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 35390 by ajfour last updated on 18/May/18

Commented by ajfour last updated on 18/May/18

Find maximum area of △ABC.  The ellipse equation is (x^2 /a^2 )+(y^2 /b^2 )=1 .

$${Find}\:{maximum}\:{area}\:{of}\:\bigtriangleup{ABC}. \\ $$$${The}\:{ellipse}\:{equation}\:{is}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:. \\ $$

Commented by MJS last updated on 18/May/18

lol. I solved this some years ago, not knowing  that Steiner solved it long before me.  all triangles with barycenter in the center  of the ellipse are of equal area.  I will post my proof but it might take some  time.  google Steiner−ellipse or look it up on   Wikipedia

$$\mathrm{lol}.\:\mathrm{I}\:\mathrm{solved}\:\mathrm{this}\:\mathrm{some}\:\mathrm{years}\:\mathrm{ago},\:\mathrm{not}\:\mathrm{knowing} \\ $$$$\mathrm{that}\:\mathrm{Steiner}\:\mathrm{solved}\:\mathrm{it}\:\mathrm{long}\:\mathrm{before}\:\mathrm{me}. \\ $$$$\mathrm{all}\:\mathrm{triangles}\:\mathrm{with}\:\mathrm{barycenter}\:\mathrm{in}\:\mathrm{the}\:\mathrm{center} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{ellipse}\:\mathrm{are}\:\mathrm{of}\:\mathrm{equal}\:\mathrm{area}. \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{post}\:\mathrm{my}\:\mathrm{proof}\:\mathrm{but}\:\mathrm{it}\:\mathrm{might}\:\mathrm{take}\:\mathrm{some} \\ $$$$\mathrm{time}. \\ $$$$\mathrm{google}\:{Steiner}−{ellipse}\:\mathrm{or}\:\mathrm{look}\:\mathrm{it}\:\mathrm{up}\:\mathrm{on}\: \\ $$$${Wikipedia} \\ $$

Commented by MJS last updated on 18/May/18

btw I′m still engaged with that other problem...

$$\mathrm{btw}\:\mathrm{I}'\mathrm{m}\:\mathrm{still}\:\mathrm{engaged}\:\mathrm{with}\:\mathrm{that}\:\mathrm{other}\:\mathrm{problem}... \\ $$

Commented by ajfour last updated on 18/May/18

so kind of you Sir!

$${so}\:{kind}\:{of}\:{you}\:{Sir}! \\ $$

Commented by MJS last updated on 18/May/18

start with a circle with radius a  x^2 +(y^∗ )^2 =a^2   insert an equilateral triangle A^∗ B^∗ C^∗  with  A^∗ = ((x_A ),((y_A ^∗ =(a/b)y_A )) ) ; B^∗ = ((x_B ),((y_B ^∗ =(a/b)y_B )) ) ;C^∗ = ((x_C ),((y_C ^∗ =(a/b)y_C )) )  now transform y=(b/a)y^∗   the circle becomes the ellipse       with area_(ell) =(b/a)×area_(circle) =abπ  the equilateral triangle becomes the given one       with area_△ =(b/a)×area_(equi△)   it should be easy to see that the barycenter  stays in place and all possible triangles  have the same area.    I will post some equations for calculating  both directions: find smallest ellipse to a  given triangle and find the angle and ratio  to transform a equilateral triangle into  any triangle... just give me some time to  find these formulas within the chaos in my  library...

$$\mathrm{start}\:\mathrm{with}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{radius}\:{a} \\ $$$${x}^{\mathrm{2}} +\left({y}^{\ast} \right)^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\mathrm{insert}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle}\:{A}^{\ast} {B}^{\ast} {C}^{\ast} \:\mathrm{with} \\ $$$${A}^{\ast} =\begin{pmatrix}{{x}_{{A}} }\\{{y}_{{A}} ^{\ast} =\frac{{a}}{{b}}{y}_{{A}} }\end{pmatrix}\:;\:{B}^{\ast} =\begin{pmatrix}{{x}_{{B}} }\\{{y}_{{B}} ^{\ast} =\frac{{a}}{{b}}{y}_{{B}} }\end{pmatrix}\:;{C}^{\ast} =\begin{pmatrix}{{x}_{{C}} }\\{{y}_{{C}} ^{\ast} =\frac{{a}}{{b}}{y}_{{C}} }\end{pmatrix} \\ $$$$\mathrm{now}\:\mathrm{transform}\:{y}=\frac{{b}}{{a}}{y}^{\ast} \\ $$$$\mathrm{the}\:\mathrm{circle}\:\mathrm{becomes}\:\mathrm{the}\:\mathrm{ellipse} \\ $$$$\:\:\:\:\:\mathrm{with}\:\mathrm{area}_{\mathrm{ell}} =\frac{{b}}{{a}}×\mathrm{area}_{\mathrm{circle}} ={ab}\pi \\ $$$$\mathrm{the}\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{becomes}\:\mathrm{the}\:\mathrm{given}\:\mathrm{one} \\ $$$$\:\:\:\:\:\mathrm{with}\:\mathrm{area}_{\bigtriangleup} =\frac{{b}}{{a}}×\mathrm{area}_{\mathrm{equi}\bigtriangleup} \\ $$$$\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see}\:\mathrm{that}\:\mathrm{the}\:\mathrm{barycenter} \\ $$$$\mathrm{stays}\:\mathrm{in}\:\mathrm{place}\:\mathrm{and}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{triangles} \\ $$$$\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{area}. \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{post}\:\mathrm{some}\:\mathrm{equations}\:\mathrm{for}\:\mathrm{calculating} \\ $$$$\mathrm{both}\:\mathrm{directions}:\:\mathrm{find}\:\mathrm{smallest}\:\mathrm{ellipse}\:\mathrm{to}\:\mathrm{a} \\ $$$$\mathrm{given}\:\mathrm{triangle}\:\mathrm{and}\:\mathrm{find}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{and}\:\mathrm{ratio} \\ $$$$\mathrm{to}\:\mathrm{transform}\:\mathrm{a}\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{into} \\ $$$$\mathrm{any}\:\mathrm{triangle}...\:\mathrm{just}\:\mathrm{give}\:\mathrm{me}\:\mathrm{some}\:\mathrm{time}\:\mathrm{to} \\ $$$$\mathrm{find}\:\mathrm{these}\:\mathrm{formulas}\:\mathrm{within}\:\mathrm{the}\:\mathrm{chaos}\:\mathrm{in}\:\mathrm{my} \\ $$$$\mathrm{library}... \\ $$

Commented by ajfour last updated on 18/May/18

A=(1/2)×a(√3)×(((3a)/2))((b/a))       =((3(√3)ab)/4) .  Understood Sir, thank you !

$${A}=\frac{\mathrm{1}}{\mathrm{2}}×{a}\sqrt{\mathrm{3}}×\left(\frac{\mathrm{3}{a}}{\mathrm{2}}\right)\left(\frac{{b}}{{a}}\right) \\ $$$$\:\:\:\:\:=\frac{\mathrm{3}\sqrt{\mathrm{3}}\boldsymbol{{ab}}}{\mathrm{4}}\:. \\ $$$$\boldsymbol{{U}}{nderstood}\:\boldsymbol{{S}}{ir},\:{thank}\:{you}\:! \\ $$

Answered by rahul 19 last updated on 18/May/18

((3(√3))/4) ab .

$$\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}\:{ab}\:. \\ $$

Commented by rahul 19 last updated on 19/May/18

 Got the same answer by assuming  triangle to be isosceles !:)

$$\:{Got}\:{the}\:{same}\:{answer}\:{by}\:{assuming} \\ $$$$\left.{triangle}\:{to}\:{be}\:{isosceles}\:!:\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com