If $\frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C}=2$ , then $\triangle ABC$ is a right triangle

Question

Show that the triangle whose angles satisfy the equality $$\frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C}=2$$ is right-angled.

I've tried many times, but was unsuccessful.

Answer 1

Hint: Show that \begin{eqnarray*} 1-\cos^2 A-\cos^2 B-\cos^2C+ 2\cos A \cos B \cos C =0 \end{eqnarray*} for any triangle.

Answer 2

COMMENT.- First note that if $C=90^{\circ}$ then you have $\dfrac{1+1}{1+0}=2$. To verify that necessarily your equality implies $C=90^{\circ}$ you can consider that $\sin C=\sin (180^{\circ}-(A+B))=\sin(A+B)$ and $\cos C= \cos (180^{\circ}-(A+B))=-\cos(A+B)$ so you get $$\sin^2 A+\sin^2 B+\sin^2 (A+B)=2(\cos^2A+\cos ^2 B+\cos^2(A+B))$$ You have to know simple formulas of trigonometry to finish.

EDITION.-Sorry, dear friend, your problem is not as simple as I had thought. But you can stay in the same direction as suggested. This way you get both $$\sin^2 A+\sin^2 B+\sin^2 (A+B) = 2\qquad (*)$$ and the similar with cosines what gives $1$ instead of $2$. Taking $(*)$ you can try to prove this equality is verified if and only if when $A + B =\dfrac{\pi}{2}$ (the "if" is obvious and we need the "only if"). So put $A + B = \dfrac{\pi}{2}\pm h$ with $h\ne 0$ and look at the functions $$\sin^2(x)+\sin^2(\dfrac{\pi}{2}\pm h-x)+\sin^2(\dfrac{\pi}{2}\pm h),\quad0\lt x\lt\dfrac{\pi}{2}$$ which becomes for $h$ positive and $h$ negative respectively $$f_1(x)=\sin^2(x)+\cos^2(x-h)+\cos^2(h)\qquad (1)\\f_2(x)=\sin^2(x)+\cos^2(x+h)+\cos^2(h)\qquad (2)$$ Now you can prove that $$\begin{equation}f_1(x)\begin{cases}\lt2\text{ when } 0\lt x\lt h\\=2 \text { when }x=h \space\text {discarded because } B\ne\dfrac{\pi}{2} \\\gt 2\text { when }h\lt x\lt\dfrac{\pi}{2}\\ \end{cases}\end{equation}$$

and that $f_2(x)$ is smaller than $2$ on $0\lt x\lt \dfrac{\pi}{2}$.

This requires elementary calculus that I guess you know how to handle.

Answer 3

\begin{align} \frac{\sin^2 A + \sin^2 B + \sin^2 C}{\cos^2 A + \cos^2 B + \cos^2 C} &=2 . \end{align}

There are known identities for any $\triangle ABC$ with sides $a,b,c$, angles $A,B,C$ radius of inscribed circle $r$, radius of circumscribed circle $R$ and semiperimeter $\rho=\tfrac12(a+b+c)$:

\begin{align} a^2+b^2+c^2 &= 2\rho^2-8r\,R-2r^2 ,\\ \cos A \cos B \cos C &=\frac{\rho^2-(r+2R)^2}{4R^2} . \end{align}

\begin{align} {\sin^2 A + \sin^2 B + \sin^2 C} &=2(\cos^2 A + \cos^2 B + \cos^2 C) ,\\ {\sin^2 A + \sin^2 B + \sin^2 C} &=2(1-\sin^2 A + 1-\sin^2 B + 1-\sin^2 C) ,\\ 3(\sin^2 A + \sin^2 B + \sin^2 C) &=6 ,\\ \sin^2 A + \sin^2 B + \sin^2 C &=2 ,\\ 4R^2\sin^2 A + 4R^2\sin^2 B + 4R^2\sin^2 C &=2\cdot4R^2 ,\\ a^2+b^2+c^2&=8R^2 ,\\ 2\rho^2-8r\,R-2r^2&=8R^2 ,\\ \rho^2-(r+2R)^2&=0 ,\\ \frac{\rho^2-(r+2R)^2}{4R^2} =0 . \end{align}

And since

\begin{align} \cos A \cos B \cos C &=\frac{\rho^2-(r+2R)^2}{4R^2} , \end{align}

one of $\cos A,\cos B,\cos C$ must be 0, hence, one of the angles must be $\tfrac\pi2$.

Answer 4

I know I am coming two years late, but I happened to find a rather nice geometric solution to this problem, so I am putting it here.

First off, reduce the problem to the more "user-friendly" equivalent

$$\Sigma \sin^2 A = 2$$

Using the law of sines, this becomes

$$a^2 + b^2 + c^2 = 8R^2$$

where $R$ is the circumradius of the triangle. By comparing this second equation to the law of cosines, we get

$$c^2 + 2ab \cos C = 4R^2$$

Since $c^2 \le 4R^2$, this means that the triangle must be acute. (This step may not actually be required -- I haven't checked.)

Now we perform the usual trick: inscribe $\triangle ABC$ in a circle of radius $\frac{1}{2}$ (this is purely for bookkeeping purposes -- the argument would be quite similar without this step), so that the statement becomes

$$AB^2 + BC^2 + CA^2 = 2$$

Now let (WLOG) $AB \ge$ other two sides.

Unfortunately, I am not fully familiar with methods of including diagrams, so this is going to be a little painful. Anyway, put $C'$ on the circle so that $CC'$ is a diameter (then, since $\triangle ABC$ is acute, $C'$ and $C$ will lie on opposite sides of $AB$). Then we get a Pythagorean relation free (recall that the diameter is 1):

$$BC^2 + CC'^2 + C'B^2 = 2$$ $$\implies AB^2 + CA^2 = CC'^2 + C'B^2$$

But since $CC'^2 = CA^2 + AC'^2$, this becomes

$$AB^2 = AC'^2 + C'B^2$$

This is Pythagoras again, so $\angle AC'B = 90$, and since the quadrilateral $ACBC'$ is cyclic, that leads us straight to $\angle ACB = 90$ as desired.