Art of Problem Solving

AoPS Online

MIE 2015/Day 1/Problem 6

Revision as of 11:51, 11 January 2018 by Alevini98 (talk | contribs) (Added solution)

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem 6

Let $(a,b,c)$ be a geometric progression and $\left(\log\left(\frac{5c}{a}\right),\log\left(\frac{3b}{5c}\right),\log\left(\frac{a}{3b}\right)\right)$ be an arithmetic progression, both in these order, so we can say that $a$ , $b$ and $c$ :

(a) are the sides of an obtusangle triangle.

(b) are the sides of an acutangle triangle that's not equilateral.

(c) are the sides of an equilateral triangle.

(d) are the sides of a right triangle.

(e) can't be the sides of a triangle.

Solution

$(a,b,c)\implies\boxed{b^2=ac}$

$\left(\log\frac{5c}{a},\log\frac{3b}{5c},\log\frac{a}{3b}\right)\implies\log\frac{3b}{5c}-\log\frac{5c}{a}=\log\frac{a}{3b}-\log\frac{3b}{5c}$

$\log\frac{3ab}{25c^2}=\log\frac{5ac}{9b^2}$

$\frac{3b}{25c^2}=\frac{5c}{9b^2}$

$3^3b^3=5^3c^3$

$\boxed{3b=5c}$

$3^2b^2=5^2c^2$

$3^2ac=5^2c^2$

$\boxed{9a=25c}$

$9a=25c$

$9a=5\cdot3b$

$\boxed{3a=5b}$

So we got these three relations

$\begin{cases}3b=5c\\9a=25c\\3a=5b\end{cases}$

By these equations we can see that $a > b > c$ , so it can't be an equilateral triangle.

First, we must see if $a$ , $b$ and $c$ are the sides of a tringle.

$|a-b| < c < a+b$

$\left|a-\frac{3}{5}a\right| < \frac{9}{25}a < a+\frac{3}{5}a$

$\frac{2}{5} < \frac{9}{25} < \frac{7}{5}$

So, $a$ , $b$ and $c$ can't be the sides of a triangle. $\boxed{\text{E}}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=MIE_2015/Day_1/Problem_6&oldid=89667"

© 2024 AoPS Incorporated