# Triangles with Sides in Geometric Progression

### Problem

A triangle has sides in geometric progression. What could be the range of the factor of the progression?

### Hint

The only condition imposed on three successive terms of a geometric progressions is that they may serve the sides of a triangle. In all likelihood, the triangle inequality may play an important role in solving the problem.

### Solution

For some $q,a\gt 0,$ the three numbers in a geometric progression could be written as $a,$ $aq,$ and $aq^2.$ It is obvious that a specific value of $a$ is of no consequence in the problem, reflecting the fact that similar triangles contribute the same value of $q$ to the solution of the problem.

The sought values of $q$ then should satisfy three inequalities:

$q^{2}\lt 1+q,$

$q \lt 1+q^{2},$ and

$1 \lt q+q^{2}.$

The first one is equivalent to $q^{2}-1-q\lt 0,$ in which the quadratic polynomial on the left has two roots: $-1/\phi$ and $\phi,$ where $\displaystyle\phi=\frac{1+\sqrt{5}}{2}$ is the golden ratio. Since we are only interested in positive solutions of the inequality, we find that $q$ should satisfy $0\lt q\lt\phi.$

The second inequality is equivalent to $q^{2}-q+1\gt 0,$ i.e., $(q-\frac{1}{2})^{2}+\frac{3}{4}\gt 0,$ which is always true.

The third inequality is equivalent to $q^{2}+q-1\gt 0,$ which, for $q\gt 0,$ is satisfied when $\displaystyle q\gt\frac{1}{\phi}=\frac{-1+\sqrt{5}}{2}.$

Combining what we found about the possible values of $q$ that satisfy all three inequalities gives the answer:

$\displaystyle\frac{1}{\phi}\lt q\lt\phi.$

### Acknowledgment

This problem is an abbreviation of a problem from a 2009 Chinese Olympiad which reads:

Suppose that the sides $a,$ $b,$ $c$ of $\Delta ABC,$ corresponding to the angles $A,$ $B,$ $C,$ constitute a geometric series. Find the range of

$\displaystyle\frac{\mbox{sin}(A)\mbox{cot}(C)+\mbox{cos}(A)}{\mbox{sin}(B)\mbox{cot}(C)+\mbox{cos}(B)}.$

To solve the problem one had to demonstrate the knowledge of basic trigonometry:

$\begin{align}\displaystyle &\frac{\mbox{sin}(A)\mbox{cot}(C)+\mbox{cos}(A)}{\mbox{sin}(B)\mbox{cot}(C)+\mbox{cos}(B)} =\\ &\frac{\mbox{sin}(A)\mbox{cos}(C)+\mbox{cos}(A)\mbox{sin}(C)}{\mbox{sin}(B)\mbox{cos}(C)+\mbox{cos}(B)\mbox{sin}(C)} =\\ &\frac{\mbox{sin}(A+C)}{\mbox{sin}(B+C)} =\\ &\frac{\mbox{sin}(180^{\circ}-B)}{\mbox{sin}(180^{\circ}-A)} =\\ &\frac{\mbox{sin}(B)}{\mbox{sin}(A)} = \frac{b}{a} = q, \end{align} $

which reduces the problem to solving the triangle inequalities as above.

### References

- Xiong Bin, Lee Peng Yee,
*Mathematical Olympiads in China (2009-2010). Problems and Solutions*, World Scientific, 2013, pp 7-8

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