Conditie echivalenta in triunghi

[opincariumihai]

Aratati ca un triunghi are un unghi de \( 30^0 \) daca si numai daca intre elementele sale exista relatia \( a^2+b^2+c^2-4\sqrt{3}S=2R^2. \)

Mihai Opincariu

[Virgil Nicula]

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

Asadar, \( 30^{\circ}\ \in\ \{\ A\ ,\ B\ ,\ C\ \}\ \Longleftrightarrow\ f\left(2-\sqrt 3\right)\ =\ 0\ \Longleftrightarrow\ \underline{\overline{\left\|\ p\ =\ R\ +\ \left(2+\sqrt 3\right)\ r\ \right\|}} \) .

Intr-adevar, \( t\equiv\tan 15^{\circ}=2-\sqrt 3 \) si \( t^2+1=4t \) si

\( pt(t^2+1)=4Rt^2+r (t^2+1)\Longleftrightarrow 4pt^2=4Rt^2+4rt\Longleftrightarrow pt=Rt+r\Longleftrightarrow p=R+r\left(2+\sqrt 3\right). \)

Se arata usor ca \( a^2+b^2+c^2-4\sqrt{3}S=2R^2\Longleftrightarrow p^2-2r\sqrt 3\cdot p-(R^2+4Rr+r^2)=0 \). Rezolvam ecuatia de gradul doi in raport cu variabila \( p \). Discriminantul redus este \( \Delta =(R+2r)^2 \) si singura radacina pozitiva este \( p=R+(2+\sqrt 3)r \).

Asadar avem \( a^2+b^2+c^2-4\sqrt{3}S=2R^2\ \Longleftrightarrow\ p=R+(2+\sqrt 3)\cdot r\ \Longleftrightarrow\ 30^{\circ}\in \{A,B,C\} \).

\( \odot \) Si acum o solutie foarte simpla pentru problema propusa la nivelul clasei a VII - a !

\( A=30^{\circ}\ \Longrightarrow\ \left\|\begin{array}{ccc}
a=2R\sin A & \Longrightarrow & a=R\\\\\\\\
bc\sin A=2S & \Longrightarrow & bc=4S\\\\\\\\
a^2=b^2+c^2-2bc\cdot\cos A & \Longrightarrow & a^2=b^2+c^2-bc\sqrt 3\end{array}\right\|\ \Longrightarrow \)

\( \underline {a^2+b^2+c^2-4S\sqrt 3}=R^2+\left(b^2+c^2-bc\sqrt 3\right)=R^2+a^2=\underline {2R^2} \) .

\( \odot \) Remarca. \( A\in\left(\ 0\ ,\ \frac {\pi}{6}\ \right]\ \Longrightarrow\ b^2+c^2\ \ge\ a^2+4S\sqrt 3 \) .

\( \odot \) Iata o problema similara: \( \underline{\overline{\left\|\ 60^{\circ}\in\{A,B,C\}\ \Longleftrightarrow\ a^2+b^2+c^2-4S\sqrt 3=2(R-2r)(3R+2r)\ \right\|}} \).

\( \odot \) Incercati o generalizare pastrand expresia \( a^2+b^2+c^2-4S\sqrt 3 \).

\( \odot \) Observatie. \( p=(p-a)+a=r\cdot\cot\frac A2+2R\sin A=\frac rx+\frac {4Rx}{1+x^2} \) , unde am notat \( x=\tan\frac A2 \) . Asadar,

\( p=\frac rx+\frac {4Rx}{1+x^2}\ \Longleftrightarrow\ px\left(1+x^2\right)=r\left(1+x^2\right)+4Rx^2\ \Longleftrightarrow\ \underline{\overline {\left\|\ p\cdot x^3-(4R+r)\cdot x^2+p\cdot x-r=0\ \right\|}} \) .

[Virgil Nicula]

\( \odot \) Incercati o generalizare pastrand expresia \( a^2+b^2+c^2-4S\sqrt 3 \) .

Iata si o generalizare care evidentiaza esential (la limita !) situatia particulara \( A=30^{\circ} \) :

\( \underline{\overline{\left\|\ a^2+b^2+c^2-4S\sqrt 3=8R^2\sin^2A+\left(\cos A-\sqrt 3\sin A\right)\left(\frac {4r^2}{1-\cos A}+8Rr\right)\ \right\|}} \).

[Mateescu Constantin]

Iata si o generalizare care evidentiaza esential (la limita !) situatia particulara \( A=30^{\circ} \) :

\( \underline{\overline{\left\|\ a^2+b^2+c^2-4S\sqrt 3=8R^2\sin^2A+\left(\cos A-\sqrt 3\sin A\right)\left(\frac {4r^2}{1-\cos A}+8Rr\right)\ \right\|}} \).

\( \left\|\ 8R^2\sin^2 A=8R^2\cdot\frac{a^2}{4R^2}=2a^2\ \right\|\ (1) \)

\( \left\|\ \cos A-\sqrt 3\sin A=\frac{b^2+c^2-a^2}{2bc}-\frac{2S\sqrt 3}{bc}=\frac{b^2+c^2-a^2-4S\sqrt 3}{2bc}\ \right\|\ (2) \)

\( \left\|\ \frac{4r^2}{1-\cos A}+8Rr=\frac{4\cdot\frac{S^2}{p^2}}{1-\frac{b^2+c^2-a^2}{2bc}}+8\cdot\frac{abc}{4S}\cdot \frac{S}{p}=\frac{\frac{4(p-a)(p-b)(p-c)}{p}}{\frac{a^2-(b-c)^2}{2bc}}+\frac{2abc}{p}\ \right\| \)

\( \Longleftrightarrow\ \left\|\ \frac{4r^2}{1-\cos A}+8Rr=\frac{4(p-a)(p-b)(p-c)}{p}\ \cdot\ \frac{2bc}{4(p-b)(p-c)}+\frac{2abc}{p}=2bc\ \right\|\ (3) \)

\( \Longrightarrow^{(1)\wedge (2)\wedge (3)}\ RHS=2a^2+b^2+c^2-a^2-4S\sqrt 3=a^2+b^2+c^2-4S\sqrt 3=LHS \)