Menu Close

Question-37447




Question Number 37447 by ajfour last updated on 13/Jun/18
Commented by ajfour last updated on 13/Jun/18
Find the maximum elliptic area  (in the shown orientation)   cicumscribed by the isosceles  triangle.
$${Find}\:{the}\:{maximum}\:{elliptic}\:{area} \\ $$$$\left({in}\:{the}\:{shown}\:{orientation}\right)\: \\ $$$${cicumscribed}\:{by}\:{the}\:{isosceles} \\ $$$${triangle}. \\ $$
Commented by MrW3 last updated on 13/Jun/18
incircle is the max. ellipse inscribed  in a triangle with side lengthes a,b,c.  r=(1/2)(√(((−a+b+c)(a−b+c)(a+b−c))/((a+b+c))))  A=πr^2 =((π(−a+b+c)(a−b+c)(a+b−c))/(4(a+b+c)))    A=((π(−p+p+q)(p−p+q)(p+p−q))/(4(p+p+q)))  ⇒A=((π(2p−q)q^2 )/(4(2p+q)))
$${incircle}\:{is}\:{the}\:{max}.\:{ellipse}\:{inscribed} \\ $$$${in}\:{a}\:{triangle}\:{with}\:{side}\:{lengthes}\:{a},{b},{c}. \\ $$$${r}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}{\left({a}+{b}+{c}\right)}} \\ $$$${A}=\pi{r}^{\mathrm{2}} =\frac{\pi\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}{\mathrm{4}\left({a}+{b}+{c}\right)} \\ $$$$ \\ $$$${A}=\frac{\pi\left(−{p}+{p}+{q}\right)\left({p}−{p}+{q}\right)\left({p}+{p}−{q}\right)}{\mathrm{4}\left({p}+{p}+{q}\right)} \\ $$$$\Rightarrow{A}=\frac{\pi\left(\mathrm{2}{p}−{q}\right){q}^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{2}{p}+{q}\right)} \\ $$
Commented by ajfour last updated on 13/Jun/18
Thank you Sir, but my solution  fetches some other conclusion.
$${Thank}\:{you}\:{Sir},\:{but}\:{my}\:{solution} \\ $$$${fetches}\:{some}\:{other}\:{conclusion}. \\ $$
Commented by MJS last updated on 13/Jun/18
incircle is not the greatest ellipse inside a  triangle and circumcircle is not the smallest  ellipse surrounding it. see my answer below
$$\mathrm{incircle}\:\mathrm{is}\:\mathrm{not}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{ellipse}\:\mathrm{inside}\:\mathrm{a} \\ $$$$\mathrm{triangle}\:\mathrm{and}\:\mathrm{circumcircle}\:\mathrm{is}\:\mathrm{not}\:\mathrm{the}\:\mathrm{smallest} \\ $$$$\mathrm{ellipse}\:\mathrm{surrounding}\:\mathrm{it}.\:\mathrm{see}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{below} \\ $$
Commented by MrW3 last updated on 13/Jun/18
Thanks for the information!
$${Thanks}\:{for}\:{the}\:{information}! \\ $$
Answered by ajfour last updated on 13/Jun/18
let altitude of triangle be h.  Also let BC be x axis with mid  point of BC as origin, the  vertical as y axis.  Since AC is tangent to ellipse,      h−b=(√(a^2 m^2 +b^2 ))   where m=(h/(q/2))    ⇒  a^2 =((h^2 −2bh)/m^2 )   ...(i)  A_(ellipse) ^2 =π^2 a^2 b^2        let  b=x ,then using (i)  A_(ellipse) ^2 =π^2 (((h^2 −2hx)/m^2 ))x^2   (dA_(ellipse) ^2 /dx) = ((π^2 h)/m^2 )(hx^2 −2x^3 )  for max. area we have       2hx=6x^2   ⇒    b=x= (h/3)  and using (i)           a^2 =[h^2 −2((h/3))h]/m^2           a= (h/(m(√3)))  ⇒ unless m=(√3)   , max. area  ellipse inscribed in triangle is  not a circle.  A_(ellipse, max) =𝛑((h/(m(√3))))((h/3))  and with m=(h/(q/2))   A_(ellipse , max) =((𝛑hq)/(6(√3))) = ((𝛑q(√(p^2 −(q^2 /4))))/(6(√3))) .
$${let}\:{altitude}\:{of}\:{triangle}\:{be}\:\boldsymbol{{h}}. \\ $$$${Also}\:{let}\:{BC}\:{be}\:{x}\:{axis}\:{with}\:{mid} \\ $$$${point}\:{of}\:{BC}\:{as}\:{origin},\:{the} \\ $$$${vertical}\:{as}\:{y}\:{axis}. \\ $$$${Since}\:{AC}\:{is}\:{tangent}\:{to}\:{ellipse}, \\ $$$$\:\:\:\:{h}−{b}=\sqrt{{a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} }\: \\ $$$${where}\:{m}=\frac{{h}}{{q}/\mathrm{2}} \\ $$$$\:\:\Rightarrow\:\:{a}^{\mathrm{2}} =\frac{{h}^{\mathrm{2}} −\mathrm{2}{bh}}{{m}^{\mathrm{2}} }\:\:\:…\left({i}\right) \\ $$$${A}_{{ellipse}} ^{\mathrm{2}} =\pi^{\mathrm{2}} {a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\:\:\:\:\:{let}\:\:{b}={x}\:,{then}\:{using}\:\left({i}\right) \\ $$$${A}_{{ellipse}} ^{\mathrm{2}} =\pi^{\mathrm{2}} \left(\frac{{h}^{\mathrm{2}} −\mathrm{2}{hx}}{{m}^{\mathrm{2}} }\right){x}^{\mathrm{2}} \\ $$$$\frac{{dA}_{{ellipse}} ^{\mathrm{2}} }{{dx}}\:=\:\frac{\pi^{\mathrm{2}} {h}}{{m}^{\mathrm{2}} }\left({hx}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{3}} \right) \\ $$$${for}\:{max}.\:{area}\:{we}\:{have} \\ $$$$\:\:\:\:\:\mathrm{2}{hx}=\mathrm{6}{x}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\:\boldsymbol{{b}}=\boldsymbol{{x}}=\:\frac{\boldsymbol{{h}}}{\mathrm{3}}\:\:{and}\:{using}\:\left({i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{{a}}^{\mathrm{2}} =\left[\boldsymbol{{h}}^{\mathrm{2}} −\mathrm{2}\left(\frac{{h}}{\mathrm{3}}\right){h}\right]/{m}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{a}}=\:\frac{\boldsymbol{{h}}}{\boldsymbol{{m}}\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\:{unless}\:{m}=\sqrt{\mathrm{3}}\:\:\:,\:{max}.\:{area} \\ $$$${ellipse}\:{inscribed}\:{in}\:{triangle}\:{is} \\ $$$${not}\:{a}\:{circle}. \\ $$$${A}_{{ellipse},\:{max}} =\boldsymbol{\pi}\left(\frac{\boldsymbol{{h}}}{\boldsymbol{{m}}\sqrt{\mathrm{3}}}\right)\left(\frac{\boldsymbol{{h}}}{\mathrm{3}}\right) \\ $$$${and}\:{with}\:{m}=\frac{{h}}{{q}/\mathrm{2}}\: \\ $$$${A}_{{ellipse}\:,\:{max}} =\frac{\boldsymbol{\pi{hq}}}{\mathrm{6}\sqrt{\mathrm{3}}}\:=\:\frac{\boldsymbol{\pi{q}}\sqrt{\boldsymbol{{p}}^{\mathrm{2}} −\frac{\boldsymbol{{q}}^{\mathrm{2}} }{\mathrm{4}}}}{\mathrm{6}\sqrt{\mathrm{3}}}\:. \\ $$$$ \\ $$
Commented by MrW3 last updated on 13/Jun/18
Nice working!
$${Nice}\:{working}! \\ $$
Answered by MJS last updated on 13/Jun/18
I once promised to give you some formulas  for the ellipse−triangle−problem  triangle abc  smallest ellipse surrounding the triangle  a=((√(a^2 +b^2 +c^2 +2(√(a^2 b^2 +a^2 c^2 +b^2 c^2 −δ^2 ))))/3)  b=((√(a^2 +b^2 +c^2 −2(√(a^2 b^2 +a^2 c^2 +b^2 c^2 −δ^2 ))))/3)  e=(√(a^2 −b^2 ))=((2(√(a^2 b^2 +a^2 c^2 +b^2 c^2 −δ^2 )))/3)  area=abπ=δ((√3)/9)  δ=(√((a+b+c)(a+b−c)(a+c−b)(b+c−a)))  the greatest ellipse inside the same triangle  a_2 =(a/2); b_2 =(b/2); e_2 =(e/2); area=((abπ)/4)=δ((√3)/(36))  the center of both ellipses is the barycenter  of the triangle  for an equilateral triangle these ellipses are  the circumcircle and the incircle    the area of the greatest inner ellipse of the  triangle ppq is ((√3)/(36))q(√(4p^2 −q^2 ))
$$\mathrm{I}\:\mathrm{once}\:\mathrm{promised}\:\mathrm{to}\:\mathrm{give}\:\mathrm{you}\:\mathrm{some}\:\mathrm{formulas} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{ellipse}−\mathrm{triangle}−\mathrm{problem} \\ $$$$\mathrm{triangle}\:{abc} \\ $$$$\mathrm{smallest}\:\mathrm{ellipse}\:\mathrm{surrounding}\:\mathrm{the}\:\mathrm{triangle} \\ $$$$\mathfrak{a}=\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} −\delta^{\mathrm{2}} }}}{\mathrm{3}} \\ $$$$\mathfrak{b}=\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} −\delta^{\mathrm{2}} }}}{\mathrm{3}} \\ $$$$\mathfrak{e}=\sqrt{\mathfrak{a}^{\mathrm{2}} −\mathfrak{b}^{\mathrm{2}} }=\frac{\mathrm{2}\sqrt{{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} −\delta^{\mathrm{2}} }}{\mathrm{3}} \\ $$$$\mathrm{area}=\mathfrak{ab}\pi=\delta\frac{\sqrt{\mathrm{3}}}{\mathrm{9}} \\ $$$$\delta=\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}+{c}−{b}\right)\left({b}+{c}−{a}\right)} \\ $$$$\mathrm{the}\:\mathrm{greatest}\:\mathrm{ellipse}\:\mathrm{inside}\:\mathrm{the}\:\mathrm{same}\:\mathrm{triangle} \\ $$$$\mathfrak{a}_{\mathrm{2}} =\frac{\mathfrak{a}}{\mathrm{2}};\:\mathfrak{b}_{\mathrm{2}} =\frac{\mathfrak{b}}{\mathrm{2}};\:\mathfrak{e}_{\mathrm{2}} =\frac{\mathfrak{e}}{\mathrm{2}};\:\mathrm{area}=\frac{\mathfrak{ab}\pi}{\mathrm{4}}=\delta\frac{\sqrt{\mathrm{3}}}{\mathrm{36}} \\ $$$$\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{both}\:\mathrm{ellipses}\:\mathrm{is}\:\mathrm{the}\:\mathrm{barycenter} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{triangle} \\ $$$$\mathrm{for}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{these}\:\mathrm{ellipses}\:\mathrm{are} \\ $$$$\mathrm{the}\:\mathrm{circumcircle}\:\mathrm{and}\:\mathrm{the}\:\mathrm{incircle} \\ $$$$ \\ $$$$\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{inner}\:\mathrm{ellipse}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{triangle}\:{ppq}\:\mathrm{is}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{36}}{q}\sqrt{\mathrm{4}{p}^{\mathrm{2}} −{q}^{\mathrm{2}} } \\ $$
Commented by MJS last updated on 13/Jun/18
put the triangle like this  A= ((((a^2 −b^2 −3c^2 )/(6c))),((−(δ/(6c)))) )  B= ((((a^2 −b^2 +3c^2 )/(6c))),((−(δ/(6c)))) )  C= ((((−a^2 +b^2 )/(3c))),((δ/(3c))) )  the barycenter is  ((0),(0) )  the center and radius of the incircle  C_I = (((((a−b)(a+b−3c))/(6c))),((((2c−a−b)δ)/(6(a+b+c)c))) )  r_I =(δ/(2(a+b+c)))  the center and radius of the circumcircle  C_C = ((((a^2 −b^2 )/(6c))),(((((a^2 +b^2 −c^2 )c)/(2δ))−(δ/(6c)))) )  R_C =((abc)/δ)  the equation of the smallest surrounding  ellipse  δ^2 x^2 +2(a^2 −b^2 )δxy+((a^2 −b^2 )^2 +3c^4 )y^2 −((c^2 δ^2 )/3)=0r  the equation of the greatest inner ellipse  4δ^2 x^2 +8(a^2 −b^2 )δxy+4((a^2 −b^2 )^2 +3c^4 )y^2 −((c^2 δ^2 )/3)=0
$$\mathrm{put}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{like}\:\mathrm{this} \\ $$$${A}=\begin{pmatrix}{\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −\mathrm{3}{c}^{\mathrm{2}} }{\mathrm{6}{c}}}\\{−\frac{\delta}{\mathrm{6}{c}}}\end{pmatrix} \\ $$$${B}=\begin{pmatrix}{\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{3}{c}^{\mathrm{2}} }{\mathrm{6}{c}}}\\{−\frac{\delta}{\mathrm{6}{c}}}\end{pmatrix} \\ $$$${C}=\begin{pmatrix}{\frac{−{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{3}{c}}}\\{\frac{\delta}{\mathrm{3}{c}}}\end{pmatrix} \\ $$$$\mathrm{the}\:\mathrm{barycenter}\:\mathrm{is}\:\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{the}\:\mathrm{center}\:\mathrm{and}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{incircle} \\ $$$${C}_{{I}} =\begin{pmatrix}{\frac{\left({a}−{b}\right)\left({a}+{b}−\mathrm{3}{c}\right)}{\mathrm{6}{c}}}\\{\frac{\left(\mathrm{2}{c}−{a}−{b}\right)\delta}{\mathrm{6}\left({a}+{b}+{c}\right){c}}}\end{pmatrix} \\ $$$${r}_{{I}} =\frac{\delta}{\mathrm{2}\left({a}+{b}+\mathrm{c}\right)} \\ $$$$\mathrm{the}\:\mathrm{center}\:\mathrm{and}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circumcircle} \\ $$$${C}_{{C}} =\begin{pmatrix}{\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{6}{c}}}\\{\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right){c}}{\mathrm{2}\delta}−\frac{\delta}{\mathrm{6}{c}}}\end{pmatrix} \\ $$$${R}_{{C}} =\frac{{abc}}{\delta} \\ $$$$\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{smallest}\:\mathrm{surrounding} \\ $$$$\mathrm{ellipse} \\ $$$$\delta^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\delta{xy}+\left(\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{3}{c}^{\mathrm{4}} \right){y}^{\mathrm{2}} −\frac{{c}^{\mathrm{2}} \delta^{\mathrm{2}} }{\mathrm{3}}=\mathrm{0r} \\ $$$$\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{inner}\:\mathrm{ellipse} \\ $$$$\mathrm{4}\delta^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{8}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\delta{xy}+\mathrm{4}\left(\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{3}{c}^{\mathrm{4}} \right){y}^{\mathrm{2}} −\frac{{c}^{\mathrm{2}} \delta^{\mathrm{2}} }{\mathrm{3}}=\mathrm{0} \\ $$
Commented by ajfour last updated on 13/Jun/18
Thank you very much Sir.
$${Thank}\:{you}\:{very}\:{much}\:{Sir}. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *