Question-75296

Question Number 75296 by ajfour last updated on 09/Dec/19

Commented by ajfour last updated on 09/Dec/19

If perimeter of △PQR is p, find maximum area of △PQR in terms of a,b,c,p. (p<a+b+c)

$${If}\:{perimeter}\:{of}\:\bigtriangleup{PQR}\:{is}\:{p},\:{find} \\ $$$${maximum}\:{area}\:{of}\:\bigtriangleup{PQR}\:{in} \\ $$$${terms}\:{of}\:{a},{b},{c},{p}.\:\:\:\left({p}<{a}+{b}+{c}\right)\: \\ $$

Commented by mr W last updated on 09/Dec/19

p_(min) =((a^2 (b^2 +c^2 −a^2 )+b^2 (c^2 +a^2 −b^2 )+c^2 (a^2 +b^2 −c^2 ))/(2abc)) p_(mac) =a+b+c p_(min) <p<p_(max)

$${p}_{{min}} =\frac{{a}^{\mathrm{2}} \left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)+{b}^{\mathrm{2}} \left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)+{c}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)}{\mathrm{2}{abc}} \\ $$$${p}_{{mac}} ={a}+{b}+{c} \\ $$$${p}_{{min}} <{p}<{p}_{{max}} \\ $$

Answered by mr W last updated on 10/Dec/19

Commented by mr W last updated on 10/Dec/19

p=(√((1−β)^2 b^2 +γ^2 c^2 −2(1−β)γbc cos A)) +(√((1−γ)^2 c^2 +α^2 a^2 −2(1−γ)αca cos B)) +(√((1−α)^2 a^2 +β^2 b^2 −2(1−α)βab cos C)) A_(ΔPQR) =A_(ΔABC) −(1/2)(1−β)γbc sin A −(1/2)(1−γ)αca sin B −(1/2)(1−α)βab sin C A_(ΔPQR) =A_(ΔABC) −(1/2)A A=(1−β)γbc sin A+(1−γ)αca sin B+(1−α)βab sin C ...

$${p}=\sqrt{\left(\mathrm{1}−\beta\right)^{\mathrm{2}} {b}^{\mathrm{2}} +\gamma^{\mathrm{2}} {c}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}−\beta\right)\gamma{bc}\:\mathrm{cos}\:{A}} \\ $$$$\:\:\:\:\:\:+\sqrt{\left(\mathrm{1}−\gamma\right)^{\mathrm{2}} {c}^{\mathrm{2}} +\alpha^{\mathrm{2}} {a}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}−\gamma\right)\alpha{ca}\:\mathrm{cos}\:{B}} \\ $$$$\:\:\:\:\:\:+\sqrt{\left(\mathrm{1}−\alpha\right)^{\mathrm{2}} {a}^{\mathrm{2}} +\beta^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}−\alpha\right)\beta{ab}\:\mathrm{cos}\:{C}} \\ $$$${A}_{\Delta{PQR}} ={A}_{\Delta{ABC}} −\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\beta\right)\gamma{bc}\:\mathrm{sin}\:{A} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\gamma\right)\alpha{ca}\:\mathrm{sin}\:{B} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\alpha\right)\beta{ab}\:\mathrm{sin}\:{C} \\ $$$${A}_{\Delta{PQR}} ={A}_{\Delta{ABC}} −\frac{\mathrm{1}}{\mathrm{2}}{A} \\ $$$${A}=\left(\mathrm{1}−\beta\right)\gamma{bc}\:\mathrm{sin}\:{A}+\left(\mathrm{1}−\gamma\right)\alpha{ca}\:\mathrm{sin}\:{B}+\left(\mathrm{1}−\alpha\right)\beta{ab}\:\mathrm{sin}\:{C} \\ $$$$… \\ $$

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