Fie patrulaterul convex inscriptibil \( ABCD \) . Notam \( AB=a \) , \( BC=b \) , \( CD=c \) , \( DA=d \) si
\( p=\frac {a+b+c+d}{2} \) . Aratati ca \( a+b=c+d\ \Longleftrightarrow\ \frac {ab}{cd}=\tan^2\frac B2 \)si relatia \( \frac {p+a}{p-a}\cdot\frac {p+b}{p-b}\cdot\frac {p+c}{p-c}\cdot\frac {p+d}{p-d}\ \ge\ 81 \) .
Inegalitate intr-un patrulater inscriptibil.
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Virgil Nicula
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Mateescu Constantin
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Avem \( \tan^2 \frac B2=\frac{1-\cos B}{1+\cos B},\ B\in\left\(0,\ \pi\right\) \) si cum \( \cos B=-\cos D\ (\angle B+\angle D=180^{\circ}) \)
\( \Longrightarrow \tan^2 \frac B2=\frac{1+\cos D}{1+\cos B} \)
Din teorema cosinusului aplicata in \( \triangle ACD \) si \( \triangle ABC \) avem:
\( \cos D=\frac{c^2+d^2-AD^2}{2cd},\ \cos B=\frac{a^2+b^2-AD^2}{2cd} \)
\( \Longrightarrow \tan^2 \frac B2=\frac{ab}{cd}\ \cdot\ \frac{(c+d)^2-AC^2}{(a+b)^2-AC^2} \)
\( \tan^2\frac B2=\frac{ab}{cd}\ \Longleftrightarrow \frac{ab}{cd}\ \cdot\ \frac{(c+d)^2-AC^2}{(a+b)^2-AC^2}=\frac{ab}{cd}\ \Longleftrightarrow a+b=c+d \)
\( (p-a)(p-b)(p-c)\le \left\(\frac{p-a+p-b+p-c}{3}\right\)^3 \)
\( \Longleftrightarrow (p-a)(p-b)(p-c)\le \left\(\frac{p+d}{3}\right\)^3 \)
Analog obtinem si \( (p-a)(p-b)(p-d)\le \left\(\frac{p+c}{3}\right\)^3,\ (p-a)(p-c)(p-d)\le \left\(\frac{p+b}{3}\right\)^3,\ (p-b)(p-c)(p-d)\le \left\(\frac{p+a}{3}\right\)^3 \)
Prin inmultirea celor 4 inegalitati se obtine:
\( \frac{(p+a)^3(p+b)^3(p+c)^3(p+d)^3}{3^{12}}\ge (p-a)^3(p-b)^3(p-c)^3(p-d)^3 \)
\( \Longleftrightarrow \frac{(p+a)(p+b)(p+c)(p+d)}{(p-a)(p-b)(p-c)(p-d)}\ge 81 \), care se mai poate scrie \( \frac{(p+a)(p+b)(p+c)(p+d)}{S^2}\ge 81 \), tinand cont ca \( ABCD \) este inscriptibil.
\( \Longrightarrow \tan^2 \frac B2=\frac{1+\cos D}{1+\cos B} \)
Din teorema cosinusului aplicata in \( \triangle ACD \) si \( \triangle ABC \) avem:
\( \cos D=\frac{c^2+d^2-AD^2}{2cd},\ \cos B=\frac{a^2+b^2-AD^2}{2cd} \)
\( \Longrightarrow \tan^2 \frac B2=\frac{ab}{cd}\ \cdot\ \frac{(c+d)^2-AC^2}{(a+b)^2-AC^2} \)
\( \tan^2\frac B2=\frac{ab}{cd}\ \Longleftrightarrow \frac{ab}{cd}\ \cdot\ \frac{(c+d)^2-AC^2}{(a+b)^2-AC^2}=\frac{ab}{cd}\ \Longleftrightarrow a+b=c+d \)
Din \( AM-GM \) avem:Virgil Nicula wrote:\( \frac {p+a}{p-a}\cdot\frac {p+b}{p-b}\cdot\frac {p+c}{p-c}\cdot\frac {p+d}{p-d}\ \ge\ 81 \)
\( (p-a)(p-b)(p-c)\le \left\(\frac{p-a+p-b+p-c}{3}\right\)^3 \)
\( \Longleftrightarrow (p-a)(p-b)(p-c)\le \left\(\frac{p+d}{3}\right\)^3 \)
Analog obtinem si \( (p-a)(p-b)(p-d)\le \left\(\frac{p+c}{3}\right\)^3,\ (p-a)(p-c)(p-d)\le \left\(\frac{p+b}{3}\right\)^3,\ (p-b)(p-c)(p-d)\le \left\(\frac{p+a}{3}\right\)^3 \)
Prin inmultirea celor 4 inegalitati se obtine:
\( \frac{(p+a)^3(p+b)^3(p+c)^3(p+d)^3}{3^{12}}\ge (p-a)^3(p-b)^3(p-c)^3(p-d)^3 \)
\( \Longleftrightarrow \frac{(p+a)(p+b)(p+c)(p+d)}{(p-a)(p-b)(p-c)(p-d)}\ge 81 \), care se mai poate scrie \( \frac{(p+a)(p+b)(p+c)(p+d)}{S^2}\ge 81 \), tinand cont ca \( ABCD \) este inscriptibil.