Progressive Triangles

The sides are in geometric progression; say they're $\frac{a}{r} ,a, ar$ .

The angles are in arithmetic progression, and sum to $180°$ . Note this means the angle opposite the side of length $a$ must be $60°$ .

By the cosine rule, $a^2=\frac{a^2}{r^2}+a^2 r^2 - 2a^2 \cos{60}$ .

Dividing by $a^2$ and substituting $\cos 60=\frac12$ , we get $1=\frac{1}{r^2}+r^2-1$ , or $\left(r-\frac{1}{r} \right)^2=0$ . Hence $r=1$ is the only solution, that is the only such triangle is the unit equilateral triangle.

Alternative (longer) solution below:

Let the sides of the triangle be $a \le b \le c$ and the angles opposite these sides be $A\le B\le C$ .

Since the angles are in arithmetic progression, and they sum to $180°$ , they must be $60-\theta,60,60+\theta$ for some non-negative $\theta<60$ .

Since the sides are in geometric progression, they satisfy $\frac{a}{b}=\frac{b}{c}$ .

By the sine rule, $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ . With a little rearrangement and cancellation, this becomes $\sin{(60-\theta)} \sin{(60+\theta)}=\sin^2 60=\frac34$ .

Expanding, $\begin{aligned} \sin{(60-\theta)} \sin{(60+\theta)} &=(\sin{60} \cos{\theta}-\sin{\theta} \cos{60})(\sin{60} \cos{\theta}+\sin{\theta} \cos{60}) \\ &=\sin^2 {60} \cos^2 {\theta}-\sin^2 {\theta} \cos^2 {60} \\ &=\frac14 \left( 3\cos^2 {\theta}- \sin^2 {\theta} \right) \\ &=\frac14 (1+2\cos {2\theta})\\ &=\frac34 \end{aligned}$

so that finally $\cos{2\theta}=1$ and the unique solution is $\theta=0$ corresponding to an equilateral triangle; from the remaining given condition, the only such triangle is the unit equilateral triangle.

Let $ABC$ be the triangle with $a=BC$ , $b=CA$ and $c=AB$ and $\angle A \le \angle B \le \angle C$ . Since the angles are in arithmetic progression, we have

$\angle B = \frac{{\angle A + \angle C}}{2} = \frac{{180^\circ - \angle B}}{2} \Rightarrow \angle B = 60^\circ$ .

Now, the relation $\angle A \le \angle B \le \angle C$ yields $a \le b \le c$ and since the sides are in geometric progression, we get

$\begin{array}{l} {b^2} = a \cdot c \Rightarrow {\sin ^2}B = \sin A \cdot \sin C & {\color{#20A900}{\text{(by sine rule)}}}\\ \Rightarrow {\sin ^2}60^\circ = \sin A \cdot \sin C\\ \Rightarrow \sin A \cdot \sin C = \frac{3}{4}\\ \Rightarrow \cos \left( {A - C} \right) - \cos \left( {A + C} \right) = \frac{3}{2}\\ \Rightarrow \cos \left( {A - C} \right) + \cos B = \frac{3}{2}\\ \Rightarrow \cos \left( {A - C} \right) = \frac{3}{2} - \cos 60^\circ \\ \Rightarrow \cos \left( {A - C} \right) = 1\\ \Rightarrow \angle \left( {A - C} \right) = 0 & {\color{#20A900}{\text{(since }} - 180^\circ < \angle \left( {A - C} \right) < 180^\circ )}\\ \Rightarrow \angle A = \angle C \end{array}$

This result leads to a unique solution, namely the (unit side) equilateral triangle.

The correct answer is ${\color{#D61F06}{\text{one}}}$ .

Since the angles are in an arithmetic progression, we can label them $a - d$ , $a$ , and $a + d$ . The angle sum of a triangle is $180°$ , so $(a - d) + a + (a + d) = 180°$ , which solves to $a = 60°$ , its medium angle.

Since the sides are in a geometric progression, we can label them $\frac{x}{r}$ , $x$ , and $rx$ .

The medium side $x$ must be across its medium angle $a = 60°$ , so by the law of cosines, $x^2 = (\frac{x}{r})^2 + (rx)^2 - \frac{1}{2} \cdot \frac{x}{r} \cdot rx \cdot \cos 60°$ , which rearranges to $0 = (r^2 - 1)^2$ and has one real solution $r = 1$ .

Since $r = 1$ , the three sides must all be the same, so it must be an equilateral triangle (and therefore $d = 0$ ), and there is only one possible equilateral triangle with at least one unit side.