Question-15006

Question Number 15006 by ajfour last updated on 06/Jun/17

Commented by RasheedSoomro last updated on 07/Jun/17

Which app do you use to draw so nice figures ?!

$$\mathrm{Which}\:\boldsymbol{\mathrm{app}}\:\:\mathrm{do}\:\mathrm{you}\:\mathrm{use}\:\mathrm{to}\:\mathrm{draw}\:\mathrm{so} \\ $$$$\mathrm{nice}\:\mathrm{figures}\:?! \\ $$

Commented by ajfour last updated on 06/Jun/17

solution to Q.14940 sides of △AOB divided in 3 parts (special case): [image got posted as question several times; i too need to complain this issue..]

$${solution}\:\:{to}\:{Q}.\mathrm{14940}\:\:\: \\ $$$${sides}\:{of}\:\bigtriangleup{AOB}\:{divided}\:{in}\:\mathrm{3}\:{parts} \\ $$$$\left({special}\:{case}\right): \\ $$$$\left[{image}\:{got}\:{posted}\:{as}\:{question}\:{several}\right. \\ $$$$\left.{times};\:{i}\:{too}\:{need}\:{to}\:{complain}\:{this}\:{issue}..\right] \\ $$

Commented by ajfour last updated on 07/Jun/17

$${Lekh}\:{diagram} \\ $$

Commented by Tinkutara last updated on 07/Jun/17

Please tell me also Sir! Which app is this?

$$\mathrm{Please}\:\mathrm{tell}\:\mathrm{me}\:\mathrm{also}\:\mathrm{Sir}!\:\mathrm{Which}\:\mathrm{app}\:\mathrm{is} \\ $$$$\mathrm{this}? \\ $$

Commented by RasheedSoomro last updated on 07/Jun/17

TO WHOM WHO INTREST IN GEOMETRY A free geometry book in PDF. The book is treasure of theorems. For the book and about the book see the image below

$$\mathrm{TO}\:\:\mathrm{WHOM}\:\:\:\:\mathrm{WHO}\:\:\:\:\mathrm{INTREST}\:\:\:\:\mathrm{IN}\:\:\:\mathrm{GEOMETRY} \\ $$$$\mathrm{A}\:\mathrm{free}\:\mathrm{geometry}\:\mathrm{book}\:\mathrm{in}\:\mathrm{PDF}. \\ $$$$\mathrm{The}\:\mathrm{book}\:\mathrm{is}\:\mathrm{treasure}\:\mathrm{of}\:\mathrm{theorems}. \\ $$$$\mathrm{For}\:\mathrm{the}\:\mathrm{book}\:\mathrm{and}\:\mathrm{about}\:\mathrm{the}\:\mathrm{book} \\ $$$$\mathrm{see}\:\mathrm{the}\:\mathrm{image}\:\mathrm{below} \\ $$

Commented by RasheedSoomro last updated on 07/Jun/17

Commented by Tinkutara last updated on 07/Jun/17

I found PDF of this book on net but this book does not contain any questions (problems). It only contains figures. How to find its question?

$$\mathrm{I}\:\mathrm{found}\:\mathrm{PDF}\:\mathrm{of}\:\mathrm{this}\:\mathrm{book}\:\mathrm{on}\:\mathrm{net}\:\mathrm{but} \\ $$$$\mathrm{this}\:\mathrm{book}\:\mathrm{does}\:\mathrm{not}\:\mathrm{contain}\:\mathrm{any}\:\mathrm{questions} \\ $$$$\left(\mathrm{problems}\right).\:\mathrm{It}\:\mathrm{only}\:\mathrm{contains}\:\mathrm{figures}. \\ $$$$\mathrm{How}\:\mathrm{to}\:\mathrm{find}\:\mathrm{its}\:\mathrm{question}? \\ $$

Commented by RasheedSoomro last updated on 07/Jun/17

Statements of theorems (data and what to prove both) can be understand from figures!

$$\mathrm{Statements}\:\mathrm{of}\:\mathrm{theorems}\:\left(\mathrm{data}\:\mathrm{and}\:\mathrm{what}\:\mathrm{to}\right. \\ $$$$\left.\mathrm{prove}\:\mathrm{both}\right)\:\mathrm{can}\:\mathrm{be}\:\mathrm{understand}\:\mathrm{from}\:\:\mathrm{figures}! \\ $$

Answered by ajfour last updated on 06/Jun/17

eqn. of UT: (x_a /a)−(y_b /b)=1 (triangular coordinates) eqn. of ST: (x_a /(2a))+(y/b)=1 solving them to get T≡(((4a)/3), (b/3)) eqn. of RS: (x_a /a)+(y_b /(2b))=1 solving with that for ST to get S≡(((2a)/3), ((2b)/3)) Area_(△red) =((9ap)/2)−3ap=((3ap)/2) Area_(△SaT) = (1/2)(2a−a)(y_S −y_T ) = (1/2)(a)(((2p)/3)−(p/3))=((ap)/6) Area_(PQRSTU) =Area_(△red) −3Area_(SaT) =((3ap)/2)−3(((ap)/6))=ap total area of △AOB=((9ap)/2) so, ((Area_(PQRSTU) )/(Area_(△AOB) )) =((ap)/((9ap/2))) = (2/9) .

$${eqn}.\:{of}\:{UT}: \\ $$$$\:\frac{{x}_{{a}} }{{a}}−\frac{{y}_{{b}} }{{b}}=\mathrm{1}\:\:\:\:\:\left({triangular}\:{coordinates}\right) \\ $$$${eqn}.\:{of}\:{ST}: \\ $$$$\:\frac{{x}_{{a}} }{\mathrm{2}{a}}+\frac{{y}}{{b}}=\mathrm{1} \\ $$$${solving}\:{them}\:{to}\:{get}\:\:\boldsymbol{{T}}\equiv\left(\frac{\mathrm{4}{a}}{\mathrm{3}},\:\frac{{b}}{\mathrm{3}}\right) \\ $$$$\:{eqn}.\:{of}\:{RS}: \\ $$$$\:\frac{{x}_{{a}} }{{a}}+\frac{{y}_{{b}} }{\mathrm{2}{b}}=\mathrm{1} \\ $$$${solving}\:{with}\:{that}\:{for}\:{ST}\:{to}\:{get} \\ $$$$\:\:\:\:\boldsymbol{{S}}\equiv\left(\frac{\mathrm{2}{a}}{\mathrm{3}},\:\frac{\mathrm{2}{b}}{\mathrm{3}}\right) \\ $$$${Area}_{\bigtriangleup{red}} =\frac{\mathrm{9}{ap}}{\mathrm{2}}−\mathrm{3}{ap}=\frac{\mathrm{3}{ap}}{\mathrm{2}} \\ $$$${Area}_{\bigtriangleup{SaT}} =\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{a}−{a}\right)\left({y}_{{S}} −{y}_{{T}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({a}\right)\left(\frac{\mathrm{2}{p}}{\mathrm{3}}−\frac{{p}}{\mathrm{3}}\right)=\frac{\boldsymbol{{ap}}}{\mathrm{6}} \\ $$$${Area}_{{PQRSTU}} ={Area}_{\bigtriangleup{red}} −\mathrm{3}{Area}_{{SaT}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}{ap}}{\mathrm{2}}−\mathrm{3}\left(\frac{{ap}}{\mathrm{6}}\right)={ap} \\ $$$${total}\:{area}\:{of}\:\bigtriangleup{AOB}=\frac{\mathrm{9}{ap}}{\mathrm{2}} \\ $$$${so},\:\frac{{Area}_{{PQRSTU}} }{{Area}_{\bigtriangleup{AOB}} }\:=\frac{\boldsymbol{{ap}}}{\left(\mathrm{9}\boldsymbol{{ap}}/\mathrm{2}\right)}\:=\:\frac{\mathrm{2}}{\mathrm{9}}\:. \\ $$

Commented by mrW1 last updated on 06/Jun/17

$$\mathrm{very}\:\mathrm{nice}\:\mathrm{sir}!\:\mathrm{just}\:\mathrm{go}\:\mathrm{on}. \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 06/Jun/17

$${nice}\:{solution}\:{mr}\:{Ajfour}!\:{thanks}. \\ $$$${i}\:{love}\:{this}. \\ $$

Commented by ajfour last updated on 06/Jun/17

$${thanks}\:{sir},\:{symmetry}\:{is}\:{a} \\ $$$${beautiful}\:{property}\:.. \\ $$

Commented by mrW1 last updated on 07/Jun/17

mr ajfour: you only need to replace 2a with (n−1)a and 2b with (n−1)b, then you can get the general solution for case n. To make the thing a little bit easier, you can put point A on the y−axis, since this will not change the areas as we know.

$$\mathrm{mr}\:\mathrm{ajfour}: \\ $$$$\mathrm{you}\:\mathrm{only}\:\mathrm{need}\:\mathrm{to}\:\mathrm{replace}\:\mathrm{2a}\:\mathrm{with}\:\left(\mathrm{n}−\mathrm{1}\right)\mathrm{a} \\ $$$$\mathrm{and}\:\mathrm{2b}\:\mathrm{with}\:\left(\mathrm{n}−\mathrm{1}\right)\mathrm{b},\:\mathrm{then}\:\mathrm{you}\:\mathrm{can} \\ $$$$\mathrm{get}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{for}\:\mathrm{case}\:\mathrm{n}. \\ $$$$\mathrm{To}\:\mathrm{make}\:\mathrm{the}\:\mathrm{thing}\:\mathrm{a}\:\mathrm{little}\:\mathrm{bit}\:\mathrm{easier}, \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{put}\:\mathrm{point}\:\mathrm{A}\:\mathrm{on}\:\mathrm{the}\:\mathrm{y}−\mathrm{axis}, \\ $$$$\mathrm{since}\:\mathrm{this}\:\mathrm{will}\:\mathrm{not}\:\mathrm{change}\:\mathrm{the}\:\mathrm{areas} \\ $$$$\mathrm{as}\:\mathrm{we}\:\mathrm{know}. \\ $$

Commented by ajfour last updated on 07/Jun/17

$${All}\:{right}\:\mathcal{S}{ir},\:{shall}\:{attempt}.. \\ $$

Commented by mrW1 last updated on 07/Jun/17

$$\mathbb{G}\mathrm{reat}\:\mathrm{job}\:\mathrm{sir}!\:\mathbb{F}\mathrm{antastic}! \\ $$

Commented by ajfour last updated on 07/Jun/17

Got the answer Sir, for n≥2 and n∈N ((Area_(PQRST) )/(Area_(△AOB) )) = ((2(n^2 −3n+3)^2 )/(n^3 (2n−3))) .

$${Got}\:{the}\:{answer}\:\mathcal{S}{ir}, \\ $$$$\:\:\:{for}\:{n}\geqslant\mathrm{2}\:{and}\:{n}\in\mathbb{N} \\ $$$$\:\frac{{Area}_{{PQRST}} }{{Area}_{\bigtriangleup{AOB}} }\:=\:\frac{\mathrm{2}\left(\boldsymbol{{n}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{n}}+\mathrm{3}\right)^{\mathrm{2}} }{\boldsymbol{{n}}^{\mathrm{3}} \left(\mathrm{2}\boldsymbol{{n}}−\mathrm{3}\right)}\:. \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 08/Jun/17

$${mr}\:{Ajfour}!\:{please}\:{post}\:{your}\:{solution}\:{her}. \\ $$

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