Question Number 16068 by Tinkutara last updated on 21/Jun/17
$$\mathrm{Let}\:{ABCD}\:\mathrm{be}\:\mathrm{a}\:\mathrm{convex}\:\mathrm{quadrilateral} \\ $$$$\mathrm{and}\:\mathrm{let}\:\mathrm{E}\:\mathrm{and}\:\mathrm{F}\:\mathrm{be}\:\mathrm{the}\:\mathrm{points}\:\mathrm{of} \\ $$$$\mathrm{intersections}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lines}\:{AB},\:{CD}\:\mathrm{and} \\ $$$${AD},\:{BC},\:\mathrm{respectively}.\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{midpoints}\:\mathrm{of}\:\mathrm{the}\:\mathrm{segments}\:{AC},\:{BD}, \\ $$$$\mathrm{and}\:{EF}\:\mathrm{are}\:\mathrm{collinear}. \\ $$
Answered by ajfour last updated on 30/Jun/17
![z_I =(z_1 /2) ; z_G =((mz_2 +z_1 +k(z_2 −z_1 ))/2) ; z_G −z_I =(−(k/2))z_1 +(((m+k)/2))z_2 z_D =z_1 +λ(z_1 −mz_2 ) also z_D =μ[z_1 +k(z_2 −z_1 )] z_D −z_D =0 ⇒ (1+λ−μ+μk)z_1 −(λm+μk)z_2 =0 ⇒ 1+λ−μ(1−k)=0 and μ=−((λm)/k) so 1+λ+((λm(1−k))/k)=0 k+λ(k+m−mk)=0 or λ=(k/(mk−(m+k))) z_D = z_1 +(k/([mk−(m+k)]))(z_1 −mz_2 ) =((mkz_1 −mz_1 −kz_1 +kz_1 −mkz_2 )/(mk−(m+k))) =((m(k−1)z_1 −mkz_2 )/(mk−(m+k))) z_H =((z_2 +z_D )/2) ⇒ z_H −z_I =((z_2 +z_D −z_1 )/2) z_H −z_I =((z_2 −z_1 )/2)−((m(k−1)z_1 −mkz_2 )/(2(mk−m−k))) =(((mk−m−k)(z_1 −z_2 )−mk(z_1 −z_2 )+mz_1 )/(2(mk−m−k))) =((−kz_1 +(m+k)z_2 )/(2c)) ∀ c=mk−m−k z_H −z_I =(1/c)[−(k/2)z_1 +(((m+k)/2))z_2 ] =((z_G −z_I )/c) ⇒ z_H ,z_G , and z_I are collinear.](https://www.tinkutara.com/question/Q17058.png)
$$\mathrm{z}_{\boldsymbol{\mathrm{I}}} =\frac{\mathrm{z}_{\mathrm{1}} }{\mathrm{2}}\:;\:\:\:\mathrm{z}_{\mathrm{G}} =\frac{\mathrm{mz}_{\mathrm{2}} +\mathrm{z}_{\mathrm{1}} +\mathrm{k}\left(\mathrm{z}_{\mathrm{2}} −\mathrm{z}_{\mathrm{1}} \right)}{\mathrm{2}}\:; \\ $$$$\mathrm{z}_{\mathrm{G}} −\mathrm{z}_{\mathrm{I}} =\left(−\frac{\mathrm{k}}{\mathrm{2}}\right)\mathrm{z}_{\mathrm{1}} +\left(\frac{\mathrm{m}+\mathrm{k}}{\mathrm{2}}\right)\mathrm{z}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{z}_{\mathrm{D}} =\mathrm{z}_{\mathrm{1}} +\lambda\left(\mathrm{z}_{\mathrm{1}} −\mathrm{mz}_{\mathrm{2}} \right) \\ $$$$\:\mathrm{also}\:\:\:\:\mathrm{z}_{\mathrm{D}} =\mu\left[\mathrm{z}_{\mathrm{1}} +\mathrm{k}\left(\mathrm{z}_{\mathrm{2}} −\mathrm{z}_{\mathrm{1}} \right)\right] \\ $$$$\:\mathrm{z}_{\mathrm{D}} −\mathrm{z}_{\mathrm{D}} =\mathrm{0}\:\Rightarrow\: \\ $$$$\left(\mathrm{1}+\lambda−\mu+\mu\mathrm{k}\right)\mathrm{z}_{\mathrm{1}} −\left(\lambda\mathrm{m}+\mu\mathrm{k}\right)\mathrm{z}_{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{1}+\lambda−\mu\left(\mathrm{1}−\mathrm{k}\right)=\mathrm{0}\:\mathrm{and}\:\mu=−\frac{\lambda\mathrm{m}}{\mathrm{k}} \\ $$$$\mathrm{so}\:\mathrm{1}+\lambda+\frac{\lambda\mathrm{m}\left(\mathrm{1}−\mathrm{k}\right)}{\mathrm{k}}=\mathrm{0} \\ $$$$\mathrm{k}+\lambda\left(\mathrm{k}+\mathrm{m}−\mathrm{mk}\right)=\mathrm{0} \\ $$$$\:\mathrm{or}\:\:\lambda=\frac{\mathrm{k}}{\mathrm{mk}−\left(\mathrm{m}+\mathrm{k}\right)} \\ $$$$\mathrm{z}_{\mathrm{D}} =\:\mathrm{z}_{\mathrm{1}} +\frac{\mathrm{k}}{\left[\mathrm{mk}−\left(\mathrm{m}+\mathrm{k}\right)\right]}\left(\mathrm{z}_{\mathrm{1}} −\mathrm{mz}_{\mathrm{2}} \right) \\ $$$$\:\:\:\:=\frac{\mathrm{mkz}_{\mathrm{1}} −\mathrm{mz}_{\mathrm{1}} −\mathrm{kz}_{\mathrm{1}} +\mathrm{kz}_{\mathrm{1}} −\mathrm{mkz}_{\mathrm{2}} }{\mathrm{mk}−\left(\mathrm{m}+\mathrm{k}\right)} \\ $$$$\:\:\:=\frac{\mathrm{m}\left(\mathrm{k}−\mathrm{1}\right)\mathrm{z}_{\mathrm{1}} −\mathrm{mkz}_{\mathrm{2}} }{\mathrm{mk}−\left(\mathrm{m}+\mathrm{k}\right)} \\ $$$$\mathrm{z}_{\mathrm{H}} =\frac{\mathrm{z}_{\mathrm{2}} +\mathrm{z}_{\mathrm{D}} }{\mathrm{2}}\:\Rightarrow\:\:\mathrm{z}_{\mathrm{H}} −\mathrm{z}_{\boldsymbol{\mathrm{I}}} =\frac{\mathrm{z}_{\mathrm{2}} +\mathrm{z}_{\mathrm{D}} −\mathrm{z}_{\mathrm{1}} }{\mathrm{2}} \\ $$$$\mathrm{z}_{\mathrm{H}} −\mathrm{z}_{\boldsymbol{\mathrm{I}}} =\frac{\mathrm{z}_{\mathrm{2}} −\mathrm{z}_{\mathrm{1}} }{\mathrm{2}}−\frac{\mathrm{m}\left(\mathrm{k}−\mathrm{1}\right)\mathrm{z}_{\mathrm{1}} −\mathrm{mkz}_{\mathrm{2}} }{\mathrm{2}\left(\mathrm{mk}−\mathrm{m}−\mathrm{k}\right)} \\ $$$$\:=\frac{\left(\mathrm{mk}−\mathrm{m}−\mathrm{k}\right)\left(\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} \right)−\mathrm{mk}\left(\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} \right)+\mathrm{mz}_{\mathrm{1}} }{\mathrm{2}\left(\mathrm{mk}−\mathrm{m}−\mathrm{k}\right)} \\ $$$$=\frac{−\mathrm{kz}_{\mathrm{1}} +\left(\mathrm{m}+\mathrm{k}\right)\mathrm{z}_{\mathrm{2}} }{\mathrm{2c}}\:\:\:\forall\:\mathrm{c}=\mathrm{mk}−\mathrm{m}−\mathrm{k} \\ $$$$\mathrm{z}_{\mathrm{H}} −\mathrm{z}_{\boldsymbol{\mathrm{I}}} =\frac{\mathrm{1}}{\mathrm{c}}\left[−\frac{\mathrm{k}}{\mathrm{2}}\mathrm{z}_{\mathrm{1}} +\left(\frac{\mathrm{m}+\mathrm{k}}{\mathrm{2}}\right)\mathrm{z}_{\mathrm{2}} \right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{z}_{\mathrm{G}} −\mathrm{z}_{\boldsymbol{\mathrm{I}}} }{\mathrm{c}} \\ $$$$\Rightarrow\:\:\mathrm{z}_{\mathrm{H}} ,\mathrm{z}_{\mathrm{G}} ,\:\mathrm{and}\:\mathrm{z}_{\boldsymbol{\mathrm{I}}} \:\mathrm{are}\:\mathrm{collinear}. \\ $$
Commented by ajfour last updated on 30/Jun/17
$$\mathrm{let}\:\mathrm{me}\:\mathrm{see}\:\mathrm{your}\:\mathrm{book}'\mathrm{s}\:\mathrm{solution}.. \\ $$
Commented by ajfour last updated on 30/Jun/17
Answered by Tinkutara last updated on 30/Jun/17
![Let P, Q, and R be the midpoints of AC, BD, and EF. (Figure). Denote by S the area of ABCD. As we have seen the locus of the points M in the interior of ABCD for which [MAB] + [MCD] = (1/2) S is a segment. We see that P and Q belong to this segment. Indeed, [PAB] + [PCD] = (1/2)[ABC] + (1/2)[ACD] = (1/2) S. and [QAD] + [QCD] = (1/2)[ABD] + (1/2)[BCD] = (1/2) S. Now we have [RAB] = (1/2)[FAB], since the distance from F to AB is twice the distance from R to AB. Similarly, [RCD] = (1/2)[FCD]. We obtain [RAB] − [RCD] = (1/2)[FAB] − [FCD] = (1/2) S. Taking into account the observation in the solution to Problem 16064, it follows that P, Q and R are collinear.](https://www.tinkutara.com/question/Q17063.png)
$$\mathrm{Let}\:{P},\:{Q},\:\mathrm{and}\:{R}\:\mathrm{be}\:\mathrm{the}\:\mathrm{midpoints}\:\mathrm{of} \\ $$$${AC},\:{BD},\:\mathrm{and}\:{EF}.\:\left(\mathrm{Figure}\right).\:\mathrm{Denote}\:\mathrm{by} \\ $$$${S}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:{ABCD}.\:\mathrm{As}\:\mathrm{we}\:\mathrm{have}\:\mathrm{seen} \\ $$$$\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{the}\:\mathrm{points}\:{M}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interior} \\ $$$$\mathrm{of}\:{ABCD}\:\mathrm{for}\:\mathrm{which} \\ $$$$\left[{MAB}\right]\:+\:\left[{MCD}\right]\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:{S} \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{segment}.\:\mathrm{We}\:\mathrm{see}\:\mathrm{that}\:\mathrm{P}\:\mathrm{and}\:\mathrm{Q} \\ $$$$\mathrm{belong}\:\mathrm{to}\:\mathrm{this}\:\mathrm{segment}.\:\mathrm{Indeed}, \\ $$$$\left[{PAB}\right]\:+\:\left[{PCD}\right]\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[{ABC}\right]\:+\:\frac{\mathrm{1}}{\mathrm{2}}\left[{ACD}\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:{S}. \\ $$$$\mathrm{and} \\ $$$$\left[{QAD}\right]\:+\:\left[{QCD}\right]\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[{ABD}\right]\:+\:\frac{\mathrm{1}}{\mathrm{2}}\left[{BCD}\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:{S}. \\ $$$$\mathrm{Now}\:\mathrm{we}\:\mathrm{have}\:\left[{RAB}\right]\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[{FAB}\right], \\ $$$$\mathrm{since}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{from}\:{F}\:\mathrm{to}\:{AB}\:\mathrm{is} \\ $$$$\mathrm{twice}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{from}\:{R}\:\mathrm{to}\:{AB}. \\ $$$$\mathrm{Similarly},\:\left[{RCD}\right]\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[{FCD}\right]. \\ $$$$\mathrm{We}\:\mathrm{obtain} \\ $$$$\left[{RAB}\right]\:−\:\left[{RCD}\right]\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[{FAB}\right]\:−\:\left[{FCD}\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:{S}. \\ $$$$\mathrm{Taking}\:\mathrm{into}\:\mathrm{account}\:\mathrm{the}\:\mathrm{observation}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{Problem}\:\mathrm{16064},\:\mathrm{it} \\ $$$$\mathrm{follows}\:\mathrm{that}\:{P},\:{Q}\:\mathrm{and}\:{R}\:\mathrm{are}\:\mathrm{collinear}. \\ $$
Commented by Tinkutara last updated on 30/Jun/17
