}\Delta_{{ABC}} ^{\mathrm{3}} }{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\checkmark \\ $$
Commented by Tawa11 last updated on 08/Oct/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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