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Question-16194




Question Number 16194 by RasheedSoomro last updated on 18/Jun/17
Commented by RasheedSoomro last updated on 18/Jun/17
A,B and C are three non-collinear points.  Also they are centers of three circles having  radii a,b and c respectively. D,E & F have been   taken on three circles respectively in such a way  that △DEF is an equilateral triangle.  (i) Draw any such triangle using Euclidean tools.  (ii)(Using Euclidean tools) Draw such triangle of           maximum size and find out its area.  (ii)(Using Euclidean tools) Draw such triangle of           minimum size and find out its area.  Take  AB=l,BC=m and CA=n.
$$\mathrm{A},\mathrm{B}\:\mathrm{and}\:\mathrm{C}\:\mathrm{are}\:\mathrm{three}\:\mathrm{non}-\mathrm{collinear}\:\mathrm{points}. \\ $$$$\mathrm{Also}\:\mathrm{they}\:\mathrm{are}\:\mathrm{centers}\:\mathrm{of}\:\mathrm{three}\:\mathrm{circles}\:\mathrm{having} \\ $$$$\mathrm{radii}\:{a},{b}\:\mathrm{and}\:{c}\:\mathrm{respectively}.\:\mathrm{D},\mathrm{E}\:\&\:\mathrm{F}\:\mathrm{have}\:\mathrm{been}\: \\ $$$$\mathrm{taken}\:\mathrm{on}\:\mathrm{three}\:\mathrm{circles}\:\mathrm{respectively}\:\mathrm{in}\:\mathrm{such}\:\mathrm{a}\:\mathrm{way} \\ $$$$\mathrm{that}\:\bigtriangleup\mathrm{DEF}\:\mathrm{is}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle}. \\ $$$$\left(\mathrm{i}\right)\:\mathrm{Draw}\:\mathrm{any}\:\mathrm{such}\:\mathrm{triangle}\:\mathrm{using}\:\mathrm{Euclidean}\:\mathrm{tools}. \\ $$$$\left(\mathrm{ii}\right)\left(\mathrm{Using}\:\mathrm{Euclidean}\:\mathrm{tools}\right)\:\mathrm{Draw}\:\mathrm{such}\:\mathrm{triangle}\:\mathrm{of}\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{maximum}\:\mathrm{size}\:\mathrm{and}\:\mathrm{find}\:\mathrm{out}\:\mathrm{its}\:\mathrm{area}. \\ $$$$\left(\mathrm{ii}\right)\left(\mathrm{Using}\:\mathrm{Euclidean}\:\mathrm{tools}\right)\:\mathrm{Draw}\:\mathrm{such}\:\mathrm{triangle}\:\mathrm{of} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{minimum}\:\mathrm{size}\:\mathrm{and}\:\mathrm{find}\:\mathrm{out}\:\mathrm{its}\:\mathrm{area}. \\ $$$$\mathrm{Take}\:\:\mathrm{AB}={l},\mathrm{BC}={m}\:\mathrm{and}\:\mathrm{CA}={n}. \\ $$
Commented by RasheedSoomro last updated on 18/Jun/17
I think the question is clear now.  In question of Mr Amir (Behi) the circles  were concentric, whereas here circles have  different centers.
$$\mathrm{I}\:\mathrm{think}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{clear}\:\mathrm{now}. \\ $$$$\mathrm{In}\:\mathrm{question}\:\mathrm{of}\:\mathrm{Mr}\:\mathrm{Amir}\:\left(\mathrm{Behi}\right)\:\mathrm{the}\:\mathrm{circles} \\ $$$$\mathrm{were}\:\mathrm{concentric},\:\mathrm{whereas}\:\mathrm{here}\:\mathrm{circles}\:\mathrm{have} \\ $$$$\mathrm{different}\:\mathrm{centers}. \\ $$
Commented by mrW1 last updated on 20/Jun/17
This is a very interesting and challenging  question, sir. Its solution would be  not so easy I think.  For a fixed point on one of the three  circles we can find the equilateral  triangle(s) in the same way as I  described in Q15969. Then we change  the position of the fixed point along  the first circle, the size of the equilateral  triangle(s) will change correspondingly.  Among them to find the minimal and  maximal one will be very hard.
$$\mathrm{This}\:\mathrm{is}\:\mathrm{a}\:\mathrm{very}\:\mathrm{interesting}\:\mathrm{and}\:\mathrm{challenging} \\ $$$$\mathrm{question},\:\mathrm{sir}.\:\mathrm{Its}\:\mathrm{solution}\:\mathrm{would}\:\mathrm{be} \\ $$$$\mathrm{not}\:\mathrm{so}\:\mathrm{easy}\:\mathrm{I}\:\mathrm{think}. \\ $$$$\mathrm{For}\:\mathrm{a}\:\mathrm{fixed}\:\mathrm{point}\:\mathrm{on}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{three} \\ $$$$\mathrm{circles}\:\mathrm{we}\:\mathrm{can}\:\mathrm{find}\:\mathrm{the}\:\mathrm{equilateral} \\ $$$$\mathrm{triangle}\left(\mathrm{s}\right)\:\mathrm{in}\:\mathrm{the}\:\mathrm{same}\:\mathrm{way}\:\mathrm{as}\:\mathrm{I} \\ $$$$\mathrm{described}\:\mathrm{in}\:\mathrm{Q15969}.\:\mathrm{Then}\:\mathrm{we}\:\mathrm{change} \\ $$$$\mathrm{the}\:\mathrm{position}\:\mathrm{of}\:\mathrm{the}\:\mathrm{fixed}\:\mathrm{point}\:\mathrm{along} \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{circle},\:\mathrm{the}\:\mathrm{size}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equilateral} \\ $$$$\mathrm{triangle}\left(\mathrm{s}\right)\:\mathrm{will}\:\mathrm{change}\:\mathrm{correspondingly}. \\ $$$$\mathrm{Among}\:\mathrm{them}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{minimal}\:\mathrm{and} \\ $$$$\mathrm{maximal}\:\mathrm{one}\:\mathrm{will}\:\mathrm{be}\:\mathrm{very}\:\mathrm{hard}. \\ $$
Commented by RasheedSoomro last updated on 19/Jun/17
THANKS  SIR!  Is there any analytical approach for finding  maximum/minimum area of the triangle?
$$\mathcal{THANKS}\:\:\mathcal{SIR}! \\ $$$$\mathrm{Is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{analytical}\:\mathrm{approach}\:\mathrm{for}\:\mathrm{finding} \\ $$$$\mathrm{maximum}/\mathrm{minimum}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}? \\ $$
Commented by mrW1 last updated on 19/Jun/17
Commented by mrW1 last updated on 20/Jun/17
This should be an animated GIF image.   Unfortunately it is not supported by  the forum. It shows how the equi−  lateral triangles change when the point  D varies, see also single pictures later  for different positions of point D.    The position of point D can be  expressed as a variable, for example  θ. The size of the triangle(s) can also  be expressed as a function of θ:  Side length of triangle = s (s_1  and s_2 )  s=f(θ)  The max. and min. of s can be obtained  by (ds/dθ)=0.  The theory is easy, the doing is not, but  it′s generally possible.
$$\mathrm{This}\:\mathrm{should}\:\mathrm{be}\:\mathrm{an}\:\mathrm{animated}\:\mathrm{GIF}\:\mathrm{image}.\: \\ $$$$\mathrm{Unfortunately}\:\mathrm{it}\:\mathrm{is}\:\mathrm{not}\:\mathrm{supported}\:\mathrm{by} \\ $$$$\mathrm{the}\:\mathrm{forum}.\:\mathrm{It}\:\mathrm{shows}\:\mathrm{how}\:\mathrm{the}\:\mathrm{equi}− \\ $$$$\mathrm{lateral}\:\mathrm{triangles}\:\mathrm{change}\:\mathrm{when}\:\mathrm{the}\:\mathrm{point} \\ $$$$\mathrm{D}\:\mathrm{varies},\:\mathrm{see}\:\mathrm{also}\:\mathrm{single}\:\mathrm{pictures}\:\mathrm{later} \\ $$$$\mathrm{for}\:\mathrm{different}\:\mathrm{positions}\:\mathrm{of}\:\mathrm{point}\:\mathrm{D}. \\ $$$$ \\ $$$$\mathrm{The}\:\mathrm{position}\:\mathrm{of}\:\mathrm{point}\:\mathrm{D}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{expressed}\:\mathrm{as}\:\mathrm{a}\:\mathrm{variable},\:\mathrm{for}\:\mathrm{example} \\ $$$$\theta.\:\mathrm{The}\:\mathrm{size}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\left(\mathrm{s}\right)\:\mathrm{can}\:\mathrm{also} \\ $$$$\mathrm{be}\:\mathrm{expressed}\:\mathrm{as}\:\mathrm{a}\:\mathrm{function}\:\mathrm{of}\:\theta: \\ $$$$\mathrm{Side}\:\mathrm{length}\:\mathrm{of}\:\mathrm{triangle}\:=\:\mathrm{s}\:\left(\mathrm{s}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{s}_{\mathrm{2}} \right) \\ $$$$\mathrm{s}=\mathrm{f}\left(\theta\right) \\ $$$$\mathrm{The}\:\mathrm{max}.\:\mathrm{and}\:\mathrm{min}.\:\mathrm{of}\:\mathrm{s}\:\mathrm{can}\:\mathrm{be}\:\mathrm{obtained} \\ $$$$\mathrm{by}\:\frac{\mathrm{ds}}{\mathrm{d}\theta}=\mathrm{0}. \\ $$$$\mathrm{The}\:\mathrm{theory}\:\mathrm{is}\:\mathrm{easy},\:\mathrm{the}\:\mathrm{doing}\:\mathrm{is}\:\mathrm{not},\:\mathrm{but} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{generally}\:\mathrm{possible}. \\ $$
Commented by mrW1 last updated on 19/Jun/17
Commented by mrW1 last updated on 19/Jun/17
Commented by mrW1 last updated on 19/Jun/17
Commented by mrW1 last updated on 19/Jun/17
Commented by mrW1 last updated on 19/Jun/17
Commented by mrW1 last updated on 19/Jun/17
Commented by mrW1 last updated on 19/Jun/17
Commented by RasheedSoomro last updated on 20/Jun/17
Oh Sir, lot of  THANKS!!
$$\boldsymbol{\mathrm{Oh}}\:\boldsymbol{\mathrm{Sir}},\:\boldsymbol{\mathrm{lot}}\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathcal{T}}\mathcal{HANKS}!! \\ $$
Commented by mrW1 last updated on 20/Jun/17
Commented by mrW1 last updated on 20/Jun/17
When we display the size of the equi−  lateral triangles corresponding to  point D as AS_1  and AS_2 , we will get  two curves for point S_1  and S_2 ,  which show the change of the size of  the triangles when point D changes.  Then we can get the min. and max.  of them, see diagram.  AN=min. side length of triangle  AM=max. side length of triangle
$$\mathrm{When}\:\mathrm{we}\:\mathrm{display}\:\mathrm{the}\:\mathrm{size}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equi}− \\ $$$$\mathrm{lateral}\:\mathrm{triangles}\:\mathrm{corresponding}\:\mathrm{to} \\ $$$$\mathrm{point}\:\mathrm{D}\:\mathrm{as}\:\mathrm{AS}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{AS}_{\mathrm{2}} ,\:\mathrm{we}\:\mathrm{will}\:\mathrm{get} \\ $$$$\mathrm{two}\:\mathrm{curves}\:\mathrm{for}\:\mathrm{point}\:\mathrm{S}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{S}_{\mathrm{2}} , \\ $$$$\mathrm{which}\:\mathrm{show}\:\mathrm{the}\:\mathrm{change}\:\mathrm{of}\:\mathrm{the}\:\mathrm{size}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{triangles}\:\mathrm{when}\:\mathrm{point}\:\mathrm{D}\:\mathrm{changes}. \\ $$$$\mathrm{Then}\:\mathrm{we}\:\mathrm{can}\:\mathrm{get}\:\mathrm{the}\:\mathrm{min}.\:\mathrm{and}\:\mathrm{max}. \\ $$$$\mathrm{of}\:\mathrm{them},\:\mathrm{see}\:\mathrm{diagram}. \\ $$$$\mathrm{AN}=\mathrm{min}.\:\mathrm{side}\:\mathrm{length}\:\mathrm{of}\:\mathrm{triangle} \\ $$$$\mathrm{AM}=\mathrm{max}.\:\mathrm{side}\:\mathrm{length}\:\mathrm{of}\:\mathrm{triangle} \\ $$
Commented by RasheedSoomro last updated on 20/Jun/17
One thing to understand:  S1 and S2 are points and also lengths?
$$\mathrm{One}\:\mathrm{thing}\:\mathrm{to}\:\mathrm{understand}: \\ $$$$\mathrm{S1}\:\mathrm{and}\:\mathrm{S2}\:\mathrm{are}\:\mathrm{points}\:\mathrm{and}\:\mathrm{also}\:\mathrm{lengths}? \\ $$
Commented by mrW1 last updated on 20/Jun/17
AS_1 ^(−) =length s_1   AS_2 ^(−) =length s_2
$$\overline {\mathrm{AS}_{\mathrm{1}} }=\mathrm{length}\:\mathrm{s}_{\mathrm{1}} \\ $$$$\overline {\mathrm{AS}_{\mathrm{2}} }=\mathrm{length}\:\mathrm{s}_{\mathrm{2}} \\ $$
Commented by mrW1 last updated on 20/Jun/17
Commented by mrW1 last updated on 20/Jun/17
Commented by RasheedSoomro last updated on 21/Jun/17
TH𝛂nks for your deeply involvement  in my question!  TH𝛂nks for so much labour!
$$\boldsymbol{\mathcal{TH}\alpha{nks}}\:{for}\:{your}\:{deeply}\:{involvement} \\ $$$${in}\:{my}\:{question}! \\ $$$$\boldsymbol{\mathcal{TH}\alpha{nks}}\:{for}\:{so}\:{much}\:{labour}! \\ $$

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