Prove or disprove the implication:
$a^2\cdot \tan(B-C)+ b^2\cdot \tan(C-A)+ c^2\cdot \tan(A-B)=0 \implies$ $ ABC$ is an isosceles triangle.
I tried to break down the left hand side in factors, but all efforts were in vain. Does anyone have a suggestion?
Thank you very much!
I will follow the method suggested by Blue in his comment to show that the implication does not hold.
The sides $a$, $b$, $c$ are proportional to $\sin A$, $\sin B$ and $\sin C$, respectively, by the sine law, so the equation is equivalent to $S = 0$, where $$ S = \sin^2 A \tan(B-C) + \sin^2 B \tan(C-A) + \sin^2 C \tan(A - B).$$ Letting $x = \tan A$, $y = \tan B$, and $z = \tan C$, we can rewrite $S$ in terms of $x$, $y$, $z$; for example, we have $\sin^2 A = x^2/(1 + x^2)$ and $\tan(B - C) = (y-z)/(1 + yz)$. (We will ignore for now the special case where $ABC$ is a right triangle, and one of $x$, $y$, $z$, is correspondingly infinite.) Thus $$ S = \frac{x^2}{1 + x^2} \frac{y-z}{1 + yz} + \frac{y^2}{1 + y^2} \frac{z-x}{1 + zx} + \frac{z^2}{1 + z^2} \frac{x-y}{1 + xy.}$$ Now we note that $S = 0$ when any two of $x, y, z$ are equal, so we expect the numerator of $S$ to be divisible by $(y-z)(z-x)(x-y)$. This is indeed the case (I used a computer for this part, although it is possible to do long division first by $y-z$, then by $z-x$ and finally by $x-y$): $$S = \frac{(y-z)(z-x)(x-y)(2x^2 y^2 z^2 + x^2 y^2 + y^2 z^2 + z^2 x^2 - 1)}{(1 + yz)(1 + zx)(1 + xy)(1 + x^2)(1 + y^2)(1 + z^2)}.$$ Now that we have recovered the factors $y-z$, $z-x$ and $x - y$, we can split off factors equal to the tangents that originally appeared in $S$, so that $$S = \frac{y-z}{1 + yz} \frac{z-x}{1 + zx} \frac{x-y}{1 + xy} T = \tan(B-C)\tan(C-A)\tan(A-B) T,$$ where $$ T = \frac{2x^2 y^2 z^2 + x^2 y^2 + y^2 z^2 + z^2 x^2 - 1}{(1 + x^2)(1 + y^2)(1 + z^2)}.$$ To simplify $T$, we write $X = 1 + x^2$, $Y = 1 + y^2$, $Z = 1 + z^2$, so that $X = 1 + \tan^2 A = \sec^2 A$, and similarly, $Y = \sec^2 B$, $Z = \sec^2 C$. Then by replacing $x^2$, $y^2$ and $z^2$ with $X-1$, $Y-1$, $Z-1$, respectively, where they appear in $T$, we obtain $$ \begin{align} T &= \frac{2(X-1)(Y-1)(Z-1) + (X-1)(Y-1) + (Y-1)(Z-1) + (Z-1)(X-1) - 1}{XYZ} \\ &= \frac{2XYZ - XY - YZ - ZX}{XYZ} \\ &= 2 - \frac{1}{X} - \frac{1}{Y} - \frac{1}{Z} \\ &= 2 - \cos^2 A - \cos^2 B - \cos^2 C. \end{align} $$ We therefore have the following identity: $$S = \tan(B-C)\tan(C-A)\tan(A-B) (2 - \cos^2 A - \cos^2 B - \cos^2 C),$$ which corresponds more or less to Blue's factorization. Interestingly, we never used the fact that $A$, $B$ and $C$ were the angles of a triangle. The identity extends to the case where one of the angles is a right angle by continuity.
Now the only question becomes whether it is possible to find a scalene triangle for which $$\cos^2 A + \cos^2 B + \cos^2 C = 2.$$ If we assume that $A < B < C$, then this amounts to showing that the function $$f(A,B) = \cos^2 A + \cos^2 B + \cos^2 (A + B)$$ takes the value $2$ somewhere in the interior of the triangle $\Delta$ bounded by the lines $A = 0$, $B = A$ and $A + 2B = \pi$. At the vertex $(0,0)$ of $\Delta$, we find $f(0,0) = 3$, so by continuity, at points in the interior of $\Delta$ near $(0,0)$, we must have $f(A,B) \approx 3$. Likewise, at the vertex corresponding to an equilateral triangle we have $f(\pi/3, \pi/3) = 3/4$, so at some points in the interior of $\Delta$, we must have $f(A, B) \approx 3/4$. Since the interior of $\Delta$ is a connected set, by the intermediate value theorem, we must have $f(A, B) = 2$ for some choice of $(A, B)$ in the interior of $\Delta$, which therefore corresponds to a scalene triangle.
Question: Was it predictable somehow, before the factoring step, that the numerator of $T$ would be expressible in terms of $x^2$, $y^2$ and $z^2$? That is what allowed the method to work as smoothly as it did.
The claim holds if we replace the function $\tan(B-C)$ (and cyclic permutations) with $\tan^*(B-C)=\frac{\sin(B-C)}{\cos(B+C)}$: Since: $$\tan^*(B-C)=\frac{\sin B \cos C - \sin C \cos B}{\cos B \cos C - \sin B \sin C}$$ and $2ab\cos(C) = a^2+b^2-c^2, 2R\sin A=a, 2ab\sin C=4\Delta$, we have: $$2R\tan^*(B-C)= \frac{2abc(a^2+b^2-c^2-a^2-c^2+b^2)}{(a^2+c^2-b^2)(a^2+b^2-c^2)-16\Delta^2},$$ but Heron's formula gives $16\Delta^2 = 4b^2c^2-(b^2+c^2-a^2)^2$, from which $$2R\tan^*(B-C) = \frac{4abc(b^2-c^2)}{a^4-(b^2-c^2)^2-4b^2c^2+(b^2+c^2-a^2)^2}$$ follows. This gives: $$2R\tan^*(B-C) = \frac{2abc(b^2-c^2)}{a^2(a^2-b^2-c^2)}=\frac{c^2-b^2}{a \cos A}$$ and: $$2Ra^2\tan^*(B-C) = \frac{a(c^2-b^2)}{\cos A} = 2R(c^2-b^2)\tan(A),$$ $$ a^2 \tan^*(B-C) = (c^2-b^2)\tan A = (c^2-b^2)\frac{4\Delta}{b^2+c^2-a^2}.\tag{1}$$ So we have, in terms of side lengths, $$ \sum_{cyc}(c^2-b^2)(a^2+c^2-b^2)(a^2+b^2-c^2) = 0, \tag{2} $$ or: $$ \sum_{cyc}(c^2-b^2)(a^4-b^4-c^4+2b^2c^2) = 0,$$ $$ \sum_{cyc}\left(b^6-c^6-a^4b^2+a^2b^4+3b^2c^4-3b^4c^2\right)=0, $$ $$ \sum_{cyc}\left(-a^4b^2+a^2b^4+3b^2c^4-3b^4c^2\right)=0, $$ $$ \sum_{cyc}\left(4b^2c^4-4b^4c^2\right)=0,\tag{3} $$
Now it is easy to notice that $a=\pm b,a=\pm c$ or $b=\pm c$ imply that the LHS is zero, so $(a^2-b^2)(b^2-c^2)(c^2-a^2)$ divides the LHS in $(3)$.
$$ \sum_{cyc}\left(4b^2c^4-4b^4c^2\right) = 4(a^2-b^2)(b^2-c^2)(c^2-a^2)\tag{4} $$ proves the (modified) claim.
The implication is false: Here is a contour plot of the values of the quantity $$w = a^2\tan(B-C)+b^2\tan(C-A)+c^2\tan(A-B)$$ where the vertices $B$ and $C$ are fixed at (-1,0) and (1,0), respectively, and the vertex $A$ is allowed to vary through the plane. The green contours are where $w=0$. The two prominent green circles and the green y-axis correspond to isosceles triangles. (The green x-axis corresponds to degenerate triangles.) As you can see though, there is some more green on the map, so there are definitely other zeros.
I'm not sure what the equations for the nontrivial green contours are. But I'll point out some other notable features of the graph. The red curve that appears to be a rectangular hyperbola is indeed the rectangular hyperbola $x^2 - y^2 = 1$: it's the locus of points where $B-C = \pm90^{\circ}$, i.e. $w = \infty$. (Red and blue are supposed to be the contours for $w=1$ and $w=-1$ respectively, but I guess Matlab's plotting doesn't handle the infinity that well, and just plots them in red.) The other "nice" red curves look like they might be cubic curves, and they should correspond to $w = \infty$ where $C-A = \pm90^{\circ}$ and $A-B = \pm90^{\circ}$. I haven't examined the matter closely though.
Here's one concrete example of a scalene triangle where we get $0$: Put vertex $A$ at $(3,\sqrt{-2 + 2\sqrt{17}})$, which makes a triangle with sides $a = 2, b = \sqrt{2 + 2\sqrt{17}}, c = \sqrt{14 + 2\sqrt{17}}.$ You can solve the triangle to verify.
How you can make a plot like this: View the plane as the complex plane, with the position of vertex $A$ as $z=x+iy$. Let $z_a = 2$ be the vector from vertex $B$ to vertex $C$, $z_b = -1 + z$ be the vector from vertex $C$ to vertex $A$, and $z_c = -1 - z$ be the vector from vertex $A$ to vertex $B$. Then $a = |z_a|$, and $B-C = \arg(z_b z_c / z_a^2)$ (can you see why?). In this manner, you can get $w$ as a function of $z$ to input into your software of choice.
I believe I have an example in which the identity is true but the triangle is not isosceles. All the algebra in the following was done by Maple.
Let $a=1$ and $b=2$, and take $c^2$ to be a root of the cubic $x^3-5x^2-25x+45$. The cubic has two roots between $1$ and $9$, either of which will give a valid set of sides $\{a,b,c\}$ for a triangle.
Now $c^2\ne4$ since $4$ is not a root of the cubic; so $c\ne2$; likewise $c\ne1$; so the triangle is not isosceles. Future forbidden values of $c$ are checked in the same way and will be given without comment.
Now use the cosine rule to find $\cos A$, $\cos B$ and $\cos C$ in terms of $a,b,c$. Then use the sine rule to write $\sin A$ and $\sin B$ as multiples of $\sin C$. Check that $\cos C\ne\pm1$ and so $\sin C\ne0$. Check that $\cos(B-C)$ and so on are not zero, then calculate $\tan(B-C)$ and so on, replacing $\sin^2C$ by $1-\cos^2C$; they all have a factor of $\sin C$.
Form the given expression and cancel $\sin C$; after simplification, Maple gives the numerator of the result as a constant times $$(c-1)(c-2)(c+1)(c+2)(c^6-5c^4-25c^2+45)\ ,$$ which is zero.
This also fits in with the answer given by @Blue: with the values of $a,b,c$ I have specified, the first factor in his/her product is zero and not the others.
RTP:$$ a^{2}\tan(B-C)+b^{2}\tan(C-A)+c^{2}\tan(A-B)=0$$ $\implies$ ABC is an isosceles triangle
$$ c^{2}=b^{2}({\frac{\sin C}{\sin B}})^{2} \implies b^{2}=c^{2}({\frac{\sin B}{\sin C}})^{2} $$
$$ a^{2}=c^{2}({\frac{\sin(\pi-A)}{\sin C}})^{2}=c^{2}({\frac{\sin A}{\sin C}})^{2} $$
$$ a^{2}\tan(B-C)+b^{2}\tan(C-A)+c^{2}\tan(A-B)=0 $$ $$ \implies c^{2}[({\frac{\sin A}{\sin C}})^{2}({\frac{\tan B-\tan C}{1+\tan B\tan C}})+({\frac{\sin B}{\sin C}})^{2\ }({\frac{\tan C-\tan A}{1+\tan C\tan A}})+({\frac{\tan A-\tan B}{1+\tan A\tan B}})]=0 $$
$$ \implies [({\frac{\sin A}{\sin C}})^{2}({\frac{\tan B-\tan C}{1+\tan B\tan C}})+({\frac{\sin B}{\sin C}})^{2\ }({\frac{\tan C-\tan A}{1+\tan C\tan A}})+({\frac{\tan A-\tan B}{1+\tan A\tan B}})]=0 $$
Now for any A>B>C the above equation will be false. [It is a conjecture]
Conjecture is a statement which is accepted as to be true/false until any counter example is presented.
So now the above equation is false (for A>B>C) according to conjecture.
Now there are only two possibilities either any two angles be equal or all three angles be equal.
In both the cases the above equation will be true.
and therefore the implication is proved to be right.
