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Question-16483




Question Number 16483 by RasheedSoomro last updated on 23/Jun/17
Commented by RasheedSoomro last updated on 23/Jun/17
Thanks sir for mentioning mistake.  The figure has been changed.
$$\mathrm{Thanks}\:\mathrm{sir}\:\mathrm{for}\:\mathrm{mentioning}\:\mathrm{mistake}. \\ $$$$\mathrm{The}\:\mathrm{figure}\:\mathrm{has}\:\mathrm{been}\:\mathrm{changed}. \\ $$
Commented by RasheedSoomro last updated on 23/Jun/17
There are three concentric squares.  Construct by ruler & compass   an equilateral triangle whose vertices are on  different squares and which is (i)maximum  (ii)minimum  This question is posted to show only a variant   of some past questions. If its answering is too  laborious , pl don′t solve it.
$$\mathrm{There}\:\mathrm{are}\:\mathrm{three}\:\mathrm{concentric}\:\mathrm{squares}. \\ $$$$\mathrm{Construct}\:\mathrm{by}\:\mathrm{ruler}\:\&\:\mathrm{compass}\: \\ $$$$\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle}\:\mathrm{whose}\:\mathrm{vertices}\:\mathrm{are}\:\mathrm{on} \\ $$$$\mathrm{different}\:\mathrm{squares}\:\mathrm{and}\:\mathrm{which}\:\mathrm{is}\:\left(\mathrm{i}\right)\mathrm{maximum} \\ $$$$\left(\mathrm{ii}\right)\mathrm{minimum} \\ $$$$\mathrm{This}\:\mathrm{question}\:\mathrm{is}\:\mathrm{posted}\:\mathrm{to}\:\mathrm{show}\:\mathrm{only}\:\mathrm{a}\:\mathrm{variant} \\ $$$$\:\mathrm{of}\:\mathrm{some}\:\mathrm{past}\:\mathrm{questions}.\:\mathrm{If}\:\mathrm{its}\:\mathrm{answering}\:\mathrm{is}\:\mathrm{too} \\ $$$$\mathrm{laborious}\:,\:\mathrm{pl}\:\mathrm{don}'\mathrm{t}\:\mathrm{solve}\:\mathrm{it}. \\ $$
Commented by mrW1 last updated on 23/Jun/17
the question is interesting. it′s similar  to the question with 3 parallel lines,  but these 3 lines have limited range,  so the answer is a little bit more difficult,  and there is not always a solution, I  think.
$$\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{interesting}.\:\mathrm{it}'\mathrm{s}\:\mathrm{similar} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{question}\:\mathrm{with}\:\mathrm{3}\:\mathrm{parallel}\:\mathrm{lines}, \\ $$$$\mathrm{but}\:\mathrm{these}\:\mathrm{3}\:\mathrm{lines}\:\mathrm{have}\:\mathrm{limited}\:\mathrm{range}, \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{a}\:\mathrm{little}\:\mathrm{bit}\:\mathrm{more}\:\mathrm{difficult}, \\ $$$$\mathrm{and}\:\mathrm{there}\:\mathrm{is}\:\mathrm{not}\:\mathrm{always}\:\mathrm{a}\:\mathrm{solution},\:\mathrm{I} \\ $$$$\mathrm{think}. \\ $$
Commented by RasheedSoomro last updated on 23/Jun/17
Actually the question is to determine  three points on three squares which  are pairwise equidistant from each other.  By experimenting with geogebra I thought  a possibility of existing such points.  I think that the method suggested by you  in some past questions will also be successful  here.  However to determine minimum/maximum  size-triangle by ruler & compass is doubtful.  .....  My first diagram was wrong and now I have  repaced it.  THANKS   for taking interest  Sir!
$$\mathrm{Actually}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{to}\:\mathrm{determine} \\ $$$$\mathrm{three}\:\mathrm{points}\:\mathrm{on}\:\mathrm{three}\:\mathrm{squares}\:\mathrm{which} \\ $$$$\mathrm{are}\:\mathrm{pairwise}\:\mathrm{equidistant}\:\mathrm{from}\:\mathrm{each}\:\mathrm{other}. \\ $$$$\mathrm{By}\:\mathrm{experimenting}\:\mathrm{with}\:\mathrm{geogebra}\:\mathrm{I}\:\mathrm{thought} \\ $$$$\mathrm{a}\:\mathrm{possibility}\:\mathrm{of}\:\mathrm{existing}\:\mathrm{such}\:\mathrm{points}. \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{that}\:\mathrm{the}\:\mathrm{method}\:\mathrm{suggested}\:\mathrm{by}\:\mathrm{you} \\ $$$$\mathrm{in}\:\mathrm{some}\:\mathrm{past}\:\mathrm{questions}\:\mathrm{will}\:\mathrm{also}\:\mathrm{be}\:\mathrm{successful} \\ $$$$\mathrm{here}. \\ $$$$\mathrm{However}\:\mathrm{to}\:\mathrm{determine}\:\mathrm{minimum}/\mathrm{maximum} \\ $$$$\mathrm{size}-\mathrm{triangle}\:\mathrm{by}\:\boldsymbol{\mathrm{ruler}}\:\&\:\boldsymbol{\mathrm{compass}}\:\mathrm{is}\:\mathrm{doubtful}. \\ $$$$….. \\ $$$$\mathrm{My}\:\mathrm{first}\:\mathrm{diagram}\:\mathrm{was}\:\mathrm{wrong}\:\mathrm{and}\:\mathrm{now}\:\mathrm{I}\:\mathrm{have} \\ $$$$\mathrm{repaced}\:\mathrm{it}. \\ $$$$\mathcal{THANKS}\:\:\:\mathrm{for}\:\mathrm{taking}\:\mathrm{interest}\:\:\mathcal{S}{ir}! \\ $$
Commented by mrW1 last updated on 23/Jun/17
Commented by RasheedSoomro last updated on 23/Jun/17
Special thanks for so much ELEGANT    diagrams!!!  As usual you did the best!  These min/max are only for point B?  They aren′t absolute min/max for any  point of three squares?
$$\mathcal{S}{pecial}\:{thanks}\:{for}\:{so}\:{much}\:\mathcal{ELEGANT}\:\: \\ $$$${diagrams}!!! \\ $$$$\boldsymbol{\mathrm{As}}\:\boldsymbol{\mathrm{usual}}\:\mathrm{you}\:\mathrm{did}\:\mathrm{the}\:\mathrm{best}! \\ $$$$\mathrm{These}\:\mathrm{min}/\mathrm{max}\:\mathrm{are}\:\mathrm{only}\:\mathrm{for}\:\mathrm{point}\:\mathrm{B}? \\ $$$$\mathrm{They}\:\mathrm{aren}'\mathrm{t}\:\mathrm{absolute}\:\mathrm{min}/\mathrm{max}\:\mathrm{for}\:\mathrm{any} \\ $$$$\mathrm{point}\:\mathrm{of}\:\mathrm{three}\:\mathrm{squares}? \\ $$
Commented by mrW1 last updated on 23/Jun/17
Thanks for your compliment!  To find the min. the point on the second  square should be as close to the point  on the first square (A) as possible.  To find the max. the point on the second  square should be as far from the point  on the first square (B) as possible.  So the min. and max. shown in the  figure are the absolute min. and max.  in this case. Certainly one can construct  more identical such min. and identical  such max. due to symmetry. But no  smaller or bigger equilateral triangles  can be constructed than the min. or  max. shown.
$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{compliment}! \\ $$$$\mathrm{To}\:\mathrm{find}\:\mathrm{the}\:\mathrm{min}.\:\mathrm{the}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{second} \\ $$$$\mathrm{square}\:\mathrm{should}\:\mathrm{be}\:\mathrm{as}\:\mathrm{close}\:\mathrm{to}\:\mathrm{the}\:\mathrm{point} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{first}\:\mathrm{square}\:\left(\mathrm{A}\right)\:\mathrm{as}\:\mathrm{possible}. \\ $$$$\mathrm{To}\:\mathrm{find}\:\mathrm{the}\:\mathrm{max}.\:\mathrm{the}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{second} \\ $$$$\mathrm{square}\:\mathrm{should}\:\mathrm{be}\:\mathrm{as}\:\mathrm{far}\:\mathrm{from}\:\mathrm{the}\:\mathrm{point} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{first}\:\mathrm{square}\:\left(\mathrm{B}\right)\:\mathrm{as}\:\mathrm{possible}. \\ $$$$\mathrm{So}\:\mathrm{the}\:\mathrm{min}.\:\mathrm{and}\:\mathrm{max}.\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{figure}\:\mathrm{are}\:\mathrm{the}\:\mathrm{absolute}\:\mathrm{min}.\:\mathrm{and}\:\mathrm{max}. \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}.\:\mathrm{Certainly}\:\mathrm{one}\:\mathrm{can}\:\mathrm{construct} \\ $$$$\mathrm{more}\:\mathrm{identical}\:\mathrm{such}\:\mathrm{min}.\:\mathrm{and}\:\mathrm{identical} \\ $$$$\mathrm{such}\:\mathrm{max}.\:\mathrm{due}\:\mathrm{to}\:\mathrm{symmetry}.\:\mathrm{But}\:\mathrm{no} \\ $$$$\mathrm{smaller}\:\mathrm{or}\:\mathrm{bigger}\:\mathrm{equilateral}\:\mathrm{triangles} \\ $$$$\mathrm{can}\:\mathrm{be}\:\mathrm{constructed}\:\mathrm{than}\:\mathrm{the}\:\mathrm{min}.\:\mathrm{or} \\ $$$$\mathrm{max}.\:\mathrm{shown}. \\ $$
Commented by RasheedSoomro last updated on 23/Jun/17
Sir, can it be possible to draw these min  or max by ruler & compass?
$$\mathcal{S}{ir},\:{c}\mathrm{an}\:\mathrm{it}\:\mathrm{be}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{draw}\:\mathrm{these}\:\mathrm{min} \\ $$$$\mathrm{or}\:\mathrm{max}\:\mathrm{by}\:\mathrm{ruler}\:\&\:\mathrm{compass}? \\ $$
Commented by mrW1 last updated on 23/Jun/17
Yes. The method I showed uses only  ruler and compass in mathematical  sense though I work on computer.
$$\mathrm{Yes}.\:\mathrm{The}\:\mathrm{method}\:\mathrm{I}\:\mathrm{showed}\:\mathrm{uses}\:\mathrm{only} \\ $$$$\mathrm{ruler}\:\mathrm{and}\:\mathrm{compass}\:\mathrm{in}\:\mathrm{mathematical} \\ $$$$\mathrm{sense}\:\mathrm{though}\:\mathrm{I}\:\mathrm{work}\:\mathrm{on}\:\mathrm{computer}. \\ $$
Commented by mrW1 last updated on 23/Jun/17
Commented by mrW1 last updated on 24/Jun/17
We start at point B.  On the side of third square opposite to   B we select two arbitrary points P_1  and   P_2  and construct equilateral triangles  ΔBP_1 Q_1  and ΔBP_2 Q_2 . Connecting  Q_1  and Q_2  we get a line which intersects  with the side of second square opposite  to B at point C. Using C as center and  CB as radius we draw a circle which  intersects the side of third square at  point D. ΔBCD is then the requested   equilateral triangle.  All these operations use only ruler and  compass.
$$\mathrm{We}\:\mathrm{start}\:\mathrm{at}\:\mathrm{point}\:\mathrm{B}. \\ $$$$\mathrm{On}\:\mathrm{the}\:\mathrm{side}\:\mathrm{of}\:\mathrm{third}\:\mathrm{square}\:\mathrm{opposite}\:\mathrm{to}\: \\ $$$$\mathrm{B}\:\mathrm{we}\:\mathrm{select}\:\mathrm{two}\:\mathrm{arbitrary}\:\mathrm{points}\:\mathrm{P}_{\mathrm{1}} \:\mathrm{and}\: \\ $$$$\mathrm{P}_{\mathrm{2}} \:\mathrm{and}\:\mathrm{construct}\:\mathrm{equilateral}\:\mathrm{triangles} \\ $$$$\Delta\mathrm{BP}_{\mathrm{1}} \mathrm{Q}_{\mathrm{1}} \:\mathrm{and}\:\Delta\mathrm{BP}_{\mathrm{2}} \mathrm{Q}_{\mathrm{2}} .\:\mathrm{Connecting} \\ $$$$\mathrm{Q}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{Q}_{\mathrm{2}} \:\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:\mathrm{line}\:\mathrm{which}\:\mathrm{intersects} \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{side}\:\mathrm{of}\:\mathrm{second}\:\mathrm{square}\:\mathrm{opposite} \\ $$$$\mathrm{to}\:\mathrm{B}\:\mathrm{at}\:\mathrm{point}\:\mathrm{C}.\:\mathrm{Using}\:\mathrm{C}\:\mathrm{as}\:\mathrm{center}\:\mathrm{and} \\ $$$$\mathrm{CB}\:\mathrm{as}\:\mathrm{radius}\:\mathrm{we}\:\mathrm{draw}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{which} \\ $$$$\mathrm{intersects}\:\mathrm{the}\:\mathrm{side}\:\mathrm{of}\:\mathrm{third}\:\mathrm{square}\:\mathrm{at} \\ $$$$\mathrm{point}\:\mathrm{D}.\:\Delta\mathrm{BCD}\:\mathrm{is}\:\mathrm{then}\:\mathrm{the}\:\mathrm{requested}\: \\ $$$$\mathrm{equilateral}\:\mathrm{triangle}. \\ $$$$\mathrm{All}\:\mathrm{these}\:\mathrm{operations}\:\mathrm{use}\:\mathrm{only}\:\mathrm{ruler}\:\mathrm{and} \\ $$$$\mathrm{compass}. \\ $$
Commented by RasheedSoomro last updated on 24/Jun/17
 [(T,H,A,N,K,S),(H,⌈,�,�,⌉,K),(A,∣,⌈,⌉,∣,N),(N,∣,⌊,⌋,∣,A),(K,⌊,_,_,⌋,H),(S,K,N,A,H,T) ]!  Appriciate your knowledge of geometry!  Appriciate your knowledge-sharing skill!  Appriciate your sharing-sprit!
$$\begin{bmatrix}{\mathcal{T}}&{\mathcal{H}}&{\mathcal{A}}&{\mathcal{N}}&{\mathcal{K}}&{\mathcal{S}}\\{\mathcal{H}}&{\lceil}&{}&{}&{\rceil}&{\mathcal{K}}\\{\mathcal{A}}&{\mid}&{\lceil}&{\rceil}&{\mid}&{\mathcal{N}}\\{\mathcal{N}}&{\mid}&{\lfloor}&{\rfloor}&{\mid}&{\mathcal{A}}\\{\mathcal{K}}&{\lfloor}&{\_}&{\_}&{\rfloor}&{\mathcal{H}}\\{\mathcal{S}}&{\mathcal{K}}&{\mathcal{N}}&{\mathcal{A}}&{\mathcal{H}}&{\mathcal{T}}\end{bmatrix}! \\ $$$$\mathrm{Appriciate}\:\mathrm{your}\:\mathrm{knowledge}\:\mathrm{of}\:\mathrm{geometry}! \\ $$$$\mathrm{Appriciate}\:\mathrm{your}\:\mathrm{knowledge}-\mathrm{sharing}\:\mathrm{skill}! \\ $$$$\mathrm{Appriciate}\:\mathrm{your}\:\mathrm{sharing}-\mathrm{sprit}! \\ $$$$ \\ $$

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