Difference between revisions of "MIE 2015/Day 1/Problem 6"

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===Problem 6===
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==Problem 6==
 
Let <math>(a,b,c)</math> be a geometric progression and <math>\left(\log\left(\frac{5c}{a}\right),\log\left(\frac{3b}{5c}\right),\log\left(\frac{a}{3b}\right)\right)</math> be an arithmetic progression, both in these order, so we can say that <math>a</math>, <math>b</math> and <math>c</math>:
 
Let <math>(a,b,c)</math> be a geometric progression and <math>\left(\log\left(\frac{5c}{a}\right),\log\left(\frac{3b}{5c}\right),\log\left(\frac{a}{3b}\right)\right)</math> be an arithmetic progression, both in these order, so we can say that <math>a</math>, <math>b</math> and <math>c</math>:
  
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===Solution===
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==Solution==
 
<math>(a,b,c)\implies\boxed{b^2=ac}</math>
 
<math>(a,b,c)\implies\boxed{b^2=ac}</math>
  
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So, <math>a</math>, <math>b</math> and <math>c</math> can't be the sides of a triangle. <math>\boxed{\text{E}}</math>
 
So, <math>a</math>, <math>b</math> and <math>c</math> can't be the sides of a triangle. <math>\boxed{\text{E}}</math>
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==See also==
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*[[Military Institute of Engineering]]
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*[[MIE 2016]]

Latest revision as of 11:54, 11 January 2018

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. $\boxed{\text{E}}$


See also

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