Estude! Com Prof. Caju

Problema Proposto
27 - Na circunferência de raio R se inscreve
o hexágono regular ABCDEF e no arco BC se
marca o ponto N de modo que m[tex3]\overset{\LARGE{\frown}}{BC}[/tex3] = m[tex3]\overset{\LARGE{\frown}}{NC}[/tex3] . I é o incentro do triângulo AND.
EI [tex3]\cap [/tex3]AD= {Q} [tex3]\cap [/tex3]IF n AD= {P}.
Calcular PQ , se R = (2 + [tex3]\sqrt{3}[/tex3] +[tex3]\sqrt{6}[/tex3]) cm.

[tex3]\mathsf{DN=AN = l_4=R\sqrt2\\T.Incentro \triangle ANC:\\
\frac{IN}{IO}=\frac{DN+AN}{AD}\implies\\
\frac{R-x}{x}=\frac{R\sqrt2+R\sqrt2}{2R}\rightarrow R-x = \sqrt2x\\
x(\sqrt2+1) = R \rightarrow \boxed{x = R\sqrt2-R}\\
\triangle IOP \sim IEF:\\
\frac{x}{x+\frac{R\sqrt3}{2}} = \frac{PQ}{R} \implies PQ = \frac{2Rx}{2x+R\sqrt3}\\
Substituindo~x \implies
PQ = \frac{2R^2(R\sqrt2-R)}{2(R\sqrt2)-R)+R\sqrt3}=\frac{2R(\sqrt2-1)}{2\sqrt2-2+\sqrt3}=\\
Substituindo~R:: \frac{(2+\sqrt3+\sqrt6)(2\sqrt2-2)}{2\sqrt2-2+\sqrt3}=\frac{4\sqrt2-4+2\sqrt6-2\sqrt3+2\sqrt12-2\sqrt6}{2\sqrt2-2+\sqrt3}=\\
\frac{4\sqrt2-4-2\sqrt3+4\sqrt3}{2\sqrt2-2+\sqrt3}=2\frac{(\cancel{2\sqrt2-2+\sqrt3})}{\cancel{2\sqrt2-2+\sqrt3}\\
\therefore \boxed{\color{red}PQ = 2}}} [/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{DN=AN = l_4=R\sqrt2\\T.Incentro \triangle ANC:\\
\frac{IN}{IO}=\frac{DN+AN}{AD}\implies\\
\frac{R-x}{x}=\frac{R\sqrt2+R\sqrt2}{2R}\rightarrow R-x = \sqrt2x\\
x(\sqrt2+1) = R \rightarrow \boxed{x = R\sqrt2-R}\\
\triangle IOP \sim IEF:\\
\frac{x}{x+\frac{R\sqrt3}{2}} = \frac{PQ}{R} \implies PQ = \frac{2Rx}{2x+R\sqrt3}\\
Substituindo~x \implies
PQ = \frac{2R^2(R\sqrt2-R)}{2(R\sqrt2)-R)+R\sqrt3}=\frac{2R(\sqrt2-1)}{2\sqrt2-2+\sqrt3}=\\
Substituindo~R:: \frac{(2+\sqrt3+\sqrt6)(2\sqrt2-2)}{2\sqrt2-2+\sqrt3}=\frac{4\sqrt2-4+2\sqrt6-2\sqrt3+2\sqrt12-2\sqrt6}{2\sqrt2-2+\sqrt3}=\\
\frac{4\sqrt2-4-2\sqrt3+4\sqrt3}{2\sqrt2-2+\sqrt3}=2\frac{(\cancel{2\sqrt2-2+\sqrt3})}{\cancel{2\sqrt2-2+\sqrt3}\\
\therefore \boxed{\color{red}PQ = 2}}} [/tex3]

(Solução:geobson)