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Fantezi Cebir => Analiz-Cebir => Konuyu başlatan: stuart clark - Mayıs 30, 2011, 04:01:14 öö
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Given that $a$, $b$, and $c$ are in geometric progression and the expressions$\ln(a) - \ln(2b)$, $\ln(2b) - \ln(3c)$, and $\ln(3c) - \ln(a)$ form an arithmetic progression, and also that $a$, $b$, and $c$ are the sides of a triangle, determine the type of triangle $ABC$.
Options:
(a) acute triangle
(b) obtuse triangle
(c) right triangle
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Let [tex]\mathbf{A=ln(a)-ln(2b),A+D=ln(2b)-ln(3c)}[/tex] and [tex]\mathbf{A+2D=ln(3c)-ln(a)}[/tex]
Now [tex]\mathbf{3(A+D)=ln(a)-ln(2b)+ln(2b)-ln(3c)+ln(3c)-ln(a)=0}[/tex]
So [tex]\mathbf{3.(A+D)=0\Leftrightarrow A+D=0\Leftrightarrow ln(2b)-ln(3c)=0\Leftrightarrow ln\left(\frac{2b}{3c}\right)=0\Leftrightarrow \frac{2b}{3c}=1}[/tex]
So [tex]\mathbf{c=\frac{2}{3}b}[/tex] and [tex]\mathbf{b^2=ac}[/tex] (bcz [tex]\mathbf{a,b,c}[/tex] are in [tex]\mathbf{G.P}[/tex])
So [tex]\mathbf{b=\frac{2}{3}a}[/tex] and [tex]\mathbf{c=\frac{4}{9}a}[/tex]
So [tex]\mathbf{a>b>c}[/tex]
and [tex]\mathbf{b^2+c^2=\left(\frac{2}{3}\right)^2.a^2+\left(\frac{2}{3}\right)^4.a^2=\frac{52}{81}a^2<a^2}[/tex]
So [tex]\mathbf{b^2+c^2-a^2<0}[/tex]
Now Using Cosine formula [tex]\mathbf{cos\;A=\frac{b^2+c^2-a^2}{2.b.c}<0}[/tex]
So [tex]\mathbf{A}[/tex] is an Obtuse angle.
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Solution: Since a, b, c are geometric progression, then b2 = ac. By arithmetic progression and properties of the logaritm function: 2.Ln(2b/3c) = Ln(3ac/2ab) and therefore c = 2b/3. Hence we yields: a = 3b/2. a2 > b2 + c2. So, A > 90o.
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Thanks scarface.