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Question-14502

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Question Number 14502 by ajfour last updated on 01/Jun/17
Commented by ajfour last updated on 01/Jun/17
Find x,y, and z in terms of a,b, and c using parameters h, and k.
$${Find}\:\boldsymbol{{x}},\boldsymbol{{y}},\:{and}\:\boldsymbol{{z}}\:\:{in}\:{terms}\:{of}\: \\ $$$$\boldsymbol{{a}},\boldsymbol{{b}},\:{and}\:\boldsymbol{{c}}\:{using}\:{parameters} \\ $$$$\boldsymbol{{h}},\:{and}\:\boldsymbol{{k}}.\: \\ $$
Commented by RasheedSoomro last updated on 02/Jun/17
ajfour is adventurist in mathematics!
$$\mathrm{ajfour}\:\mathrm{is}\:\mathrm{adventurist}\:\mathrm{in}\:\mathrm{mathematics}! \\ $$
Commented by chux last updated on 02/Jun/17
seriously
$$\mathrm{seriously} \\ $$
Answered by ajfour last updated on 02/Jun/17
EQUATIONS first: −−−−−−−−−−−−−−−−−− Largest equilateral triangles have all sides , s =x+y+z After the h shifting, and k shifting of the lower △ , the outer hexagon has unequal sides, namely a,b,c,a,b,c in the said order. Keeping in mind that half side length of each △ is s/2 , and that the lower ▽ has been pulled down by k and moved to the right by ′ h′. It follows from Pythagoras theorem that, k^2 +((s/2)+h)^2 =a^2 ....(i) h^2 +(((s(√3))/2)−k)^2 =b^2 ....(ii) k^2 +((s/2)−h)^2 =c^2 .....(iii) Again it can be observed in the figure that x= ((2k)/( (√3))) ...(iv) y= (s/2)+h−(x/2) ...(v) z= (s/2)−h−(x/2) ....(vi) eqn. (i)−(ii) yields 2hs = a^2 −c^2 or h = (((a^2 −c^2 ))/(2s)) ...(vii) Adding eqns. (i) and (iii) : 2k^2 +(s^2 /2)+2h^2 = a^2 +c^2 ...(I) doubling eqn. (ii) gives: 2h^2 +2k^2 +((3s^2 )/2)−2(√3)ks = 2b^2 subtracting them, we find s^2 −2(√3)ks+(a^2 +c^2 −2b^2 )=0 ⇒ k=((s^2 +(a^2 +c^2 −2b^2 ))/(2(√3)s)) ..(viii) continued.. Replacing for h and k from eqns. (vii) and (viii) in (I) : ((2(s^2 +a^2 +c^2 −2b^2 )^2 )/(12s^2 ))+(((a^2 −c^2 )^2 )/(2s^2 )) +(s^2 /2)−(a^2 +c^2 )=0 ⇒ (s^2 +a^2 +c^2 −2b^2 )^2 +3(a^2 −c^2 )^2 +3s^4 −6s^2 (a^2 +c^2 )=0 let a^2 +c^2 −2b^2 = p , a^2 −c^2 = q , and (a^2 +c^2 )=r, then, s^4 +2ps^2 +3q^2 +3s^4 −6rs^2 =0 4s^4 +(2p−6r)s^2 +3q^2 =0 s^2 =(((6r−2p)±(√((2p−6r)^2 −48q^2 )))/8) or, s^2 =(((3r−p)±(√((3r−p)^2 −12q^2 )))/4) After evaluating ′s′ we find h, and k . Then we can calculate x,y, and z.
$$\boldsymbol{{E}}{QUATIONS}\:{first}: \\ $$$$−−−−−−−−−−−−−−−−−− \\ $$$${Largest}\:{equilateral}\:{triangles}\:{have}\:\boldsymbol{{all}} \\ $$$${sides}\:,\:\:\:\:\boldsymbol{{s}}\:=\boldsymbol{{x}}+\boldsymbol{{y}}+\boldsymbol{{z}}\:\: \\ $$$$\boldsymbol{{After}}\:\boldsymbol{{the}}\:\:\boldsymbol{{h}}\:\boldsymbol{{shifting}},\:\:{and} \\ $$$$\:\boldsymbol{{k}}\:\boldsymbol{{shifting}}\:{of}\:{the} \\ $$$$\boldsymbol{{lower}}\:\bigtriangleup\:,\:{the}\:\boldsymbol{{outer}}\:\boldsymbol{{hexagon}} \\ $$$$\boldsymbol{{has}}\:\boldsymbol{{unequal}}\:\boldsymbol{{sides}},\:{namely} \\ $$$$\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}},\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}\:\:{in}\:{the}\:{said}\:{order}. \\ $$$$ \\ $$$$\boldsymbol{{K}}{eeping}\:{in}\:{mind}\:{that}\:\boldsymbol{{half}}\:\boldsymbol{{side}} \\ $$$${length}\:{of}\:{each}\:\bigtriangleup\:{is}\:\boldsymbol{{s}}/\mathrm{2}\:\:,\:{and} \\ $$$${that}\:{the}\:{lower}\:\bigtriangledown\:{has}\:{been}\:{pulled} \\ $$$${down}\:{by}\:\boldsymbol{{k}}\:{and} \\ $$$${moved}\:\:{to}\:{the}\:{right}\:{by}\:'\:\boldsymbol{{h}}'. \\ $$$$\:\:{It}\:{follows}\:{from}\:\boldsymbol{{Pythagoras}} \\ $$$$\boldsymbol{{theorem}}\:{that}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{k}}^{\mathrm{2}} +\left(\frac{\boldsymbol{{s}}}{\mathrm{2}}+\boldsymbol{{h}}\right)^{\mathrm{2}} =\boldsymbol{{a}}^{\mathrm{2}} \:\:\:\:\:….\left(\boldsymbol{{i}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{h}}^{\mathrm{2}} +\left(\frac{\boldsymbol{{s}}\sqrt{\mathrm{3}}}{\mathrm{2}}−\boldsymbol{{k}}\right)^{\mathrm{2}} =\boldsymbol{{b}}^{\mathrm{2}} \:\:\:….\left(\boldsymbol{{ii}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{k}}^{\mathrm{2}} +\left(\frac{\boldsymbol{{s}}}{\mathrm{2}}−\boldsymbol{{h}}\right)^{\mathrm{2}} =\boldsymbol{{c}}^{\mathrm{2}} \:\:…..\left(\boldsymbol{{iii}}\right) \\ $$$$\:\:\:\:\:\boldsymbol{{A}}{gain}\:{it}\:{can}\:{be}\:{observed} \\ $$$${in}\:{the}\:{figure}\:{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{x}}=\:\frac{\mathrm{2}\boldsymbol{{k}}}{\:\sqrt{\mathrm{3}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\left(\boldsymbol{{iv}}\right)\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{y}}=\:\frac{\boldsymbol{{s}}}{\mathrm{2}}+\boldsymbol{{h}}−\frac{\boldsymbol{{x}}}{\mathrm{2}}\:\:\:\:\:\:\:\:…\left(\boldsymbol{{v}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{z}}=\:\frac{\boldsymbol{{s}}}{\mathrm{2}}−\boldsymbol{{h}}−\frac{\boldsymbol{{x}}}{\mathrm{2}}\:\:\:\:\:\:\:….\left(\boldsymbol{{vi}}\right) \\ $$$$\:{eqn}.\:\left(\boldsymbol{{i}}\right)−\left(\boldsymbol{{ii}}\right)\:{yields} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\boldsymbol{{hs}}\:=\:\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:{or}\:\:\:\:\boldsymbol{{h}}\:=\:\frac{\left(\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} \right)}{\mathrm{2}\boldsymbol{{s}}}\:\:\:\:\:\:\:…\left(\boldsymbol{{vii}}\right) \\ $$$$\:\:{Adding}\:\:{eqns}.\:\left(\boldsymbol{{i}}\right)\:{and}\:\left(\boldsymbol{{iii}}\right)\:: \\ $$$$\:\:\:\:\:\:\mathrm{2}\boldsymbol{{k}}^{\mathrm{2}} +\frac{\boldsymbol{{s}}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{2}\boldsymbol{{h}}^{\mathrm{2}} =\:\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} \:\:\:\:…\left(\boldsymbol{{I}}\right) \\ $$$${doubling}\:{eqn}.\:\left(\boldsymbol{{ii}}\right)\:{gives}: \\ $$$$\:\:\:\mathrm{2}\boldsymbol{{h}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{k}}^{\mathrm{2}} +\frac{\mathrm{3}\boldsymbol{{s}}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{3}}\boldsymbol{{ks}}\:=\:\mathrm{2}\boldsymbol{{b}}^{\mathrm{2}} \\ $$$$\:\:\:\:\:{subtracting}\:{them},\:{we}\:{find}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\boldsymbol{{s}}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}\boldsymbol{{ks}}+\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{b}}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow\:\:\boldsymbol{{k}}=\frac{\boldsymbol{{s}}^{\mathrm{2}} +\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{b}}^{\mathrm{2}} \right)}{\mathrm{2}\sqrt{\mathrm{3}}\boldsymbol{{s}}}\:\:\:\:..\left(\boldsymbol{{viii}}\right) \\ $$$${continued}.. \\ $$$${Replacing}\:{for}\:\boldsymbol{{h}}\:{and}\:\boldsymbol{{k}}\:{from} \\ $$$${eqns}.\:\:\left({vii}\right)\:{and}\:\left({viii}\right)\:{in}\:\:\left(\boldsymbol{{I}}\right)\:: \\ $$$$\: \\ $$$$\:\:\frac{\mathrm{2}\left(\boldsymbol{{s}}^{\mathrm{2}} +\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{b}}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{12}\boldsymbol{{s}}^{\mathrm{2}} }+\frac{\left(\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{2}\boldsymbol{{s}}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\boldsymbol{{s}}^{\mathrm{2}} }{\mathrm{2}}−\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow\:\left(\boldsymbol{{s}}^{\mathrm{2}} +\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{b}}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{3}\left(\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{3}\boldsymbol{{s}}^{\mathrm{4}} −\mathrm{6}\boldsymbol{{s}}^{\mathrm{2}} \left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${let}\:\:\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{b}}^{\mathrm{2}} =\:\boldsymbol{{p}}\:,\: \\ $$$$\:\:\:\:\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} =\:\boldsymbol{{q}}\:,\:\:{and}\:\:\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} \right)=\boldsymbol{{r}}, \\ $$$${then},\:\boldsymbol{{s}}^{\mathrm{4}} +\mathrm{2}\boldsymbol{{ps}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{q}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{s}}^{\mathrm{4}} −\mathrm{6}\boldsymbol{{rs}}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\mathrm{4}\boldsymbol{{s}}^{\mathrm{4}} +\left(\mathrm{2}\boldsymbol{{p}}−\mathrm{6}\boldsymbol{{r}}\right)\boldsymbol{{s}}^{\mathrm{2}} +\mathrm{3}\boldsymbol{{q}}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\boldsymbol{{s}}^{\mathrm{2}} =\frac{\left(\mathrm{6}\boldsymbol{{r}}−\mathrm{2}\boldsymbol{{p}}\right)\pm\sqrt{\left(\mathrm{2}\boldsymbol{{p}}−\mathrm{6}\boldsymbol{{r}}\right)^{\mathrm{2}} −\mathrm{48}\boldsymbol{{q}}^{\mathrm{2}} }}{\mathrm{8}} \\ $$$${or},\:\:\:\boldsymbol{{s}}^{\mathrm{2}} =\frac{\left(\mathrm{3}\boldsymbol{{r}}−\boldsymbol{{p}}\right)\pm\sqrt{\left(\mathrm{3}\boldsymbol{{r}}−\boldsymbol{{p}}\right)^{\mathrm{2}} −\mathrm{12}\boldsymbol{{q}}^{\mathrm{2}} }}{\mathrm{4}} \\ $$$${After}\:{evaluating}\:\:'\boldsymbol{{s}}'\:{we}\:{find}\:\: \\ $$$$\:\boldsymbol{{h}},\:{and}\:\boldsymbol{{k}}\:. \\ $$$$\:\boldsymbol{{T}}{hen}\:{we}\:{can}\:{calculate}\:\boldsymbol{{x}},\boldsymbol{{y}},\:{and}\:\boldsymbol{{z}}. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 02/Jun/17
mr Ajfour.it is so beautiful. thank you for this nice job.
$${mr}\:{Ajfour}.{it}\:{is}\:{so}\:{beautiful}. \\ $$$${thank}\:{you}\:{for}\:{this}\:{nice}\:{job}. \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 02/Jun/17
nice!this is only for regular hexagon. mr Ajfor!please Q.14535.
$${nice}!{this}\:{is}\:{only}\:{for}\:{regular}\:{hexagon}. \\ $$$${mr}\:{Ajfor}!{please}\:{Q}.\mathrm{14535}. \\ $$

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