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Question-16789

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Question Number 16789 by ajfour last updated on 26/Jun/17
Commented by ajfour last updated on 26/Jun/17
solution to Q.16065 find locus of M such that Area(△MAB)=2Area(△MCD).
$$\:\mathrm{solution}\:\mathrm{to}\:\mathrm{Q}.\mathrm{16065} \\ $$$$\mathrm{find}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{M}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\mathrm{Area}\left(\bigtriangleup\mathrm{MAB}\right)=\mathrm{2Area}\left(\bigtriangleup\mathrm{MCD}\right). \\ $$
Answered by ajfour last updated on 26/Jun/17
Area(△MAB)=(1/2)(ay) Area(△MCD)=(1/2) determinant ((x,y,1),(b,c,1),(d,e,1)) 2×Area(△MCD)=Area(△MAB) determinant ((x,y,1),(b,c,1),(d,e,1))=((ay)/2) so a straight line.
$$\:\mathrm{Area}\left(\bigtriangleup\mathrm{MAB}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{ay}\right) \\ $$$$\:\mathrm{Area}\left(\bigtriangleup\mathrm{MCD}\right)=\frac{\mathrm{1}}{\mathrm{2}}\begin{vmatrix}{\mathrm{x}}&{\mathrm{y}}&{\mathrm{1}}\\{\mathrm{b}}&{\mathrm{c}}&{\mathrm{1}}\\{\mathrm{d}}&{\mathrm{e}}&{\mathrm{1}}\end{vmatrix} \\ $$$$\mathrm{2}×\mathrm{Area}\left(\bigtriangleup\mathrm{MCD}\right)=\mathrm{Area}\left(\bigtriangleup\mathrm{MAB}\right) \\ $$$$\:\begin{vmatrix}{\mathrm{x}}&{\mathrm{y}}&{\mathrm{1}}\\{\mathrm{b}}&{\mathrm{c}}&{\mathrm{1}}\\{\mathrm{d}}&{\mathrm{e}}&{\mathrm{1}}\end{vmatrix}=\frac{\mathrm{ay}}{\mathrm{2}} \\ $$$$\mathrm{so}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}. \\ $$
Commented by Tinkutara last updated on 26/Jun/17
Yes Sir the answer is a straight line but here it is required to draw that line/or identify it. I am giving the figure in which it is located. There is some theory part of the solution but it is not clear to me. Please explain.
$$\mathrm{Yes}\:\mathrm{Sir}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{but} \\ $$$$\mathrm{here}\:\mathrm{it}\:\mathrm{is}\:\mathrm{required}\:\mathrm{to}\:\mathrm{draw}\:\mathrm{that}\:\mathrm{line}/\mathrm{or} \\ $$$$\mathrm{identify}\:\mathrm{it}.\:\mathrm{I}\:\mathrm{am}\:\mathrm{giving}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{in} \\ $$$$\mathrm{which}\:\mathrm{it}\:\mathrm{is}\:\mathrm{located}.\:\mathrm{There}\:\mathrm{is}\:\mathrm{some} \\ $$$$\mathrm{theory}\:\mathrm{part}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{but}\:\mathrm{it}\:\mathrm{is} \\ $$$$\mathrm{not}\:\mathrm{clear}\:\mathrm{to}\:\mathrm{me}.\:\mathrm{Please}\:\mathrm{explain}. \\ $$
Commented by Tinkutara last updated on 26/Jun/17
Commented by Tinkutara last updated on 26/Jun/17
Here it is given that JK is the required locus of points M. Solution is given below.
$$\mathrm{Here}\:\mathrm{it}\:\mathrm{is}\:\mathrm{given}\:\mathrm{that}\:{JK}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{required}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{points}\:{M}.\:\mathrm{Solution}\:\mathrm{is} \\ $$$$\mathrm{given}\:\mathrm{below}. \\ $$
Commented by Tinkutara last updated on 26/Jun/17
Commented by ajfour last updated on 26/Jun/17
pretty lengthy, cannot comprehend right now.
$$\mathrm{pretty}\:\mathrm{lengthy},\:\mathrm{cannot}\:\mathrm{comprehend} \\ $$$$\mathrm{right}\:\mathrm{now}. \\ $$
Answered by mrW1 last updated on 26/Jun/17
Let AB=a and CD=b Case 1: AB//CD A_(ΔABM) =(1/2)AB×h_1 =(1/2)ah_1 A_(ΔCDM) =(1/2)CD×h_2 =(1/2)bh_2 A_(ΔABM) =2A_(ΔCDM) ⇒ah_1 =2bh_2 (h_1 /h_2 )=((2b)/a) h_1 +h_2 =h ⇒h_1 =((2b)/(a+2b))h=constant ⇒h_2 =(a/(a+2b))h=constant ⇒M lies on a straight line // to AB and CD
$$\mathrm{Let}\:\mathrm{AB}=\mathrm{a}\:\mathrm{and}\:\mathrm{CD}=\mathrm{b} \\ $$$$ \\ $$$$\mathrm{Case}\:\mathrm{1}:\:\mathrm{AB}//\mathrm{CD} \\ $$$$\mathrm{A}_{\Delta\mathrm{ABM}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{AB}×\mathrm{h}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ah}_{\mathrm{1}} \\ $$$$\mathrm{A}_{\Delta\mathrm{CDM}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{CD}×\mathrm{h}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{bh}_{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{A}_{\Delta\mathrm{ABM}} =\mathrm{2A}_{\Delta\mathrm{CDM}} \\ $$$$\Rightarrow\mathrm{ah}_{\mathrm{1}} =\mathrm{2bh}_{\mathrm{2}} \\ $$$$\frac{\mathrm{h}_{\mathrm{1}} }{\mathrm{h}_{\mathrm{2}} }=\frac{\mathrm{2b}}{\mathrm{a}} \\ $$$$\mathrm{h}_{\mathrm{1}} +\mathrm{h}_{\mathrm{2}} =\mathrm{h} \\ $$$$\Rightarrow\mathrm{h}_{\mathrm{1}} =\frac{\mathrm{2b}}{\mathrm{a}+\mathrm{2b}}\mathrm{h}=\mathrm{constant} \\ $$$$\Rightarrow\mathrm{h}_{\mathrm{2}} =\frac{\mathrm{a}}{\mathrm{a}+\mathrm{2b}}\mathrm{h}=\mathrm{constant} \\ $$$$\Rightarrow\mathrm{M}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\://\:\mathrm{to}\:\mathrm{AB} \\ $$$$\mathrm{and}\:\mathrm{CD} \\ $$
Commented by mrW1 last updated on 26/Jun/17
Commented by mrW1 last updated on 27/Jun/17
Case 2: AB and CD meets at point O, and ∠AOD=α≠0 let OM=d ∠AOM=β h_1 =d sin β h_2 =d sin (α−β) A_(ΔABM) =(1/2)ah_1 A_(ΔCDM) =(1/2)bh_2 A_(ΔABM) =2A_(ΔCDM) ⇒ah_1 =2bh_2 ⇒ad sin β=2bd sin (α−β) ⇒a sin β=2b sin (α−β) ⇒(a/(2b)) sin β=sin (α−β) with μ=(a/(2b)) ⇒μ sin β=sin α cos β+cos α sin β ⇒(μ−cos α) sin β=sin α cos β ⇒tan β=((sin α)/(μ−cos α)) ⇒β=tan^(−1) (((sin α)/(μ−cos α)))=constant ⇒M lies on a straight line which passes the point O.
$$\mathrm{Case}\:\mathrm{2}: \\ $$$$\mathrm{AB}\:\mathrm{and}\:\mathrm{CD}\:\mathrm{meets}\:\mathrm{at}\:\mathrm{point}\:\mathrm{O},\:\mathrm{and} \\ $$$$\angle\mathrm{AOD}=\alpha\neq\mathrm{0} \\ $$$$ \\ $$$$\mathrm{let}\:\mathrm{OM}=\mathrm{d} \\ $$$$\angle\mathrm{AOM}=\beta \\ $$$$\mathrm{h}_{\mathrm{1}} =\mathrm{d}\:\mathrm{sin}\:\beta \\ $$$$\mathrm{h}_{\mathrm{2}} =\mathrm{d}\:\mathrm{sin}\:\left(\alpha−\beta\right) \\ $$$$ \\ $$$$\mathrm{A}_{\Delta\mathrm{ABM}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ah}_{\mathrm{1}} \\ $$$$\mathrm{A}_{\Delta\mathrm{CDM}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{bh}_{\mathrm{2}} \\ $$$$\mathrm{A}_{\Delta\mathrm{ABM}} =\mathrm{2A}_{\Delta\mathrm{CDM}} \\ $$$$\Rightarrow\mathrm{ah}_{\mathrm{1}} =\mathrm{2bh}_{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{ad}\:\mathrm{sin}\:\beta=\mathrm{2bd}\:\mathrm{sin}\:\left(\alpha−\beta\right) \\ $$$$\Rightarrow\mathrm{a}\:\mathrm{sin}\:\beta=\mathrm{2b}\:\mathrm{sin}\:\left(\alpha−\beta\right) \\ $$$$\Rightarrow\frac{\mathrm{a}}{\mathrm{2b}}\:\mathrm{sin}\:\beta=\mathrm{sin}\:\left(\alpha−\beta\right) \\ $$$$\mathrm{with}\:\mu=\frac{\mathrm{a}}{\mathrm{2b}} \\ $$$$\Rightarrow\mu\:\mathrm{sin}\:\beta=\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta+\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta \\ $$$$\Rightarrow\left(\mu−\mathrm{cos}\:\alpha\right)\:\mathrm{sin}\:\beta=\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta \\ $$$$\Rightarrow\mathrm{tan}\:\beta=\frac{\mathrm{sin}\:\alpha}{\mu−\mathrm{cos}\:\alpha} \\ $$$$\Rightarrow\beta=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{sin}\:\alpha}{\mu−\mathrm{cos}\:\alpha}\right)=\mathrm{constant} \\ $$$$ \\ $$$$\Rightarrow\mathrm{M}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{a}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{which} \\ $$$$\mathrm{passes}\:\mathrm{the}\:\mathrm{point}\:\mathrm{O}. \\ $$
Commented by mrW1 last updated on 26/Jun/17
Commented by mrW1 last updated on 28/Jun/17
This method is valid generally for A_(ΔABM) =nA_(ΔCDM) we just need to make D′E′/E′A′=1/n
$$\mathrm{This}\:\mathrm{method}\:\mathrm{is}\:\mathrm{valid}\:\mathrm{generally}\:\mathrm{for} \\ $$$$\mathrm{A}_{\Delta\mathrm{ABM}} =\mathrm{nA}_{\Delta\mathrm{CDM}} \\ $$$$\mathrm{we}\:\mathrm{just}\:\mathrm{need}\:\mathrm{to}\:\mathrm{make}\:\mathrm{D}'\mathrm{E}'/\mathrm{E}'\mathrm{A}'=\mathrm{1}/\mathrm{n} \\ $$
Commented by mrW1 last updated on 26/Jun/17
This is how to draw the line: Make OA′=AB=a Make OD′=CD=b Make D′E′=(1/3)A′D′ i.e. A′E′=2D′E′ ⇒A_(ΔOA′E′) =2A_(ΔOD′E′) Connecting OE′ we get a line which intersects with AD at E and with BC at F. EF is the line for point M.
$$\mathrm{This}\:\mathrm{is}\:\mathrm{how}\:\mathrm{to}\:\mathrm{draw}\:\mathrm{the}\:\mathrm{line}: \\ $$$$ \\ $$$$\mathrm{Make}\:\mathrm{OA}'=\mathrm{AB}=\mathrm{a} \\ $$$$\mathrm{Make}\:\mathrm{OD}'=\mathrm{CD}=\mathrm{b} \\ $$$$\mathrm{Make}\:\mathrm{D}'\mathrm{E}'=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{A}'\mathrm{D}' \\ $$$$\mathrm{i}.\mathrm{e}.\:\mathrm{A}'\mathrm{E}'=\mathrm{2D}'\mathrm{E}' \\ $$$$\Rightarrow\mathrm{A}_{\Delta\mathrm{OA}'\mathrm{E}'} =\mathrm{2A}_{\Delta\mathrm{OD}'\mathrm{E}'} \\ $$$$ \\ $$$$\mathrm{Connecting}\:\mathrm{OE}'\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:\mathrm{line}\:\mathrm{which} \\ $$$$\mathrm{intersects}\:\mathrm{with}\:\mathrm{AD}\:\mathrm{at}\:\mathrm{E}\:\mathrm{and}\:\mathrm{with}\:\mathrm{BC} \\ $$$$\mathrm{at}\:\mathrm{F}.\:\mathrm{EF}\:\mathrm{is}\:\mathrm{the}\:\mathrm{line}\:\mathrm{for}\:\mathrm{point}\:\mathrm{M}. \\ $$
Commented by mrW1 last updated on 26/Jun/17
Commented by mrW1 last updated on 27/Jun/17
As we can see in my calculation above the line with angle β for the condition A_(ΔABM) =2A_(ΔCDM) is independent from the position of points A and D, but only dependent from a, b and α. Therefore we can move the point A and D to the point O, and the line for locus of point M (expressed by β) is the same. And in this special situation the line for locus of point M can be easily determined through point E′.
$$\mathrm{As}\:\mathrm{we}\:\mathrm{can}\:\mathrm{see}\:\mathrm{in}\:\mathrm{my}\:\mathrm{calculation}\:\mathrm{above} \\ $$$$\mathrm{the}\:\mathrm{line}\:\mathrm{with}\:\mathrm{angle}\:\beta\:\mathrm{for}\:\mathrm{the}\:\mathrm{condition} \\ $$$$\mathrm{A}_{\Delta\mathrm{ABM}} =\mathrm{2A}_{\Delta\mathrm{CDM}} \:\mathrm{is}\:\mathrm{independent}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{position}\:\mathrm{of}\:\mathrm{points}\:\mathrm{A}\:\mathrm{and}\:\mathrm{D},\:\mathrm{but} \\ $$$$\mathrm{only}\:\mathrm{dependent}\:\mathrm{from}\:\mathrm{a},\:\mathrm{b}\:\mathrm{and}\:\alpha. \\ $$$$\mathrm{Therefore}\:\mathrm{we}\:\mathrm{can}\:\mathrm{move}\:\mathrm{the}\:\mathrm{point}\:\mathrm{A} \\ $$$$\mathrm{and}\:\mathrm{D}\:\mathrm{to}\:\mathrm{the}\:\mathrm{point}\:\mathrm{O},\:\mathrm{and}\:\mathrm{the}\:\mathrm{line}\:\mathrm{for} \\ $$$$\mathrm{locus}\:\mathrm{of}\:\mathrm{point}\:\mathrm{M}\:\left(\mathrm{expressed}\:\mathrm{by}\:\beta\right) \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{same}. \\ $$$$\mathrm{And}\:\mathrm{in}\:\mathrm{this}\:\mathrm{special}\:\mathrm{situation}\:\mathrm{the}\:\mathrm{line} \\ $$$$\mathrm{for}\:\mathrm{locus}\:\mathrm{of}\:\mathrm{point}\:\mathrm{M}\:\mathrm{can}\:\mathrm{be}\:\mathrm{easily} \\ $$$$\mathrm{determined}\:\mathrm{through}\:\mathrm{point}\:\mathrm{E}'. \\ $$
Commented by mrW1 last updated on 28/Jun/17

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