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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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