Question Number 206064 by mr W last updated on 06/Apr/24
Commented by mr W last updated on 06/Apr/24
$${an}\:{unsolved}\:{old}\:{question}\:\left({Q}\mathrm{172465}\right) \\ $$
Answered by A5T last updated on 06/Apr/24
![WLOG, let N be the origin and BNE coincide with the real axis(or x axis). B(−b,0),E(b,0),N(0,0), C(x,y), D(u,v) ⇒BC=(√((x+b)^2 +y^2 ))=AB=(√(x^2 +y^2 +b^2 +2xb)) [ABC]=((√3)/4)(x^2 +y^2 +b^2 +2xb) A(((x−b−y(√3))/2),((x(√3)+b(√3)+y)/2)); F(((u+b−v(√3))/2),((u(√3)−b(√3)+v)/2)) DE=(√((u−b)^2 +v^2 ))=(√(u^2 +v^2 +b^2 −2ub)) ⇒[FED]=((√3)/4)(u^2 +v^2 +b^2 −2ub) P(((2x+b−v(√3))/4),((2y+u(√3)−b(√3)+v)/4)) M(((2u+x−b−y(√3))/4),((2v+x(√3)+b(√3)+y)/4)) ⇒[PNM]=(1/2)∣(((2x+b−v(√3))(2v+x(√3)+b(√3)+y))/(16)) −(((2y+u(√3)−b(√3)+v)(2u+x−b−y(√3)))/(16))∣ =(((√3)∣x^2 +y^2 −u^2 −v^2 +2bu+2bx∣)/(16)) ∣[ABC]−[DEF]∣=(((√3)∣x^2 +y^2 +2xb+2ub−u^2 −v^2 ∣)/4) ⇒(([PNM[)/(∣[ABC−[DEF]∣))=(((√3)/(16))/((√3)/4))=(1/4)](https://www.tinkutara.com/question/Q206076.png)
$${WLOG},\:{let}\:{N}\:{be}\:{the}\:{origin}\:{and}\:{BNE}\:{coincide} \\ $$$${with}\:{the}\:{real}\:{axis}\left({or}\:{x}\:{axis}\right). \\ $$$${B}\left(−{b},\mathrm{0}\right),{E}\left({b},\mathrm{0}\right),{N}\left(\mathrm{0},\mathrm{0}\right),\:{C}\left({x},{y}\right),\:{D}\left({u},{v}\right) \\ $$$$\Rightarrow{BC}=\sqrt{\left({x}+{b}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} }={AB}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{xb}} \\ $$$$\left[{ABC}\right]=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{xb}\right) \\ $$$${A}\left(\frac{{x}−{b}−{y}\sqrt{\mathrm{3}}}{\mathrm{2}},\frac{{x}\sqrt{\mathrm{3}}+{b}\sqrt{\mathrm{3}}+{y}}{\mathrm{2}}\right); \\ $$$${F}\left(\frac{{u}+{b}−{v}\sqrt{\mathrm{3}}}{\mathrm{2}},\frac{{u}\sqrt{\mathrm{3}}−{b}\sqrt{\mathrm{3}}+{v}}{\mathrm{2}}\right) \\ $$$${DE}=\sqrt{\left({u}−{b}\right)^{\mathrm{2}} +{v}^{\mathrm{2}} }=\sqrt{{u}^{\mathrm{2}} +{v}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ub}} \\ $$$$\Rightarrow\left[{FED}\right]=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left({u}^{\mathrm{2}} +{v}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ub}\right) \\ $$$${P}\left(\frac{\mathrm{2}{x}+{b}−{v}\sqrt{\mathrm{3}}}{\mathrm{4}},\frac{\mathrm{2}{y}+{u}\sqrt{\mathrm{3}}−{b}\sqrt{\mathrm{3}}+{v}}{\mathrm{4}}\right) \\ $$$${M}\left(\frac{\mathrm{2}{u}+{x}−{b}−{y}\sqrt{\mathrm{3}}}{\mathrm{4}},\frac{\mathrm{2}{v}+{x}\sqrt{\mathrm{3}}+{b}\sqrt{\mathrm{3}}+{y}}{\mathrm{4}}\right) \\ $$$$\Rightarrow\left[{PNM}\right]=\frac{\mathrm{1}}{\mathrm{2}}\mid\frac{\left(\mathrm{2}{x}+{b}−{v}\sqrt{\mathrm{3}}\right)\left(\mathrm{2}{v}+{x}\sqrt{\mathrm{3}}+{b}\sqrt{\mathrm{3}}+{y}\right)}{\mathrm{16}} \\ $$$$−\frac{\left(\mathrm{2}{y}+{u}\sqrt{\mathrm{3}}−{b}\sqrt{\mathrm{3}}+{v}\right)\left(\mathrm{2}{u}+{x}−{b}−{y}\sqrt{\mathrm{3}}\right)}{\mathrm{16}}\mid \\ $$$$=\frac{\sqrt{\mathrm{3}}\mid{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{u}^{\mathrm{2}} −{v}^{\mathrm{2}} +\mathrm{2}{bu}+\mathrm{2}{bx}\mid}{\mathrm{16}} \\ $$$$\mid\left[{ABC}\right]−\left[{DEF}\right]\mid=\frac{\sqrt{\mathrm{3}}\mid{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xb}+\mathrm{2}{ub}−{u}^{\mathrm{2}} −{v}^{\mathrm{2}} \mid}{\mathrm{4}} \\ $$$$\Rightarrow\frac{\left[{PNM}\left[\right.\right.}{\mid\left[{ABC}−\left[{DEF}\right]\mid\right.}=\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{16}}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by mr W last updated on 07/Apr/24
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Answered by mr W last updated on 07/Apr/24
Commented by mr W last updated on 07/Apr/24
![say BN=NE=c x_A =−c+a cos α y_A =a sin α x_D =c−b cos β y_D =b sin β x_M =((x_A +x_D )/2)=((a cos α−b cos β)/2) y_M =((y_A +y_D )/2)=((a sin α+b sin β)/2) similarly x_P =((a cos ((π/3)−α)−b cos ((π/3)−β))/2) y_P =−((a sin ((π/3)−α)+b sin ((π/3)−β))/2) [MNP]=∣(((x_M +x_P )(y_M −y_P )−x_M y_M +x_P y_P )/2)∣ [MNP]=∣((x_P y_M −x_M y_P )/2)∣ [MNP]=∣(((a cos ((π/3)−α)−b cos ((π/3)−β))(a sin α+b sin β)+(a cos α−b cos β)(a sin ((π/3)−α)+b sin ((π/3)−β)))/8)∣ [MNP]=∣((a^2 sin α cos ((π/3)−α)−ab sin α cos ((π/3)−β)+ab sin β cos ((π/3)−α)−b^2 sin β cos ((π/3)−β)+a^2 cos α sin ((π/3)−α)+ab cos α sin ((π/3)−β)−ab cos β sin ((π/3)−α)−b^2 cos β sin ((π/3)−β))/8)∣ [MNP]=∣((a^2 sin (π/3)+ab sin ((π/3)−α−β)+ab sin (α+β−(π/3))−b^2 sin (π/3))/8)∣ [MNP]=∣(((a^2 −b^2 ) sin (π/3))/8)∣ [MNP]=∣(((√3)(a^2 −b^2 ))/(16))∣ since X=(((√3)a^2 )/4), Y=(((√3)b^2 )/4) ⇒ [MNP]=((∣X−Y∣)/4) ✓](https://www.tinkutara.com/question/Q206113.png)
$${say}\:{BN}={NE}={c} \\ $$$${x}_{{A}} =−{c}+{a}\:\mathrm{cos}\:\alpha \\ $$$${y}_{{A}} ={a}\:\mathrm{sin}\:\alpha \\ $$$${x}_{{D}} ={c}−{b}\:\mathrm{cos}\:\beta \\ $$$${y}_{{D}} ={b}\:\mathrm{sin}\:\beta \\ $$$${x}_{{M}} =\frac{{x}_{{A}} +{x}_{{D}} }{\mathrm{2}}=\frac{{a}\:\mathrm{cos}\:\alpha−{b}\:\mathrm{cos}\:\beta}{\mathrm{2}} \\ $$$${y}_{{M}} =\frac{{y}_{{A}} +{y}_{{D}} }{\mathrm{2}}=\frac{{a}\:\mathrm{sin}\:\alpha+{b}\:\mathrm{sin}\:\beta}{\mathrm{2}} \\ $$$${similarly} \\ $$$${x}_{{P}} =\frac{{a}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}−\alpha\right)−{b}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}−\beta\right)}{\mathrm{2}} \\ $$$${y}_{{P}} =−\frac{{a}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\alpha\right)+{b}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\beta\right)}{\mathrm{2}} \\ $$$$\left[{MNP}\right]=\mid\frac{\left({x}_{{M}} +{x}_{{P}} \right)\left({y}_{{M}} −{y}_{{P}} \right)−{x}_{{M}} {y}_{{M}} +{x}_{{P}} {y}_{{P}} }{\mathrm{2}}\mid \\ $$$$\left[{MNP}\right]=\mid\frac{{x}_{{P}} {y}_{{M}} −{x}_{{M}} {y}_{{P}} }{\mathrm{2}}\mid \\ $$$$\left[{MNP}\right]=\mid\frac{\left({a}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}−\alpha\right)−{b}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}−\beta\right)\right)\left({a}\:\mathrm{sin}\:\alpha+{b}\:\mathrm{sin}\:\beta\right)+\left({a}\:\mathrm{cos}\:\alpha−{b}\:\mathrm{cos}\:\beta\right)\left({a}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\alpha\right)+{b}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\beta\right)\right)}{\mathrm{8}}\mid \\ $$$$\left[{MNP}\right]=\mid\frac{{a}^{\mathrm{2}} \:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}−\alpha\right)−{ab}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}−\beta\right)+{ab}\:\mathrm{sin}\:\beta\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}−\alpha\right)−{b}^{\mathrm{2}} \:\mathrm{sin}\:\beta\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}−\beta\right)+{a}^{\mathrm{2}} \:\mathrm{cos}\:\alpha\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\alpha\right)+{ab}\:\mathrm{cos}\:\alpha\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\beta\right)−{ab}\:\mathrm{cos}\:\beta\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\alpha\right)−{b}^{\mathrm{2}} \:\mathrm{cos}\:\beta\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\beta\right)}{\mathrm{8}}\mid \\ $$$$\left[{MNP}\right]=\mid\frac{{a}^{\mathrm{2}} \:\mathrm{sin}\:\frac{\pi}{\mathrm{3}}+{ab}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}−\alpha−\beta\right)+{ab}\:\mathrm{sin}\:\left(\alpha+\beta−\frac{\pi}{\mathrm{3}}\right)−{b}^{\mathrm{2}} \:\mathrm{sin}\:\frac{\pi}{\mathrm{3}}}{\mathrm{8}}\mid \\ $$$$\left[{MNP}\right]=\mid\frac{\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\:\mathrm{sin}\:\frac{\pi}{\mathrm{3}}}{\mathrm{8}}\mid \\ $$$$\left[{MNP}\right]=\mid\frac{\sqrt{\mathrm{3}}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)}{\mathrm{16}}\mid \\ $$$${since}\:{X}=\frac{\sqrt{\mathrm{3}}{a}^{\mathrm{2}} }{\mathrm{4}},\:{Y}=\frac{\sqrt{\mathrm{3}}{b}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow\:\left[{MNP}\right]=\frac{\mid{X}−{Y}\mid}{\mathrm{4}}\:\checkmark \\ $$
Answered by aleks041103 last updated on 08/Apr/24
$${Here}\:{is}\:{a}\:{vectorial}\:{way}\:{to}\:{appoach}\:{this}. \\ $$$$ \\ $$$$\boldsymbol{{Firstly}},\:{we}\:{should}\:{note}\:{that}\:{the}\:{vectors}\:{M},\:{N} \\ $$$${and}\:{P}\:\:{are}\:{linear}\:{combinations}\:{of}\:{the}\:{vectors} \\ $$$${of}\:{the}\:{initial}\:\mathrm{6}\:{vertices}\:−\:{A},\:{B},\:{C},{D},{E},{F}.\:{More}\:{precisely}\:{M},{N},{P} \\ $$$${are}\:{the}\:{average}\:{of}\:{some}\:\mathrm{2}\:{vertices}. \\ $$$${For}\:{example} \\ $$$${N}=\frac{\mathrm{1}}{\mathrm{2}}\left({B}+{E}\right) \\ $$$$ \\ $$$$\boldsymbol{{Secondly}},\:{by}\:{the}\:{first}\:{observation},\:{if}\:{we}\: \\ $$$${translate}\:\bigtriangleup{DFN}\:{by}\:{a}\:{vector}\:{v} \\ $$$${then}\:{the}\:{vertices}\:{of}\:\bigtriangleup{PMN}\:{translate}\:{by}\:\frac{\mathrm{1}}{\mathrm{2}}{v}\:{and} \\ $$$${therefore}\:{only}\:{the}\:{triangle}'{s}\:{position}\:{will}\:{change}. \\ $$$${In}\:{particular},\:{its}\:{area}\:{remains}\:{the}\:{same}. \\ $$$$ \\ $$$$\boldsymbol{{Thirdly}},\:{translate}\:{the}\:{pink}\:\bigtriangleup,\:{s}.{t}\:{F}\equiv{C}\equiv{P}\equiv{O}, \\ $$$${where}\:{O}\:{is}\:{the}\:{beginning}\:{of}\:{the}\:{coordinate} \\ $$$${system}. \\ $$$${Then} \\ $$$${M}=\frac{\mathrm{1}}{\mathrm{2}}\left({A}+{D}\right) \\ $$$${N}=\frac{\mathrm{1}}{\mathrm{2}}\left({B}+{E}\right) \\ $$$${If}\:×\:{is}\:{the}\:{vector}\:{cross}\:{product}: \\ $$$$\mathrm{2}{se}_{{z}} ={M}×{N}=\frac{\mathrm{1}}{\mathrm{4}}\left({A}+{D}\right)×\left({B}+{E}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left({A}×{B}+{A}×{E}+{D}×{B}+{D}×{E}\right) \\ $$$${A}×{B}=\mathrm{2}{xe}_{{z}} \\ $$$${E}×{D}=\mathrm{2}{ye}_{{z}} \\ $$$$\Rightarrow{se}_{{z}} =\frac{\mathrm{1}}{\mathrm{4}}\left({x}−{y}\right){e}_{{z}} + \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{{e}_{{z}} }{\mathrm{8}}\left(\mid{D}\mid\mid{B}\mid{sin}\left(\angle{DOB}\right)−\mid{A}\mid\mid{E}\mid{sin}\left(\angle{AOE}\right)\right) \\ $$$${Since}\:{the}\:{triangles}\:{are}\:{equilateral},\:\mid{A}\mid=\mid{B}\mid={a}, \\ $$$$\mid{D}\mid=\mid{E}\mid={b}. \\ $$$${Also} \\ $$$$\angle{DOB}=\angle{AOB}+\angle{DOA}= \\ $$$$=\mathrm{60}+\angle{DOA}= \\ $$$$=\angle{DOE}+\angle{DOA}=\angle{AOE} \\ $$$$ \\ $$$${Therefore},\:{finally}: \\ $$$$\Rightarrow{s}=\frac{\mathrm{1}}{\mathrm{4}}\mid{x}−{y}\mid \\ $$
Commented by mr W last updated on 08/Apr/24
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