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Question Number 15908 by mrW1 last updated on 15/Jun/17
Commented by mrW1 last updated on 15/Jun/17
On each side of a triangle an equilateral triangle is constructed. The centroids of these triangles form a new triangle. Prove that the new triangle is equilateral. What is the length of its sides if the original triangle has sides a,b,c? What is the area of the new triangle?
$$\mathrm{On}\:\mathrm{each}\:\mathrm{side}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{an}\:\mathrm{equilateral} \\ $$$$\mathrm{triangle}\:\mathrm{is}\:\mathrm{constructed}.\:\mathrm{The}\:\mathrm{centroids} \\ $$$$\mathrm{of}\:\mathrm{these}\:\mathrm{triangles}\:\mathrm{form}\:\mathrm{a}\:\mathrm{new}\:\mathrm{triangle}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{new}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{equilateral}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{its}\:\mathrm{sides}\:\mathrm{if}\:\mathrm{the} \\ $$$$\mathrm{original}\:\mathrm{triangle}\:\mathrm{has}\:\mathrm{sides}\:\mathrm{a},\mathrm{b},\mathrm{c}? \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{new}\:\mathrm{triangle}? \\ $$
Commented by ajfour last updated on 16/Jun/17
side of the equilateral triangle formed is : s=(1/( (√3)))(√(a^2 +b^2 +2abcos (C−((2π)/3)))) therefore its area is : A=(1/(4(√3))) [a^2 +b^2 +2abcos (C−((2π)/3))] where C=cos^(−1) (((a^2 +b^2 −c^2 )/(2ab))) . please confirm, Sir. answer seems somewhat messy..
$${side}\:{of}\:{the}\:{equilateral}\:{triangle} \\ $$$${formed}\:{is}\:: \\ $$$$\:\:\:{s}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}\mathrm{cos}\:\left({C}−\frac{\mathrm{2}\pi}{\mathrm{3}}\right)}\: \\ $$$${therefore}\:{its}\:{area}\:{is}\:: \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}}\:\left[{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}\mathrm{cos}\:\left({C}−\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right] \\ $$$${where}\:\:\:\:{C}=\mathrm{cos}^{−\mathrm{1}} \left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}\right)\:. \\ $$$$\:{please}\:{confirm},\:{Sir}. \\ $$$${answer}\:{seems}\:{somewhat}\:{messy}.. \\ $$
Commented by ajfour last updated on 16/Jun/17
is it not correct, sir (mrW1) ?
$${is}\:{it}\:{not}\:{correct},\:{sir}\:\left({mrW}\mathrm{1}\right)\:? \\ $$
Commented by mrW1 last updated on 16/Jun/17
I try to go on with your result: cos C=((a^2 +b^2 −c^2 )/(2ab)) sin C=((2(√(p′(p′−a)(p′−b)(p′−c))))/(ab)) with p′=((a+b+c)/2) cos (C−((2π)/3))=cos C cos ((2π)/3) + sin C sin ((2π)/3) =(1/2)(−cos C+(√3)sin C) =(1/2)[−((a^2 +b^2 −c^2 )/(2ab))+2(√3)((√(p′(p′−a)(p′−b)(p′−c)))/(ab))] =(1/(2ab))[−((a^2 +b^2 −c^2 )/2)+2(√3)(√(p′(p′−a)(p′−b)(p′−c)))] s=(1/( (√3)))(√(a^2 +b^2 +2abcos (C−((2π)/3)))) =(1/( (√3)))(√(a^2 +b^2 −((a^2 +b^2 −c^2 )/2)+2(√3)(√(p′(p′−a)(p′−b)(p′−c))))) =(1/( (√3)))(√(((a^2 +b^2 +c^2 )/2)+2(√3)(√(p′(p′−a)(p′−b)(p′−c))))) =(√(((a^2 +b^2 +c^2 )/6)+((2(√3))/3)(√(((a+b+c)(b+c−a)(c+a−b)(a+b−c))/(2×2×2×2))))) =(√((a^2 +b^2 +c^2 +(√(3(a+b+c)(b+c−a)(c+a−b)(a+b−c))))/6))
$$\mathrm{I}\:\mathrm{try}\:\mathrm{to}\:\mathrm{go}\:\mathrm{on}\:\mathrm{with}\:\mathrm{your}\:\mathrm{result}: \\ $$$$ \\ $$$$\mathrm{cos}\:{C}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$$$\mathrm{sin}\:\mathrm{C}=\frac{\mathrm{2}\sqrt{\mathrm{p}'\left(\mathrm{p}'−\mathrm{a}\right)\left(\mathrm{p}'−\mathrm{b}\right)\left(\mathrm{p}'−\mathrm{c}\right)}}{\mathrm{ab}} \\ $$$$\mathrm{with}\:\mathrm{p}'=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\left(\mathrm{C}−\frac{\mathrm{2}\pi}{\mathrm{3}}\right)=\mathrm{cos}\:\mathrm{C}\:\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{3}}\:+\:\mathrm{sin}\:\mathrm{C}\:\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{cos}\:\mathrm{C}+\sqrt{\mathrm{3}}\mathrm{sin}\:\mathrm{C}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[−\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}+\mathrm{2}\sqrt{\mathrm{3}}\frac{\sqrt{\mathrm{p}'\left(\mathrm{p}'−\mathrm{a}\right)\left(\mathrm{p}'−\mathrm{b}\right)\left(\mathrm{p}'−\mathrm{c}\right)}}{\mathrm{ab}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2ab}}\left[−\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{3}}\sqrt{\mathrm{p}'\left(\mathrm{p}'−\mathrm{a}\right)\left(\mathrm{p}'−\mathrm{b}\right)\left(\mathrm{p}'−\mathrm{c}\right)}\right] \\ $$$$ \\ $$$${s}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}\mathrm{cos}\:\left({C}−\frac{\mathrm{2}\pi}{\mathrm{3}}\right)}\: \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{3}}\sqrt{\mathrm{p}'\left(\mathrm{p}'−\mathrm{a}\right)\left(\mathrm{p}'−\mathrm{b}\right)\left(\mathrm{p}'−\mathrm{c}\right)}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\sqrt{\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{3}}\sqrt{\mathrm{p}'\left(\mathrm{p}'−\mathrm{a}\right)\left(\mathrm{p}'−\mathrm{b}\right)\left(\mathrm{p}'−\mathrm{c}\right)}} \\ $$$$=\sqrt{\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }{\mathrm{6}}+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\sqrt{\frac{\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left(\mathrm{b}+\mathrm{c}−\mathrm{a}\right)\left(\mathrm{c}+\mathrm{a}−\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}−\mathrm{c}\right)}{\mathrm{2}×\mathrm{2}×\mathrm{2}×\mathrm{2}}}} \\ $$$$=\sqrt{\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} +\sqrt{\mathrm{3}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left(\mathrm{b}+\mathrm{c}−\mathrm{a}\right)\left(\mathrm{c}+\mathrm{a}−\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}−\mathrm{c}\right)}}{\mathrm{6}}} \\ $$
Commented by mrW1 last updated on 16/Jun/17
that means your answer is correct.
$$\mathrm{that}\:\mathrm{means}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{correct}. \\ $$
Commented by mrW1 last updated on 16/Jun/17
please show how you proceeded to s=(1/( (√3)))(√(a^2 +b^2 +2abcos (C−((2π)/3))))
$$\mathrm{please}\:\mathrm{show}\:\mathrm{how}\:\mathrm{you}\:\mathrm{proceeded}\:\mathrm{to} \\ $$$${s}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}\mathrm{cos}\:\left({C}−\frac{\mathrm{2}\pi}{\mathrm{3}}\right)}\: \\ $$$$ \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/Jun/17
let:E′,P ′,D ′,be the 3rd vertex of triangles:^ CBE′,ACP′,ABD′.because of this triangles are equilateral,we can write for each of them: (ω^2 +ω+1=0,z: is a complex number) ((1/ω)=−(1+ω),(1/ω^2 )=−1−(1/ω)=−1+1+ω=ω) ΔBCE′: z_B + ωz_(E′) +ω^2 z_C =0 ΔACP′: z_A +ωz_C +ω^2 z_(P′^′ ) =0 ΔD′BA: z_(D′) +ωz_B +ω^2 z_A =0 z_D =(1/3)(z_A +z_B +z_(D′) ),cordinate of centroied. now for triangle:DEP we can write: z_D +ωz_E +ω^2 z_P =(1/3)(z_(D′) +z_B +z_A )+(ω/3)(z_B +z_(E′) +z_C )+ (ω^2 /3)(z_A +z_C +z_(P′) )=(1/3)(z_B +ωz_(E′) +ω^2 z_C )+ +(1/3)(z_(D′) +ωz_B +ω^2 z_A )+(1/3)(z_A +ωz_C +ω^2 z_P^( ′) )= =(1/3)(0+0+0)=0 . i.e:DE^Δ P is equilateral. 2)cordinate of centroid of :AB^Δ C =(1/3)(z_A +z_B +z_C ) for triangle:DE^Δ P=(1/3)(z_D +z_E +z_P )= =(1/3)((1/3)(z_(D′) +z_B +z_A )+(1/3)(z_B +z_(E′) +z_C )+(1/3)(z_A +z_C +z_(P′) ))= =(1/9)(2z_A +2z_B +2z_C +z_(E′) +z_(P′) +z_(D′) )= =(1/9)(2z_A +2z_B +2z_C −((z_B +ω^2 z_C )/ω)−((z_A +ωz_C )/ω^2 )−ωz_B −ω^2 z_A )= =(1/9)(2z_A +2z_B +2z_C +(1+ω)(z_B +ω^2 z_C )−ω(z_A +ωz_C )−ωz_B −ω^2 z_A )= =(1/9)(2z_A +2z_B +2z_C +z_B +ω^2 z_C +ωz_B +ω^3 z_C −ωz_A −ω^2 z_C −ωz_B −ω^2 z_A )= =(1/9)(3z_A +3z_B +3z_C )=(1/3)(z_A +z_B +z_C ) . it means that:triangles:AB^Δ C and DE^Δ P have the same centroied point.
$${let}:{E}',{P}\:',{D}\:',{be}\:{the}\:\mathrm{3}{rd}\:{vertex}\:{of}\:{triangles}\overset{} {:} \\ $$$${CBE}',{ACP}',{ABD}'.{because}\:{of}\:{this}\: \\ $$$${triangles}\:{are}\:{equilateral},{we}\:{can}\: \\ $$$${write}\:{for}\:{each}\:{of}\:{them}: \\ $$$$\left(\omega^{\mathrm{2}} +\omega+\mathrm{1}=\mathrm{0},{z}:\:{is}\:{a}\:{complex}\:{number}\right) \\ $$$$\left(\frac{\mathrm{1}}{\omega}=−\left(\mathrm{1}+\omega\right),\frac{\mathrm{1}}{\omega^{\mathrm{2}} }=−\mathrm{1}−\frac{\mathrm{1}}{\omega}=−\mathrm{1}+\mathrm{1}+\omega=\omega\right) \\ $$$$\Delta{BCE}':\:{z}_{{B}} +\:\omega{z}_{{E}'} +\omega^{\mathrm{2}} {z}_{{C}} =\mathrm{0} \\ $$$$\Delta{ACP}':\:{z}_{{A}} +\omega{z}_{{C}} +\omega^{\mathrm{2}} {z}_{{P}'^{'} } =\mathrm{0} \\ $$$$\Delta{D}'{BA}:\:{z}_{{D}'} +\omega{z}_{{B}} +\omega^{\mathrm{2}} {z}_{{A}} =\mathrm{0} \\ $$$${z}_{{D}} =\frac{\mathrm{1}}{\mathrm{3}}\left({z}_{{A}} +{z}_{{B}} +{z}_{{D}'} \right),{cordinate}\:{of}\:{centroied}. \\ $$$${now}\:{for}\:{triangle}:{DEP}\:{we}\:{can}\:{write}: \\ $$$${z}_{{D}} +\omega{z}_{{E}} +\omega^{\mathrm{2}} {z}_{{P}} =\frac{\mathrm{1}}{\mathrm{3}}\left({z}_{{D}'} +{z}_{{B}} +{z}_{{A}} \right)+\frac{\omega}{\mathrm{3}}\left({z}_{{B}} +{z}_{{E}'} +{z}_{{C}} \right)+ \\ $$$$\frac{\omega^{\mathrm{2}} }{\mathrm{3}}\left({z}_{{A}} +{z}_{{C}} +{z}_{{P}'} \right)=\frac{\mathrm{1}}{\mathrm{3}}\left({z}_{{B}} +\omega{z}_{{E}'} +\omega^{\mathrm{2}} {z}_{{C}} \right)+ \\ $$$$+\frac{\mathrm{1}}{\mathrm{3}}\left({z}_{{D}'} +\omega{z}_{{B}} +\omega^{\mathrm{2}} {z}_{{A}} \right)+\frac{\mathrm{1}}{\mathrm{3}}\left({z}_{{A}} +\omega{z}_{{C}} +\omega^{\mathrm{2}} {z}_{{P}^{\:\:'} } \right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{0}+\mathrm{0}+\mathrm{0}\right)=\mathrm{0}\:. \\ $$$${i}.{e}:{D}\overset{\Delta} {{E}P}\:\:{is}\:{equilateral}. \\ $$$$\left.\mathrm{2}\right){cordinate}\:{of}\:{centroid}\:{of}\::{A}\overset{\Delta} {{B}C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left({z}_{{A}} +{z}_{{B}} +{z}_{{C}} \right) \\ $$$${for}\:{triangle}:{D}\overset{\Delta} {{E}P}=\frac{\mathrm{1}}{\mathrm{3}}\left({z}_{{D}} +{z}_{{E}} +{z}_{{P}} \right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{3}}\left({z}_{{D}'} +{z}_{{B}} +{z}_{{A}} \right)+\frac{\mathrm{1}}{\mathrm{3}}\left({z}_{{B}} +{z}_{{E}'} +{z}_{{C}} \right)+\frac{\mathrm{1}}{\mathrm{3}}\left({z}_{{A}} +{z}_{{C}} +{z}_{{P}'} \right)\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{2}{z}_{{A}} +\mathrm{2}{z}_{{B}} +\mathrm{2}{z}_{{C}} +{z}_{{E}'} +{z}_{{P}'} +{z}_{{D}'} \right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{2}{z}_{{A}} +\mathrm{2}{z}_{{B}} +\mathrm{2}{z}_{{C}} −\frac{{z}_{{B}} +\omega^{\mathrm{2}} {z}_{{C}} }{\omega}−\frac{{z}_{{A}} +\omega{z}_{{C}} }{\omega^{\mathrm{2}} }−\omega{z}_{{B}} −\omega^{\mathrm{2}} {z}_{{A}} \right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{2}{z}_{{A}} +\mathrm{2}{z}_{{B}} +\mathrm{2}{z}_{{C}} +\left(\mathrm{1}+\omega\right)\left({z}_{{B}} +\omega^{\mathrm{2}} {z}_{{C}} \right)−\omega\left({z}_{{A}} +\omega{z}_{{C}} \right)−\omega{z}_{{B}} −\omega^{\mathrm{2}} {z}_{{A}} \right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{2}{z}_{{A}} +\mathrm{2}{z}_{{B}} +\mathrm{2}{z}_{{C}} +{z}_{{B}} +\omega^{\mathrm{2}} {z}_{{C}} +\omega{z}_{{B}} +\omega^{\mathrm{3}} {z}_{{C}} −\omega{z}_{{A}} −\omega^{\mathrm{2}} {z}_{{C}} −\omega{z}_{{B}} −\omega^{\mathrm{2}} {z}_{{A}} \right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\left(\mathrm{3}{z}_{{A}} +\mathrm{3}{z}_{{B}} +\mathrm{3}{z}_{{C}} \right)=\frac{\mathrm{1}}{\mathrm{3}}\left({z}_{{A}} +{z}_{{B}} +{z}_{{C}} \right)\:. \\ $$$${it}\:{means}\:{that}:{triangles}:{A}\overset{\Delta} {{B}C}\:{and}\:{D}\overset{\Delta} {{E}P} \\ $$$${have}\:{the}\:{same}\:{centroied}\:{point}. \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/Jun/17
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/Jun/17
draw outcircles of constracted triangles. let :O be the another intersection of 2 circles with AB&AC chords ,respect point:A. we have :∠AOB=∠AOC=120^• . so: ∠BOC=120^• also. i.e the circle with chord:BC ,passes from point O. but XY is perpendicular to common chord:OC,and also:XZ is perpendicular to :OB. we prove that:∠BOC=120^• and it means that: ∠X=60^• . similarly we can prove that:∠Y=∠Z=60 i.e: XY^Δ Z is equilateral. ■
$${draw}\:{outcircles}\:{of}\:{constracted}\:{triangles}. \\ $$$${let}\::{O}\:{be}\:{the}\:{another}\:{intersection}\:{of}\:\mathrm{2}\:{circles}\: \\ $$$${with}\:{AB\&AC}\:{chords}\:,{respect}\:{point}:{A}. \\ $$$${we}\:{have}\::\angle{AOB}=\angle{AOC}=\mathrm{120}^{\bullet} . \\ $$$${so}:\:\angle{BOC}=\mathrm{120}^{\bullet} {also}.\:{i}.{e}\:{the}\:{circle}\:{with} \\ $$$${chord}:{BC}\:,{passes}\:{from}\:{point}\:{O}. \\ $$$${but}\:{XY}\:{is}\:{perpendicular}\:{to}\:{common}\: \\ $$$${chord}:{OC},{and}\:{also}:{XZ}\:{is}\:{perpendicular} \\ $$$${to}\::{OB}.\:{we}\:{prove}\:{that}:\angle{BOC}=\mathrm{120}^{\bullet} \\ $$$${and}\:{it}\:{means}\:{that}:\:\angle{X}=\mathrm{60}^{\bullet} . \\ $$$${similarly}\:{we}\:{can}\:{prove}\:{that}:\angle{Y}=\angle{Z}=\mathrm{60} \\ $$$${i}.{e}:\:{X}\overset{\Delta} {{Y}Z}\:{is}\:{equilateral}.\:\blacksquare \\ $$
Commented by mrW1 last updated on 19/Jun/17
Very nice sir!
$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{sir}! \\ $$

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