toedo wrote: Given triangle $ABC$ such that angles $A$, $B$, $C$ satisfy \[ \frac{\cos A}{20}+\frac{\cos B}{21}+\frac{\cos C}{29}=\frac{29}{420} \]Prove that $ABC$ is right angled triangle Here is a rather ugly proof : Let $t=\arcsin\frac{21}{29}$ so that $\sin t=\frac{21}{29}$ and $\cos t=\frac{20}{29}$ Writing $C=\pi-A-B$, equation is $\frac{\cos A}{\cos t}+\frac{\cos B}{\sin t}-\cos (A+B)=\frac 1{\sin t\cos t}$ Which is $\left(\frac 1{\sin t}-\cos A\right)\left(\frac 1{\cos t}-\cos B\right)=\sin A\sin B$ We get that $\sin A\sin B\ne 0$ and so that : $\frac 1{\sin t}-\cos A=\alpha\sin A$ and $\frac 1{\cos t}-\cos B=\frac 1{\alpha}\sin B$ for some $\alpha>0$ Writing $u=\arcsin\frac 1{\sqrt{\alpha^2+1}}$, this becomes : $\frac{\sin u}{\sin t}=\sin(A+u)$ and $\frac{\cos u}{\cos t}=\cos(B-u)$ So $0<\sin u\le\sin t$ and $0<\cos u\le\cos t$ which implies $u=t$ And so $\boxed{(A,B,C)=\left(\frac{\pi}2-t,t,\frac{\pi}2\right)}$ Q.E.D.

Thanks pco. Maybe any other approach using some geometric construction?

$\blacklozenge$ This follows from the well-known Wolstenholme’s Inequality: Let $x,y,z$ be real numbers and a triangle $\triangle ABC$, we have the following inequality \[x^2+y^2+z^2 \ge 2yz\cos A+2zx \cos B+2xy \cos C.\]Equality holds when $x:y:z=\sin A:\sin B:\sin C$, in other words $x,y,z$ are side lengths of $\triangle ABC$. $\spadesuit \color{blue}{\textit{Proof:}}$ Using $\angle C=\pi - \angle A-\angle B$, we have \[x^2+y^2+z^2-(2yz\cos A+2zx \cos B+2xy \cos C)=(z-(x\cos B+y\cos A))^2+(x\sin B-y\sin A)^2\ge 0. \quad \square\]$\clubsuit$ Back to our problem, by setting $x=29, y=21, z=20$ in the Wolstenholme’s Inequality, we see that \[\frac{\cos A}{20}+\frac{\cos B}{21}+\frac{\cos C}{29} \le \frac{29}{420}\]and since we have equality in our problem, $\triangle ABC$ has lengths $29-21-20$ which is a right angled triangle. $\quad \blacksquare$

Here's a generalization of the above Inequality: $\spadesuit$ For a $\triangle ABC$ and real numbers $x,y,z$ and a positive integer $n$, we have \[x^2+y^2+z^2 \ge 2(-1)^{n+1}(yz \cos nA+zx\cos nB+xy\cos nC).\]

Keith50 wrote: Here's a generalization of the above Inequality: $\spadesuit$ For a $\triangle ABC$ and real numbers $x,y,z$ and a positive integer $n$, we have \[x^2+y^2+z^2 \ge 2(-1)^{n+1}(yz \cos nA+zx\cos nB+xy\cos nC).\] Could you give the proof?