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If $a, b, c$ are in geometric progression and $\log a -\log 2b, \log 2b -\log 3c, \log 3c-\log a$ are in AP. Find the type of the triangle, if it’s sides are $a, b, c$

From the given data

$$b^2=ac$$ and $$2(\log 2b-\log 3c)=\log 3c-\log a +\log a -\log 2b$$

$$\log 2b=\log 3c$$ $$2b=3c$$

How does this tell us the type of the triangle? The relations between the sides is $$2b=3c$$and $$4a=9c$$

$$2b=3c=\frac{4a}{3}$$

Aditya
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2 Answers2

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WLOG $\dfrac cb=\dfrac ba=r, b=ar; c=ar^2$

$$\log a-\log2b=\log\dfrac a{2b}=\log\dfrac1{2r}$$

Similarly, $$\log2b-\log3c=\cdots=\log\dfrac{2}{3r}$$

$$\log3c-\log a=\cdots=\log(3r^2)$$

$$\implies\log\dfrac1{2r}+\log(3r^2)=2\cdot\log\dfrac{2}{3r}$$

$$\dfrac{3r^2}{2r}=\left(\dfrac{2}{3r}\right)^2\implies r=\dfrac23$$ as $r$ is real

Now use Law of cosines to find whether it has an obtuse or a right angle

lab bhattacharjee
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  • I think the answer key has a problem here, but I got an acute triangle (taking $\frac{2a}{3}$ to be the largest side). The given answer is obtuse. Could you please verify this? – Aditya Apr 29 '20 at 16:40
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$a^2+b^2 = a^2 + (2\cdot\frac{3}{4} a)^2 = \frac{13}{4}a^2$ while $c^2 = (\frac{4}{9}a)^2 = \frac{16}{81}a^2$. Since $a^2+b^2 > c^2$, the triangle is acute from the Pythagorean inequality theorem.

Note: This inequality does not necessarily come from the cosine rule: a triangle can be continuously deformed from an acute triangle to a right triangle, and then to an obtuse triangle. Since $a^2+b^2 \ne c^2$ in this problem, the triangle must be either acute or obtuse.

When $a=b=c=1$, we have an acute, equilateral triangle, but $1^2+1^2 > 1^2$, so when $a^2+b^2 > c^2$ the triangle must be acute. See more on this post.

Toby Mak
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  • For all you know, $a$ could have been $b$, $b$ could have been $a$. What I'm implying is shouldn't we check all 3 choose 2 combinations of sums for given $a, b$ and $c$? – Rew Apr 29 '20 at 12:49