Mentionez de la bun inceput ca nu voi folosi simbolul pentru grade, ci voi scrie sin 72. (ca sa nu scriu de fiecare data "textdegree"). Si de asemenea nu voi folosi unitati de masura.
[tex]\underline{Pasul~1.~Demonstram~ca~pentagonul~este~regulat~(figura~1).} \\ \\ Cum~triunghiurile~echivalente~AED~si~CED~au~suprafata~comuna \\ \\ EUD,~rezulta~ca~A_{AUE}=A_{CUD} \Leftrightarrow \\ \\ \Leftrightarrow \frac{AU \cdot UE \cdot sin ( \angle AUE)}{2}= \frac{CU \cdot UD \cdot sin( \angle CUD) }{2} \Rightarrow AU \cdot UE =CU \cdot UD \RIghtarrow \Leftrightarrow \\ \\ \Leftrightarrow \frac{AU}{CU}= \frac{DU}{UE} \Rightarrow ~ SW~||~DE~(T.R.Thales).[/tex]
[tex]Utilizand~acelasi~ra.tio.na.ment~pentru~restul~perechilor~de~ \\ \\ triunghiuri~echivalente~alaturate,~obtinem:~ST~||~CD,~TU~||~BC, \\ \\ UV ~||~AB,~VW~||~AE~si,~respectiv~SW~||~DE. \\ \\ Deci~pentagonul~este~ \underline{regulat}.[/tex]
[tex]\underline{Pasul~2.~Determinam~sin~72.} \\ \\ Cum~sin54= \frac{\sqrt{5}+1}{4}=cos36,~deducem: \\ \\ sin36= \sqrt{1-cos^236}= \sqrt{1- \frac{6+2 \sqrt{5}}{16} } = \sqrt{ \frac{10-2 \sqrt{5}}{16} }= \frac{\sqrt{10-2 \sqrt{5}}}{4} . \\ \\ Din~sin (2a)=2sin(a)cos(a),~obtinem:~sin(36)=2 \cdot sin 18 \cdot cos 18 \Leftrightarrow \\ \\ \Leftrightarrow~ \frac{\sqrt{10-2\sqrt{5}}}{4} =2 \cdot \sqrt{1- cos^218} \cdot cos18 \Leftrightarrow [/tex]
[tex]\Leftrightarrow \frac{10-2 \sqrt{5}}{64} =cos^218(1-cos^218) \Leftrightarrow cos^418-cos^218+ \frac{5- \sqrt{5}}{32}=0. \\ \\ Notam~cos^218=t\ \textgreater \ cos^230 = \frac{3}{4}. \\ \\ t^2-t+ \frac{5- \sqrt{5}}{32}=0. \\ \\ \Delta= \frac{3+ \sqrt{5}}{8} . \\ \\ t_{1,2}= \frac{1 \pm \sqrt{\frac{3 + \sqrt{5}}{8} }}{2};~tinand~cont~de~t\ \textgreater \ \frac{3}{4},~avem~t= \frac{5+ \sqrt{5}}{8}. \\ \\ Deci~\boxed{cos18= \sqrt{ \frac{5+ \sqrt{5}}{8} }= \frac{\sqrt{10+2 \sqrt{5}}}{4}=sin72}~. [/tex]
[tex]\underline{Pasul~3.~Aflarea~ariei.(Figura~2~si~Figura~3)} \\ \\ Pentagonul~fiind~regulat,~unghiurile~A,B,C,D,E~au~masura~de~ \\ \\ 108 \textdegree.~In~Figura~3~avem~ \Delta ABE,~si~ne~propunem~sa~aflam~lungimea \\ \\ segmentului~ST=s.~Notam~AB=k. \\ \\ A_{ABE}= \frac{AB \cdot AE \cdot sin 108}{2}= \frac{k^2 \cdot sin 72}{2}=1 \Rightarrow k= \sqrt{ \frac{2}{sin72} }= \frac{2 \sqrt{2}}{ \sqrt[4]{10+2 \sqrt{5}} } .[/tex]
[tex]Aplic~T.Cosinusului~in~ \Delta ATE~(n=AT=TE): \\ \\ cos ( \angle AET)=cos 36= \frac{n^2+k^2-n^2}{2kn} = \frac{k}{2n} \Rightarrow n= \frac{k( \sqrt{5}-1)}{2} . \\ \\ Avem:~s=SE-TE=k-n= k- \frac{k( \sqrt{5}-1)}{2}=k \cdot \frac{3- \sqrt{5}}{2}= \\ \\ = \frac{3 \sqrt{2}- \sqrt{10}}{ \sqrt[4]{10+ 2 \sqrt{5}} }.~Avand~lungimea~laturii~pentagonului~hasurat~regulat, \\ \\ ii~vom~putea~calcula~aria.~:) [/tex]
