# Prove-that-ratio-of-a-regular-pentagon-diagonal-to-its-side-is-5-1-2-

Question Number 3584 by prakash jain last updated on 16/Dec/15
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{a}\:\mathrm{regular}\:\mathrm{pentagon}\:\mathrm{diagonal} \\$$$$\mathrm{to}\:\mathrm{its}\:\mathrm{side}\:\mathrm{is}\:\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}. \\$$
Commented by Yozzii last updated on 15/Dec/15
$${Its}\:{side}\:{is}\:{one}\:{of}\:{the}\:\mathrm{5}\:\:{edges}? \\$$$$\:\:\:\:\:\:\:\:\:\:{a}\:\:\:\:\:\:\:\:{b} \\$$$$\:\:\:\:\:{c}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{d} \\$$$${diagonals}={bd},{ad},{ec},{ae},{bc}? \\$$$$\\$$
Commented by prakash jain last updated on 15/Dec/15
$${yes}.\:\mathrm{Also}\:\mathrm{it}\:\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\$$
Commented by Yozzii last updated on 16/Dec/15
$${Could}\:{a}\:{general}\:{formula}\:{for}\:{the} \\$$$${value}\:{of}\:{such}\:{a}\:{ratio}\:{be}\:{found}\:{in}\:{terms} \\$$$${of}\:{n}\:{for}\:{a}\:{regular}\:{n}−{gon}? \\$$
Commented by prakash jain last updated on 16/Dec/15
$$\mathrm{A}\:\mathrm{formula}\:\mathrm{using}\:\mathrm{sin}\:\mathrm{function}\:\mathrm{is}\:\mathrm{clear}.\: \\$$$$\mathrm{A}\:\mathrm{general}\:\mathrm{algebria}\:\mathrm{formula}\:\mathrm{will}\:\mathrm{require}\:\mathrm{solving} \\$$$${n}^{{th}} \:\mathrm{root}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{as}\:\mathrm{an}\:\mathrm{algebric}\:\mathrm{formula}. \\$$
Commented by Rasheed Soomro last updated on 16/Dec/15
$${What}\:{is}\:{the}\:{diagonal}\:{of}\:{polygon}? \\$$$${If}\:\:\:{a}\:{regular}\:{polygon}\:{has}\:\:{more}\:{than}\: \\$$$${one}\:{diagonals}\:{of}\:{digferent}\:{sizes}\:{then} \\$$$${there}\:{may}\:{be}\:{more}\:{than}\:{one}\:{such}\:{ratios} \\$$$${for}\:{example}\:\frac{{D}_{\mathrm{1}} }{{s}},\frac{{D}_{\mathrm{2}} }{{s}},… \\$$
Commented by prakash jain last updated on 16/Dec/15
$$\mathrm{Yes}.\:\mathrm{There}\:\mathrm{will}\:\mathrm{diagonals}\:\mathrm{of}\:\mathrm{different}\:\mathrm{length}. \\$$
Answered by Yozzii last updated on 16/Dec/15
$${All}\:{regular}\:{n}−{gons}\:{can}\:{be}\:{inscribed} \\$$$${in}\:{a}\:{circle}. \\$$$$\mathrm{PROOF}: \\$$$${Consider}\:{the}\:{n}\:{roots}\:{of}\:{the}\:{equation} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{z}^{{n}} ={re}^{{i}\theta} \: \\$$$${where}\:{i}=\sqrt{−\mathrm{1}},{r}>\mathrm{0},{n}\in\mathbb{N}+\left\{\mathrm{0}\right\},−\pi<\theta\leqslant\pi. \\$$$${Since}\:{e}^{{i}\theta} ={e}^{{i}\left(\theta+\mathrm{2}\pi{t}\right)} \:\left({t}\in\mathbb{Z}\right),\:{because}\:{of}\:{the}\:{periodicity} \\$$$${of}\:{the}\:{sine}\:{and}\:{cosine}\:{functions}\:{which} \\$$$${arise}\:{from}\:{the}\:{form}\:{e}^{{i}\theta} ={cos}\theta+{isin}\theta, \\$$$$\Rightarrow\:{z}^{{n}} ={re}^{{i}\left(\theta+\mathrm{2}{t}\pi\right)} \Rightarrow{z}={r}^{\mathrm{1}/{n}} {e}^{{i}\frac{\theta+\mathrm{2}{t}\pi}{{n}}} . \\$$$$\\$$$${The}\:{n}\:{values}\:{of}\:{z}\:{are}\:{obtained}\:{for}\:\mathrm{0}\leqslant{t}\leqslant{n}−\mathrm{1}. \\$$$${Observe}\:{that}\:{each}\:{value}\:{of}\:{z}\:{has}\:{the} \\$$$${common}\:{modulus}\:{r}^{\mathrm{1}/{n}} .\:{On}\:{an}\:{Argand} \\$$$${diagram},\:{say}\:{the}\:{n}\:{values}\:{of}\:{z}\:{are}\:{plotted} \\$$$${with}\:{lines}\:{joining}\:{their}\:{points}\:{to}\:{the} \\$$$${origin}.\:{Suppose}\:{one}\:{point}\:{has}\:{argument} \\$$$$\alpha_{{t}} =\frac{\theta+\mathrm{2}{t}\pi}{{n}}\:{and}\:{an}\:{adjacent}\:{point}\:{has}\: \\$$$${argument}\:\alpha_{{t}+\mathrm{1}} =\frac{\theta+\mathrm{2}\left({t}+\mathrm{1}\right)\pi}{{n}}. \\$$$$\Rightarrow\:\alpha_{{t}+\mathrm{1}} −\alpha_{{t}} =\frac{\theta+\mathrm{2}{t}\pi+\mathrm{2}\pi−\theta−\mathrm{2}{t}\pi}{{n}}=\frac{\mathrm{2}\pi}{{n}}. \\$$$${The}\:{angle}\:{between}\:{the}\:{lines}\:{joining} \\$$$${consecutive}\:{points}\:{is}\:{a}\:{constant}\:\left(\frac{\mathrm{2}\pi}{{n}}\right) \\$$$${independent}\:{of}\:{t}\:{and}\:{all}\:{of}\:{these}\:{n} \\$$$${angular}\:{differences}\:{sum}\:{to}\:\mathrm{2}\pi.\:{Thus},\:{the} \\$$$${n}\:{points}\:{are}\:{regularly}\:{positioned}\:{about}\:{the} \\$$$${origin}\:{and}\:{they}\:{may}\:{be}\:{joined}\:{by}\:{lines} \\$$$${to}\:{form}\:{a}\:{regular}\:{n}−{gon}\:{figure}. \\$$$${These}\:{n}\:{points}\:{are}\:{placed}\:{about}\:{the}\: \\$$$${origin}\:{and}\:{they}\:{are}\:{at}\:{the}\:{equal}\:{distances}\:\left({r}^{\mathrm{1}/{n}} \right) \\$$$${from}\:{the}\:{origin}.\:{These}\:{points}\:{hence}, \\$$$${by}\:{definition}\:{of}\:{a}\:{circle}\:{as}\:{the}\:{locus}\:{of} \\$$$${points}\:{at}\:{a}\:{fixed}\:{distance}\:{from}\:{a}\:{fixef} \\$$$${point},\:{lie}\:{on}\:{a}\:{circle}.\:{Since}\:{these}\:{n}\:{points} \\$$$${form}\:{the}\:{corners}\:{of}\:{a}\:{regular}\:{n}−{gon}, \\$$$${it}\:{follows}\:{that}\:{such}\:{a}\:{figure}\:{can}\:{be} \\$$$${inscribed}\:{in}\:{a}\:{circle}.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Box \\$$$$\\$$$${So},\:{inscribe}\:{a}\:{pentagon}\:{ABCDE}\:{in}\:{a}\:{circle}\:{and} \\$$$${let}\:{its}\:{radius}\:{be}\:{denoted}\:{by}\:{r}.\:{The} \\$$$${angle}\:{at}\:{the}\:{centre}\:{O}\:{of}\:{the}\:{pentagon}\:{in} \\$$$${each}\:{of}\:{the}\:{five}\:{sectors}\:{is}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\equiv\mathrm{72}°. \\$$$${Denote}\:{s}\:{as}\:{the}\:{side}\:{length}\:{of}\:{the}\:{pentagon}. \\$$$${Draw}\:{a}\:{perpendicular}\:{bisector}\:{of}\:{one} \\$$$${side}\:{of}\:{the}\:{pentagon}\:\left({e}.{g}\:{AB}\right).\:{In}\:{so}\:{doing},\:{one} \\$$$${of}\:{the}\:{sectors}\:{is}\:{symmetrically}\:{bisected} \\$$$${to}\:{form}\:{two}\:{right}−{angled}\:{triangles}, \\$$$${with}\:{hypothenuses}\:{being}\:{the}\:{radius}\:{r}\:{of} \\$$$${the}\:{circle},\:{and}\:{s}\:{is}\:{halved}.\: \\$$$$\\$$
Commented by Yozzii last updated on 16/Dec/15
$${Thus},\:{in}\:{such}\:{a}\:{triangle}, \\$$$$\frac{{s}}{\mathrm{2}}={rsin}\mathrm{36}^{°} \Rightarrow{r}=\frac{{s}}{\mathrm{2}{sin}\mathrm{36}°}\:.\:\:\left(\ast\right) \\$$$${Now},\:{draw}\:{a}\:{line}\:{joining}\:{two}\:{points} \\$$$${such}\:{that}\:{a}\:{diagonal}\:{D}\:{of}\:{the}\:{pentagon} \\$$$${is}\:{obtained}\:\left({e}.{g}\:{AC}\right).\:{Draw}\:{a}\:{perpendicular} \\$$$${bisector}\:{of}\:{AC}\:{so}\:{that}\:{another}\:{right}−{angled} \\$$$${triangle}\:{is}\:{obtained}\:{as}\:{before}.\:{However}, \\$$$${the}\:{angle}\:{at}\:{O}\:{in}\:{this}\:{triangle}\:{is}\:\mathrm{72}°. \\$$$${Therefore},\:{we}\:{get} \\$$$$\frac{{D}}{\mathrm{2}}={rsin}\mathrm{72}^{°} \Rightarrow{r}=\frac{{D}}{\mathrm{2}{sin}\mathrm{72}°}\:.\:\left(\ast\ast\right) \\$$$${Equating}\:\left(\ast\right)\:{and}\:\left(\ast\ast\right)\:{we}\:{have} \\$$$$\frac{{D}}{\mathrm{2}{sin}\mathrm{72}^{°} }=\frac{{s}}{\mathrm{2}{sin}\mathrm{36}°}\Rightarrow\frac{{D}}{{s}}=\frac{{sin}\mathrm{72}^{°} }{{sin}\mathrm{36}°}=\mathrm{2}{cos}\mathrm{36}°. \\$$$$\\$$
Commented by Yozzii last updated on 16/Dec/15
$${Let}\:{f}=\mathrm{18}°\Rightarrow\:\mathrm{5}{f}=\mathrm{90}°\Rightarrow\mathrm{2}{f}=\mathrm{90}°−\mathrm{3}{f}. \\$$$$\therefore\:{sin}\mathrm{2}{f}={sin}\left(\mathrm{90}°−\mathrm{3}{f}\right) \\$$$${sin}\left(\mathrm{90}°−{x}\right)={cosx}\:{in}\:{general}. \\$$$$\therefore\:{sin}\mathrm{2}{f}={cos}\mathrm{3}{f}=\mathrm{4}{cos}^{\mathrm{3}} {f}−\mathrm{3}{cosf} \\$$$${But}\:{sin}\mathrm{2}{f}=\mathrm{2}{sinfcosf}. \\$$$$\therefore\:\mathrm{2}{sinfcosf}=\mathrm{4}{cos}^{\mathrm{3}} {f}−\mathrm{3}{cosf}. \\$$$${Now},\:{cosf}\neq\mathrm{0}\Rightarrow\mathrm{2}{sinf}=\mathrm{4}{cos}^{\mathrm{2}} {f}−\mathrm{3} \\$$$$\mathrm{2}{sinf}=\mathrm{4}\left(\mathrm{1}−{sin}^{\mathrm{2}} {f}\right)−\mathrm{3} \\$$$$\mathrm{2}{sinf}=\mathrm{4}−\mathrm{4}{sin}^{\mathrm{2}} {f}−\mathrm{3} \\$$$$\mathrm{4}{sin}^{\mathrm{2}} {f}+\mathrm{2}{sinf}−\mathrm{1}=\mathrm{0} \\$$$$\therefore{sinf}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{4}−\mathrm{4}×\mathrm{4}×\left(−\mathrm{1}\right)}}{\mathrm{8}} \\$$$${sinf}=\frac{−\mathrm{2}\pm\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{8}}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{4}}. \\$$$$\mathrm{0}<{f}<\mathrm{90}°\Rightarrow{sinf}>\mathrm{0}\Rightarrow{sinf}\neq\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{4}}. \\$$$$\therefore\:{sinf}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}} \\$$$${cos}^{\mathrm{2}} {f}=\mathrm{1}−{sin}^{\mathrm{2}} {f} \\$$$${cos}^{\mathrm{2}} {f}=\mathrm{1}−\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} =\frac{\mathrm{16}−\mathrm{5}−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{16}} \\$$$${cos}^{\mathrm{2}} {f}=\frac{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{16}}=\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{8}} \\$$$$\Rightarrow\mathrm{2}{cos}^{\mathrm{2}} {f}=\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{4}}\Rightarrow\mathrm{2}{cos}^{\mathrm{2}} {f}−\mathrm{1}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}. \\$$$${But},\:{cos}\mathrm{2}{f}={cos}\mathrm{36}°=\mathrm{2}{cos}^{\mathrm{2}} {f}−\mathrm{1}. \\$$$$\therefore{cos}\mathrm{36}°=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\Rightarrow\mathrm{2}{cos}\mathrm{36}°=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}. \\$$$${Hence}\:\frac{{D}}{{s}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare \\$$$$\\$$
Answered by Rasheed Soomro last updated on 16/Dec/15
$${Let}\:{A},{B},{C}\:{are}\:{any}\:{three}\:{consecutive}\: \\$$$${vertics},\:{then}\:\overline {{AB}}\:\:,\overline {{BC}}\:\:{are}\:{sides}\:{of}\: \\$$$${pentagon}\:{and}\:\overline {{AC}}\:{is}\:{its}\:{diagonal} \\$$$${Now}\:{let}\:{m}\overline {{AB}}={m}\overline {{BC}}={s} \\$$$${and}\:\:{m}\overline {{AC}}\:={d} \\$$$${We}\:{have}\:{to}\:{prove}\:{d}:{s}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}:\mathrm{1} \\$$$${or}\:\frac{{d}}{{s}}=\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\$$$${We}\:{know}\:{that}\:{m}\angle{ABC}=\mathrm{108}° \\$$$${By}\:{cosine}\:{law}\: \\$$$$\:\:\:\:\:\:\:\:\:{d}=\sqrt{{s}^{\mathrm{2}} +{s}^{\mathrm{2}} −\mathrm{2}\left({s}\right)\left({s}\right)\:\mathrm{cos}\:\mathrm{108}°} \\$$$$\:\:\:\:\:\:\:\:\frac{{d}}{{s}}=\frac{\sqrt{{s}^{\mathrm{2}} +{s}^{\mathrm{2}} −\mathrm{2}\left({s}\right)\left({s}\right)\:\mathrm{cos}\:\mathrm{108}°}}{{s}} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{s}\sqrt{\mathrm{2}−\mathrm{2cos}\:\mathrm{108}°}}{{s}}=\sqrt{\mathrm{2}−\mathrm{2cos}\:\mathrm{108}°} \\$$$$\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{2}−\mathrm{2}\left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{4}}\right)}=\sqrt{\frac{\mathrm{8}−\mathrm{2}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}} \\$$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}} \\$$$${Now}\:{we}\:{have}\:{to}\:{change}\:{the}\:{numerator}\:{into} \\$$$${a}+{b}\sqrt{\mathrm{5}}\:\:{form},{where}\:{a}\:,{b}\:{are}\:{rationals} \\$$$${So}\:\:{let}\:\:\:\:\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}\:\:={a}+{b}\sqrt{\mathrm{5}} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}=\left({a}+{b}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={a}^{\mathrm{2}} +\mathrm{5}{b}^{\mathrm{2}} +\mathrm{2}{ab}\sqrt{\mathrm{5}} \\$$$${a}^{\mathrm{2}} +\mathrm{5}{b}^{\mathrm{2}} =\mathrm{6}\:\:\wedge\:\:\mathrm{2}{ab}=\mathrm{2}\Rightarrow{a}=\frac{\mathrm{1}}{{b}} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\$$$${a}^{\mathrm{2}} +\mathrm{5}{b}^{\mathrm{2}} =\mathrm{6}\Rightarrow\left(\frac{\mathrm{1}}{{b}}\right)^{\mathrm{2}} +\mathrm{5}{b}^{\mathrm{2}} =\mathrm{6} \\$$$$\Rightarrow\mathrm{1}+\mathrm{5}{b}^{\mathrm{4}} =\mathrm{6}{b}^{\mathrm{2}} \Rightarrow\mathrm{5}{b}^{\mathrm{4}} −\mathrm{6}{b}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\$$$$\Rightarrow{b}^{\mathrm{2}} =\frac{\mathrm{6}\pm\sqrt{\mathrm{36}−\mathrm{20}}}{\mathrm{10}}=\frac{\mathrm{6}\pm\mathrm{4}}{\mathrm{10}}=\mathrm{1},\frac{\mathrm{1}}{\mathrm{5}}\: \\$$$$\Rightarrow{b}=\pm\mathrm{1},\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\left[{discarding}\:{as}\:{its}\:{irrational}\right] \\$$$${b}=\pm\mathrm{1}\Rightarrow{a}=\pm\mathrm{1} \\$$$${a}+{b}\sqrt{\mathrm{5}}=\mathrm{1}+\sqrt{\mathrm{5}},\mathrm{1}−\sqrt{\mathrm{5}},−\mathrm{1}+\sqrt{\mathrm{5}},−\mathrm{1}−\sqrt{\mathrm{5}} \\$$$${Last}\:{three}\:{are}\:{extraneous}\left(?\right) \\$$$$\frac{\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\$$
Commented by prakash jain last updated on 16/Dec/15
$$\frac{{d}}{{s}}=\frac{\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\: \\$$
Commented by Rasheed Soomro last updated on 16/Dec/15
$$\mathcal{TH}\alpha{n}\Bbbk\mathcal{S}! \\$$
Answered by Rasheed Soomro last updated on 16/Dec/15
$${Let}\:{A},{B},{C}\:{are}\:{three}\:{consecutive}\:{vertices} \\$$$${of}\:{a}\:{pentagon}.{Then}\:{AB}\:{and}\:{BC}\:{are}\:{sides} \\$$$${and}\:{AC}\:{is}\:{diagonal}\:{of}\:{pentagon}.\:{m}\angle{ABC}=\mathrm{108}° \\$$$$\bigtriangleup{ABC}\:{is}\:{an}\:{issoscel}\:{triangle}\:{in}\:{which}\: \\$$$$\:{let}\:{AB}={BC}={s},\:{AC}={d} \\$$$${m}\angle{ABC}=\mathrm{108}°\:,{m}\angle{CAB}={m}\angle{BCA}=\mathrm{36}° \\$$$${By}\:{sine}\:{law} \\$$$$\:\:\:\:\frac{{d}}{{sin}\:\mathrm{108}°}=\frac{{s}}{\mathrm{36}°}\Rightarrow\frac{{d}}{{s}}=\frac{{sin}\:\mathrm{108}°}{{sin}\:\mathrm{36}°} \\$$$$\:\:\:\:\:=\frac{\left(\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}\right.}{\left(\sqrt{\mathrm{5}−\sqrt{\mathrm{5}}}\right.}×\frac{\left(\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}\right.}{\left(\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}\right.}=\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{25}−\mathrm{5}}}=\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{5}}} \\$$$$\:\:\:\:\:=\frac{\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{5}}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{5}}}=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}= \\$$$$\frac{{d}}{{s}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\$$$$\\$$