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Question-226337

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Question Number 226337 by Spillover last updated on 25/Nov/25
Answered by mr W last updated on 26/Nov/25
Commented by mr W last updated on 26/Nov/25
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Commented by Spillover last updated on 27/Nov/25
very nice sketch.name of app please?
$${very}\:{nice}\:{sketch}.{name}\:{of}\:{app}\:{please}? \\ $$
Commented by mr W last updated on 27/Nov/25
for drawing such freehand sketches i usually use an app named INKredible.
$${for}\:{drawing}\:{such}\:{freehand}\:{sketches}\: \\ $$$${i}\:{usually}\:{use}\:{an}\:{app}\:{named}\: \\ $$$${INKredible}. \\ $$
Commented by fantastic2 last updated on 27/Nov/25
the app is very good you can also try lekh
$${the}\:{app}\:{is}\:{very}\:{good} \\ $$$${you}\:{can}\:{also}\:{try}\:{lekh} \\ $$
Answered by Spillover last updated on 27/Nov/25
Let P, Q ,R be the vertices of the equilateral triangle with eccentric angles α,β,γ The coordinates of the vertices P(acos α,bsin α) P(acos β,bsin β) P(acos γ,bsin γ) The centroid G(h,k) h=(a/3)(cos α+cos β+cos γ) k=(b/3)(sin α+sin β+sin γ) From properties of the eccentric angles Σcos α=((3x)/a) Σsin α=((3y)/b) 3x+acos (α+β+γ)=((3xa^2 )/(a^2 −b^2 ))+((3xb^2 )/(a^2 −b^2 )) cos (α+β+γ)=((6xb^2 )/(a(a^2 −b^2 )_(spillover) )) also sin (α+β+γ)=−((6ya^2 )/(b(a^2 −b^2 )_(spillover) )) From Σ_(i=1) ^3 x_i ^2 =((3a^2 (a^2 +3b^2 ))/(2(a^2 +b^2 )))−((3x^2 (a^2 +3b^2 ))/(a^2 −b^2 )) Σ_(i=1) ^3 y_i ^2 =((3b^2 (b^2 +3a^2 ))/(2(a^2 −b^2 )))−((3y^2 (b^2 +3a^2 ))/(a^2 −b^2 )) ((x^2 (a^2 +3b^2 )^2 )/a^2 )+((y^2 (b^2 +3a^2 )^2 )/b^2 )=(a^2 −b^2 )^2 (((a^2 +3b^2 )^2 )/(a^2 (a^2 −b^2 )^2 ))x^2 +(((b^2 +3a^2 )^2 )/(b^2 (a^2 −b^2 )^2 ))y^2 =1
$${Let}\:\:{P},\:{Q}\:,{R}\:{be}\:{the}\:{vertices}\:{of}\:{the}\:{equilateral} \\ $$$${triangle}\:{with}\:{eccentric}\:{angles}\:\alpha,\beta,\gamma \\ $$$${The}\:{coordinates}\:{of}\:{the}\:{vertices} \\ $$$${P}\left({a}\mathrm{cos}\:\alpha,{b}\mathrm{sin}\:\alpha\right) \\ $$$${P}\left({a}\mathrm{cos}\:\beta,{b}\mathrm{sin}\:\beta\right) \\ $$$${P}\left({a}\mathrm{cos}\:\gamma,{b}\mathrm{sin}\:\gamma\right) \\ $$$${The}\:{centroid}\:{G}\left({h},{k}\right) \\ $$$${h}=\frac{{a}}{\mathrm{3}}\left(\mathrm{cos}\:\alpha+\mathrm{cos}\:\beta+\mathrm{cos}\:\gamma\right) \\ $$$${k}=\frac{{b}}{\mathrm{3}}\left(\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta+\mathrm{sin}\:\:\gamma\right) \\ $$$${From}\:{properties}\:{of}\:{the}\:{eccentric}\:{angles} \\ $$$$\Sigma\mathrm{cos}\:\alpha=\frac{\mathrm{3}{x}}{{a}}\:\:\:\:\:\:\Sigma\mathrm{sin}\:\:\alpha=\frac{\mathrm{3}{y}}{{b}}\: \\ $$$$\mathrm{3}{x}+{a}\mathrm{cos}\:\left(\alpha+\beta+\gamma\right)=\frac{\mathrm{3}{xa}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }+\frac{\mathrm{3}{xb}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$$\mathrm{cos}\:\left(\alpha+\beta+\gamma\right)=\frac{\mathrm{6}{xb}^{\mathrm{2}} }{{a}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)_{{spillover}} } \\ $$$${also} \\ $$$$\mathrm{sin}\:\:\left(\alpha+\beta+\gamma\right)=−\frac{\mathrm{6}{ya}^{\mathrm{2}} }{{b}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)_{{spillover}} } \\ $$$${From} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}{x}_{{i}} ^{\mathrm{2}} =\frac{\mathrm{3}{a}^{\mathrm{2}} \left({a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} \right)}{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}−\frac{\mathrm{3}{x}^{\mathrm{2}} \left({a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} \right)}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$$\:\underset{{i}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}{y}_{{i}} ^{\mathrm{2}} =\frac{\mathrm{3}{b}^{\mathrm{2}} \left({b}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} \right)}{\mathrm{2}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}−\frac{\mathrm{3}{y}^{\mathrm{2}} \left({b}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} \right)}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$$\frac{{x}^{\mathrm{2}} \left({a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} \right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} \left({b}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\frac{\left({a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} \right)^{\mathrm{2}} }{{a}^{\mathrm{2}} \left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} }{x}^{\mathrm{2}} +\frac{\left({b}^{\mathrm{2}} +\mathrm{3}{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{{b}^{\mathrm{2}} \left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} }{y}^{\mathrm{2}} =\mathrm{1} \\ $$$$ \\ $$
Answered by Spillover last updated on 27/Nov/25
Answered by Spillover last updated on 27/Nov/25
alternative From △PQR h=((a^2 −b^2 )/(4a))(cos α+cos β+cos γ) k=((a^2 −b^2 )/(4a))(sin α+sin β+sin γ) cos (α+β+γ)=((h(a^2 +3b^2 ))/(a(a^2 −b^2 ))) sin (α+β+γ)=((k(a^2 +3b^2 ))/(b(a^2 −b^2 ))) from sin^2 θ+cos^2 θ=1_(spillover) (((h(a^2 +3b^2 ))/(a(a^2 −b^2 ))) )^2 +[((k(a^2 +3b^2 ))/(b(a^2 −b^2 ))) ]^2 =1 hence [(((x(a^2 +3b^2 ))/(a(a^2 −b^2 ))))^2 +[(((y(a^2 +3b^2 ))/(b(a^2 −b^2 )))]^2 =1
$${alternative} \\ $$$${From}\:\bigtriangleup{PQR} \\ $$$${h}=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{4}{a}}\left(\mathrm{cos}\:\alpha+\mathrm{cos}\:\beta+\mathrm{cos}\:\gamma\right) \\ $$$${k}=\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{4}{a}}\left(\mathrm{sin}\:\:\alpha+\mathrm{sin}\:\:\beta+\mathrm{sin}\:\gamma\right) \\ $$$$\mathrm{cos}\:\left(\alpha+\beta+\gamma\right)=\frac{{h}\left({a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} \right)}{{a}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)} \\ $$$$\mathrm{sin}\:\:\left(\alpha+\beta+\gamma\right)=\frac{{k}\left({a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} \right)}{{b}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)} \\ $$$${from}\:\mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{cos}\:^{\mathrm{2}} \theta=\mathrm{1}_{{spillover}} \\ $$$$\left(\frac{{h}\left({a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} \right)}{{a}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}\:\right)^{\mathrm{2}} +\left[\frac{{k}\left({a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} \right)}{{b}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}\:\right]^{\mathrm{2}} =\mathrm{1} \\ $$$${hence} \\ $$$$\left[\left(\frac{{x}\left({a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} \right)}{{a}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}\right)^{\mathrm{2}} +\left[\left(\frac{{y}\left({a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} \right)}{{b}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}\right]^{\mathrm{2}} =\mathrm{1}\right.\right. \\ $$$$\:\: \\ $$

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