A=60 <=> IH=IO

[Virgil Nicula]

Sa se arate ca in \( \triangle ABC \) avem echivalenta \( \overline{\underline{\left\|\ A=60^{\circ}\ \Longleftrightarrow\ IH=IO\ \right\|}} \) si in acest caz \( \underline{\overline{\left\|\ \sin (\widehat {IOH}) = \frac 12\cdot\sqrt {1 - \frac {2r}{R}}\ \right\|}} \) .

[Mateescu Constantin]

\( \overline{\underline{\left\|\ A=60^{\circ}\ \Longleftrightarrow\ IH=IO\ \right\|}} \)

Ecuatia \( f(x)= p\cdot x^3-(4R+r)\cdot x^2+p\cdot x-r=0 \) are solutiile \( \left\{\ \tan\frac A2\ ,\ \tan\frac B2\ ,\ \tan\frac C2\ \right\} \).

Asadar, \( A=60^{\circ}\ \Longleftrightarrow\ f\left(\frac{\sqrt 3}{3}\right)\ =\ 0\ \Longleftrightarrow\ \overline{\underline{\left\|{\ p=\sqrt{3}(R+r)\ \right\| \)

Dar in orice triunghi sunt adevarate relatiile \( \overline{\underline{\left\|\begin{array}{cc}
OI^2 & = & R^2-2Rr \\\\\\\
IH^2 & = & 4R(R+r)+3r^2-p^2\end{array}\right\| \) .

\( \Longrightarrow\ IH=IO\ \Longleftrightarrow\ R^2-2Rr=4R(R+r)+3r^2-p^2 \)

\( \Longleftrightarrow\ p^2=3R^2+6Rr+3r^2\ \Longleftrightarrow\ \overline{\underline{\left\|{\ p=\sqrt{3}(R+r)\ \right\| \)

Prin urmare avem \( \overline{\underline{\left\|\ A=60^{\circ}\ \Longleftrightarrow\ IH=IO\ \right\|}} \).

\( \underline{\overline{\left\|\ \sin (\widehat {IOH}) = \frac 12\cdot\sqrt {1 - \frac {2r}{R}}\ \right\|}} \)

Avem \( \angle BHC=\angle BIC=\angle BOC=120^{\circ} \), adica punctele \( B,\ H,\ I,\ O,\ C \) sunt conciclice \( (*) \)

Din teorema sinusurilor aplicata in \( \triangle IBO\ \Longrightarrow\ \frac{IO}{\sin(\angle IBO)}=\frac{OB}{\sin(\angle BIO)} \)

\( \Longleftrightarrow\ \sin(\angle IBO)=^{(*)} \sin(\angle IHO)=\sin(\angle IOH)=\frac{\sqrt{R^2-2Rr}\cdot\sin(\angle BIO)}{R} \)

\( \Longleftrightarrow^{(*)}\ \sin(\angle IOH)=\frac{\sqrt{R^2-2Rr}\cdot\sin(\angle BCO)}{R}=\frac 12\cdot \frac{\sqrt{R^2-2Rr}}{R} \)

\( \Longleftrightarrow\ \underline{\overline{\left\|\ \sin (\angle {IOH}) = \frac 12\cdot\sqrt {1 - \frac {2r}{R}}\ \right\|}} \)