Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

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

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com