Triangles with Sides in Geometric Progression


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


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.


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:

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


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


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.


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

Fibonacci Numbers

  1. Ceva's Theorem: A Matter of Appreciation
  2. When the Counting Gets Tough, the Tough Count on Mathematics
  3. I. Sharygin's Problem of Criminal Ministers
  4. Single Pile Games
  5. Take-Away Games
  6. Number 8 Is Interesting
  7. Curry's Paradox
  8. A Problem in Checker-Jumping
  9. Fibonacci's Quickies
  10. Fibonacci Numbers in Equilateral Triangle
  11. Binet's Formula by Inducion
  12. Binet's Formula via Generating Functions
  13. Generating Functions from Recurrences
  14. Cassini's Identity
  15. Fibonacci Idendtities with Matrices
  16. GCD of Fibonacci Numbers
  17. Binet's Formula with Cosines
  18. Lame's Theorem - First Application of Fibonacci Numbers

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