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Question Number 14306 by ajfour last updated on 30/May/17

Commented by prakash jain last updated on 30/May/17

Are a,b and c sides of triangle?

$$\mathrm{Are}\:{a},{b}\:\mathrm{and}\:{c}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{triangle}? \\ $$

Commented by ajfour last updated on 30/May/17

yes.sides of △ABC.

$${yes}.{sides}\:{of}\:\bigtriangleup{ABC}. \\ $$

Commented by ajfour last updated on 30/May/17

Find x,y, and z in terms of a, b, and c ,(whicb are real).

$${Find}\:\boldsymbol{{x}},\boldsymbol{{y}},\:{and}\:\boldsymbol{{z}}\:{in}\:{terms}\:{of} \\ $$$$\boldsymbol{{a}},\:\boldsymbol{{b}},\:{and}\:\boldsymbol{{c}}\:,\left({whicb}\:{are}\:{real}\right). \\ $$

Answered by ajfour last updated on 30/May/17

Assume C as the origin witb side a along x axis. also A(α,β) and M(h,k) α^2 +β^2 =b^2 (a−α)^2 +(−β)^2 =c^2 subtracting: 2aα−a^2 =b^2 −c^2 α=((a^2 +b^2 −c^2 )/(2a)) β^ =±(√(b^2 −α^2 )) let m=(k/h) slope of CM=m=(k/h) slope of RC=−(1/(slope of AM)) =−((α−h)/(β−k)) ∠ between CM and RC =(π/6) So, tan (π/6)=((m+(((α−h)/(β−k))))/(1−(((α−h)/(β−k)))))=(1/(√3)) as m=k/h (((√3)k)/h)+(√3)(((α−h)/(β−k)))=1−(((α−h)/(β−k))) or, (√3)k^2 −k(𝛂+𝛃(√3)) =−h^2 (√3)+𝛂(√3)h−𝛃h ...(i) A similar second case, slope of BQ=−(1/m)=−(h/k) slope of BM=(k/(h−a)) ∠ between BM and BQ =(π/6) So, (((k/(h−a))+(1/m))/(1−((k/(h−a)))(1/m)))=(1/(√3)) witb m=(k/h) , this leads to (√3)k^2 +ak=−h^2 (√3)+ah(√3) ...(ii) (ii)−(i) gives: k(a+𝛂+𝛃(√3))=h(a(√3)−𝛂(√3)+𝛃h) m=(k/h)=(((a−𝛂)(√3)+𝛃)/(a+𝛂+𝛃(√3))) replacing k=mh in (ii) (√3)(1+m^2 )h^2 =(a(√3)−m)h h shouldn′t be zero, so h=((a((√3)−m))/((√3)(1+m^2 ))) and k=mh now, y^2 =h^2 +k^2 (see figure) z^2 =(𝛂−h)^2 +(𝛃−k)^2 x^2 =(a−h)^2 +k^2 . ............................................. if a=2, b=3, and c=4 we have 𝛂=((a^2 +b^2 −c^2 )/(2a))=−(3/4) positive 𝛃=(√(b^2 −α^2 ))=((√(135))/4) m=(((a−𝛂)(√3)+𝛃)/(a+𝛂+𝛃(√3)))=1.222 h=((a((√3)−m))/((√3)(1+m^2 )))=0.236 k=mh=0.29 y=(√(h^2 +k^2 )) ≈ 0.374 z=(√((𝛂−h)^2 +(𝛃−k)^2 )) ≈ 2.79 x=(√((a−h)^2 +k^2 )) ≈ 1.79 then x^2 +y^2 +xy=a^2 =4 y^2 +z^2 +yz=b^2 =9 z^2 +x^2 +xz=c^2 =16 i have verified ....

$${Assume}\:{C}\:{as}\:{the}\:\:{origin}\:{witb} \\ $$$${side}\:\boldsymbol{{a}}\:{along}\:{x}\:{axis}. \\ $$$${also}\:\:{A}\left(\alpha,\beta\right)\:\:{and}\:{M}\left({h},{k}\right) \\ $$$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$$\left({a}−\alpha\right)^{\mathrm{2}} +\left(−\beta\right)^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$${subtracting}: \\ $$$$\mathrm{2}{a}\alpha−{a}^{\mathrm{2}} ={b}^{\mathrm{2}} −{c}^{\mathrm{2}} \\ $$$$\alpha=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{a}}\: \\ $$$$\beta^{\:} =\pm\sqrt{{b}^{\mathrm{2}} −\alpha^{\mathrm{2}} } \\ $$$${let}\:\:{m}=\frac{{k}}{{h}} \\ $$$${slope}\:{of}\:\:{CM}={m}=\frac{{k}}{{h}} \\ $$$${slope}\:{of}\:{RC}=−\frac{\mathrm{1}}{{slope}\:{of}\:{AM}} \\ $$$$\:\:\:\:=−\frac{\alpha−{h}}{\beta−{k}} \\ $$$$\angle\:{between}\:{CM}\:{and}\:{RC}\:=\frac{\pi}{\mathrm{6}} \\ $$$${So},\:\:\mathrm{tan}\:\frac{\pi}{\mathrm{6}}=\frac{{m}+\left(\frac{\alpha−{h}}{\beta−{k}}\right)}{\mathrm{1}−\left(\frac{\alpha−{h}}{\beta−{k}}\right)}=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$$${as}\:{m}={k}/{h} \\ $$$$\frac{\sqrt{\mathrm{3}}{k}}{{h}}+\sqrt{\mathrm{3}}\left(\frac{\alpha−{h}}{\beta−{k}}\right)=\mathrm{1}−\left(\frac{\alpha−{h}}{\beta−{k}}\right) \\ $$$${or}, \\ $$$$\sqrt{\mathrm{3}}\boldsymbol{{k}}^{\mathrm{2}} −\boldsymbol{{k}}\left(\boldsymbol{\alpha}+\boldsymbol{\beta}\sqrt{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:\:=−\boldsymbol{{h}}^{\mathrm{2}} \sqrt{\mathrm{3}}+\boldsymbol{\alpha}\sqrt{\mathrm{3}}\boldsymbol{{h}}−\boldsymbol{\beta{h}}\:\:\:...\left(\boldsymbol{{i}}\right) \\ $$$${A}\:{similar}\:{second}\:{case}, \\ $$$${slope}\:{of}\:{BQ}=−\frac{\mathrm{1}}{{m}}=−\frac{{h}}{{k}} \\ $$$${slope}\:{of}\:{BM}=\frac{{k}}{{h}−{a}} \\ $$$$\angle\:{between}\:{BM}\:{and}\:{BQ}\:=\frac{\pi}{\mathrm{6}} \\ $$$${So},\:\:\frac{\frac{{k}}{{h}−{a}}+\frac{\mathrm{1}}{{m}}}{\mathrm{1}−\left(\frac{{k}}{{h}−{a}}\right)\frac{\mathrm{1}}{{m}}}=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$$${witb}\:{m}=\frac{{k}}{{h}}\:,\:{this}\:{leads}\:{to} \\ $$$$\sqrt{\mathrm{3}}\boldsymbol{{k}}^{\mathrm{2}} +\boldsymbol{{ak}}=−\boldsymbol{{h}}^{\mathrm{2}} \sqrt{\mathrm{3}}+\boldsymbol{{ah}}\sqrt{\mathrm{3}}\:\:\:...\left(\boldsymbol{{ii}}\right) \\ $$$$\left(\boldsymbol{{ii}}\right)−\left(\boldsymbol{{i}}\right)\:{gives}: \\ $$$$\boldsymbol{{k}}\left(\boldsymbol{{a}}+\boldsymbol{\alpha}+\boldsymbol{\beta}\sqrt{\mathrm{3}}\right)=\boldsymbol{{h}}\left(\boldsymbol{{a}}\sqrt{\mathrm{3}}−\boldsymbol{\alpha}\sqrt{\mathrm{3}}+\boldsymbol{\beta{h}}\right) \\ $$$$\boldsymbol{{m}}=\frac{\boldsymbol{{k}}}{\boldsymbol{{h}}}=\frac{\left(\boldsymbol{{a}}−\boldsymbol{\alpha}\right)\sqrt{\mathrm{3}}+\boldsymbol{\beta}}{\boldsymbol{{a}}+\boldsymbol{\alpha}+\boldsymbol{\beta}\sqrt{\mathrm{3}}}\: \\ $$$${replacing}\:{k}={mh}\:{in}\:\left({ii}\right) \\ $$$$\sqrt{\mathrm{3}}\left(\mathrm{1}+{m}^{\mathrm{2}} \right){h}^{\mathrm{2}} =\left({a}\sqrt{\mathrm{3}}−{m}\right){h} \\ $$$${h}\:{shouldn}'{t}\:{be}\:{zero},\:{so} \\ $$$$\:\:\boldsymbol{{h}}=\frac{\boldsymbol{{a}}\left(\sqrt{\mathrm{3}}−\boldsymbol{{m}}\right)}{\sqrt{\mathrm{3}}\left(\mathrm{1}+\boldsymbol{{m}}^{\mathrm{2}} \right)}\:\:\boldsymbol{{and}}\:\boldsymbol{{k}}=\boldsymbol{{mh}} \\ $$$${now}, \\ $$$$\:\:\:\:\:\boldsymbol{{y}}^{\mathrm{2}} =\boldsymbol{{h}}^{\mathrm{2}} +\boldsymbol{{k}}^{\mathrm{2}} \:\:\:\:\:\:\left({see}\:{figure}\right) \\ $$$$\:\:\:\:\:\boldsymbol{{z}}^{\mathrm{2}} =\left(\boldsymbol{\alpha}−\boldsymbol{{h}}\right)^{\mathrm{2}} +\left(\boldsymbol{\beta}−\boldsymbol{{k}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\boldsymbol{{x}}^{\mathrm{2}} =\left(\boldsymbol{{a}}−\boldsymbol{{h}}\right)^{\mathrm{2}} +\boldsymbol{{k}}^{\mathrm{2}} \:. \\ $$$$............................................. \\ $$$${if}\:\:{a}=\mathrm{2},\:{b}=\mathrm{3},\:{and}\:{c}=\mathrm{4}\: \\ $$$${we}\:{have}\:\:\boldsymbol{\alpha}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{a}}=−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${positive}\:\boldsymbol{\beta}=\sqrt{{b}^{\mathrm{2}} −\alpha^{\mathrm{2}} }=\frac{\sqrt{\mathrm{135}}}{\mathrm{4}} \\ $$$$\boldsymbol{{m}}=\frac{\left(\boldsymbol{{a}}−\boldsymbol{\alpha}\right)\sqrt{\mathrm{3}}+\boldsymbol{\beta}}{\boldsymbol{{a}}+\boldsymbol{\alpha}+\boldsymbol{\beta}\sqrt{\mathrm{3}}}=\mathrm{1}.\mathrm{222} \\ $$$$\boldsymbol{{h}}=\frac{\boldsymbol{{a}}\left(\sqrt{\mathrm{3}}−\boldsymbol{{m}}\right)}{\sqrt{\mathrm{3}}\left(\mathrm{1}+\boldsymbol{{m}}^{\mathrm{2}} \right)}=\mathrm{0}.\mathrm{236} \\ $$$$\boldsymbol{{k}}=\boldsymbol{{mh}}=\mathrm{0}.\mathrm{29} \\ $$$$\boldsymbol{{y}}=\sqrt{\boldsymbol{{h}}^{\mathrm{2}} +\boldsymbol{{k}}^{\mathrm{2}} }\:\approx\:\mathrm{0}.\mathrm{374} \\ $$$$\boldsymbol{{z}}=\sqrt{\left(\boldsymbol{\alpha}−\boldsymbol{{h}}\right)^{\mathrm{2}} +\left(\boldsymbol{\beta}−\boldsymbol{{k}}\right)^{\mathrm{2}} }\:\approx\:\mathrm{2}.\mathrm{79} \\ $$$$\boldsymbol{{x}}=\sqrt{\left(\boldsymbol{{a}}−\boldsymbol{{h}}\right)^{\mathrm{2}} +\boldsymbol{{k}}^{\mathrm{2}} }\:\approx\:\mathrm{1}.\mathrm{79} \\ $$$$\boldsymbol{{then}}\: \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{xy}}=\boldsymbol{{a}}^{\mathrm{2}} =\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{z}}^{\mathrm{2}} +\boldsymbol{{yz}}=\boldsymbol{{b}}^{\mathrm{2}} =\mathrm{9} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{{z}}^{\mathrm{2}} +\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{xz}}=\boldsymbol{{c}}^{\mathrm{2}} =\mathrm{16}\: \\ $$$${i}\:{have}\:{verified}\:.... \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

Commented by ajfour last updated on 30/May/17

x^2 +y^2 +xy=a^2 ⇒ (y+(x/2))^2 +(((x(√3))/2))^2 =a^2 y+(x/2)=asin θ ; ((x(√3))/2)=acos θ from other two equations z+(y/2)=bsin φ ; ((y(√3))/2)=bcos φ x+(z/2)=csin ψ ; ((z(√3))/2)=ccos ψ . hence the geometrical idea .. of a prior question of Algebra.

$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}={a}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\left({y}+\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{x}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\:\:{y}+\frac{{x}}{\mathrm{2}}={a}\mathrm{sin}\:\theta\:\:;\:\frac{{x}\sqrt{\mathrm{3}}}{\mathrm{2}}={a}\mathrm{cos}\:\theta \\ $$$$\:{from}\:{other}\:{two}\:{equations} \\ $$$$\:\:{z}+\frac{{y}}{\mathrm{2}}={b}\mathrm{sin}\:\phi\:\:;\:\:\frac{{y}\sqrt{\mathrm{3}}}{\mathrm{2}}={b}\mathrm{cos}\:\phi \\ $$$$\:\:{x}+\frac{{z}}{\mathrm{2}}={c}\mathrm{sin}\:\psi\:\:;\:\frac{{z}\sqrt{\mathrm{3}}}{\mathrm{2}}={c}\mathrm{cos}\:\psi\:. \\ $$$${hence}\:{the}\:{geometrical}\:{idea}\:.. \\ $$$${of}\:{a}\:{prior}\:{question}\:{of}\:{Algebra}. \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 30/May/17

ok mr Ajfour !it is amazing. it is a perfect method .i love it. thank you so much.

$${ok}\:{mr}\:{Ajfour}\:!{it}\:{is}\:{amazing}. \\ $$$${it}\:{is}\:{a}\:{perfect}\:{method}\:.{i}\:{love}\:{it}. \\ $$$${thank}\:{you}\:{so}\:{much}. \\ $$

Commented by ajfour last updated on 30/May/17

thanks for the appreciation Sir. i have not however considered the special cases when h=0, k=0, a^2 +b^2 =c^2 , ...

$${thanks}\:{for}\:{the}\:{appreciation}\:{Sir}. \\ $$$${i}\:{have}\:{not}\:{however}\:{considered} \\ $$$${the}\:\:{special}\:{cases}\:{when} \\ $$$${h}=\mathrm{0},\:{k}=\mathrm{0},\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} ,\:... \\ $$

Commented by prakash jain last updated on 30/May/17

For a=2,b=3,c=4 I got 4 solution set. x=d,y=e,z=f is solution than x=−d,y=−e,z=−f is also a solution.

$$\mathrm{For}\:{a}=\mathrm{2},{b}=\mathrm{3},{c}=\mathrm{4} \\ $$$$\mathrm{I}\:\mathrm{got}\:\mathrm{4}\:{solution}\:{set}. \\ $$$${x}={d},{y}={e},{z}={f}\:\mathrm{is}\:\mathrm{solution} \\ $$$$\mathrm{than}\:{x}=−{d},{y}=−{e},{z}=−{f}\:\mathrm{is}\:\mathrm{also} \\ $$$$\mathrm{a}\:\mathrm{solution}. \\ $$

Commented by ajfour last updated on 30/May/17

any integer ones ?

$${any}\:{integer}\:{ones}\:? \\ $$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 31/May/17

hello mr Ajfour. this is a nice way to solve this Q.but: 1)is point :M, unique? 2)∡QBP =^? (π/6) 3)i think if M ,be a point of intersection of some segments of ΔABC,such that midans,bisects,....we can solve this very easy by using relations about a triangle segments. 4)a very spicial case in a triangle, when M ,is the intersection of midans of ΔABC and:∠AMB=∠BMC=∠CMA.

$${hello}\:{mr}\:{Ajfour}. \\ $$$${this}\:{is}\:{a}\:{nice}\:{way}\:{to}\:{solve}\:{this}\:{Q}.{but}: \\ $$$$\left.\mathrm{1}\right){is}\:\:{point}\::{M},\:{unique}? \\ $$$$\left.\mathrm{2}\right)\measuredangle{QBP}\:\:\overset{?} {=}\:\:\frac{\pi}{\mathrm{6}}\:\: \\ $$$$\left.\mathrm{3}\right){i}\:{think}\:{if}\:{M}\:,{be}\:{a}\:{point}\:{of}\:{intersection} \\ $$$${of}\:{some}\:{segments}\:{of}\:\Delta{ABC},{such}\:{that} \\ $$$${midans},{bisects},....{we}\:{can}\:{solve}\:{this} \\ $$$${very}\:{easy}\:{by}\:{using}\:{relations}\:{about} \\ $$$${a}\:{triangle}\:{segments}. \\ $$$$\left.\mathrm{4}\right){a}\:{very}\:{spicial}\:{case}\:{in}\:{a}\:{triangle}, \\ $$$${when}\:{M}\:,{is}\:{the}\:{intersection}\:{of}\:{midans} \\ $$$${of}\:\Delta{ABC}\:{and}:\angle{AMB}=\angle{BMC}=\angle{CMA}. \\ $$

Commented by prakash jain last updated on 30/May/17

No integer solution. all 4 I wrote in question 14211. There are only 4 solutions for a=2,b=3,c=4 One of solutions matches with yours. I followed simplification to quadractic in one variable approach.

$$\mathrm{No}\:\mathrm{integer}\:\mathrm{solution}. \\ $$$$\mathrm{all}\:\mathrm{4}\:\mathrm{I}\:\mathrm{wrote}\:\mathrm{in}\:\mathrm{question}\:\mathrm{14211}. \\ $$$$\mathrm{There}\:\mathrm{are}\:\mathrm{only}\:\mathrm{4}\:\mathrm{solutions}\:\mathrm{for} \\ $$$${a}=\mathrm{2},{b}=\mathrm{3},{c}=\mathrm{4} \\ $$$$\mathrm{One}\:\mathrm{of}\:\mathrm{solutions}\:\mathrm{matches}\:\mathrm{with} \\ $$$$\mathrm{yours}.\:\mathrm{I}\:\mathrm{followed}\:\mathrm{simplification} \\ $$$$\mathrm{to}\:\mathrm{quadractic}\:\mathrm{in}\:\mathrm{one}\:\mathrm{variable} \\ $$$$\mathrm{approach}. \\ $$

Commented by mrW1 last updated on 31/May/17

Mr. ajfour: Great job done!

$${Mr}.\:{ajfour}:\:{Great}\:{job}\:{done}! \\ $$

Commented by Tawa11 last updated on 15/Jan/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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