WakeUp wrote: The measure of the angles of a triangle are in arithmetic progression and the lengths of its altitudes are as well. Show that such a triangle is equilateral. Let $ABC$ be the triangle with $h_a, h_b, h_c$ as altitudes that satisfy, $A+C=2B;$ and $h_a+h_c=2h_b.$ From the first condition it is obvious that $\angle B=60^{\circ}.$ Using $h_a=\frac{2\Delta}{a}$ where $\Delta$ is the area of $ABC$, we get $ab+ca=2ca \iff b=\frac{2ca}{c+a}.$ But, we have $\cos B=\frac{a^2+c^2-b^2}{2ca}=\cos 60^{\circ}=\frac 12;$ Implying that $b^2+ca=c^2+a^2.$ But, note that $b=\frac{2ca}{c+a}\stackrel{\text{HM-GM}}{\leq}\sqrt{ca};$ So $c^2+a^2\geq 2ca\geq ca+b^2;$ ie equality must occur in all these inequalities. Therefore $a=b=c.\Box$

Suppose that: $\widehat{A} \le \widehat{B} \le \widehat{C};$ and because: "the angles of a triangle are in arithmetic progression and the lengths of its altitudes are as well" $\Rightarrow $ $\begin{cases} & \math{} \frac{\left (h_a + h_c \right )}{2}=h_b ;(1) \\ & \math{} \frac{\left (\widehat{A}+ \widehat{C} \right )}{2}=\widehat{B};(2)\\ & \math{} \widehat{A} + \widehat{B} + \widehat{C}=180^0 ;(3) \\ \end{cases}$ + From: $(2);(3):\Rightarrow \widehat{B}=60^0;$ + From:$ (1):\frac{\left (h_a + h_c \right )}{2}=h_b $ $\Rightarrow \frac{h_a}{h_b}+\frac{h_c}{h_b}=2\Rightarrow \frac{b}{a}+\frac{b}{c}=2$ $\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{b} \overset{AM-GM}{\rightarrow}\ge \frac{9}{a+b+c} ;(1')$ $\Rightarrow a+c \ge 2b\Rightarrow \boxed{\sin A +\sin C \ge \sqrt{3}};(b=2R\sin B;\widehat{B}=60^0); (*);$ Derived from the following $\forall \bigtriangleup ABC $, we have: $\sin A +\sin B +\sin C \le \frac{3\sqrt{3}}{2}\Rightarrow \boxed{\sin A +\sin C \le \sqrt{3}};(**)$ From $(*);(**):\Rightarrow \sin A +\sin C = \sqrt{3}\overset{(1')}{\rightarrow} a=b=c.$

Here 's mine: Let sides be $a-d,a,a+d$ and corresponding altitudes be $h-e,h,h+d$. Clearly $\dfrac{(a-d)(h-e)}{2} = \dfrac{ah}{2} = \dfrac{(a+d)(h+d)}{2} = \Delta$.This leads to following equalities : $ae + hd + de = 0 ; ae + hd - ed = 0$.Implies $de = 0$ and $ae + hd = 0$. Now take 2 cases : 1) $d = 0$ then putting in second equation $ae = 0$ since a is non zero hence $e = 0$. 2) $e=0$ then putting in second equation $hd=0$ since h is non zero hence $d=0$. Either way we get that $d,e = 0$ and Triangle is equilateral.

@above, the angles are in arithmetic progression in the problem, and not the sides.

Yea sorry the problem I did from another site had sides in AP. The same problem from the same contest IberoAmerican 1988 P-1 !!!