$a,b,c$ are the sides and $A,B,C$ are the angles of a triangle. If the roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ are equal then,

Question

$a,b,c$ are the sides of a $\triangle ABC$ and $A,B,C$ are the respective angles. If the roots of the equation $a(b-c)x^2+b(c-a)x+c(a-b)=0$ are equal then $\sin^2 \bigl(\frac{A}{2}\bigr), \sin^2 \bigl(\frac{B}{2}\bigr)$ and $\sin^2\bigl(\frac{C}{2}\bigr)$ are in which type of progression?(Arithmetic,geometric,harmonic)

My try,

I immediately noticed that $1$ was a root and since the product of the roots was $1$ I got that $a,b,c$ were in harmonic progression. I replaced the reciprocals of $a,b,c$ by $p,q,r$. Wrote $\sin^2\bigl({\theta\over 2}\bigr)$ as $\frac{1-cos(\theta)}{2}$ and replaced $\cos$ by the sides but did not succeed. I also tried by replacing it with $\frac{(s-b)(s-c)}{bc}$ but still could not get the required answer.

If I assume the result I am getting the answer by simplifying the expression. Please give an elegant solution.

Answer 1

As you noticed, the roots are,

$$x = 1\hspace{1 cm}\mathrm{ or }\hspace{1 cm}x = \frac{(a-b)c}{a(b-c)}$$

The roots are equal,$$1 = \frac{(a-b)c}{a(b-c)}.$$

Therefore, we can find

$$b = \frac{2ac}{a+c}.$$

Now let's talk about trigonometry. The half angle formula for sine is

$$\sin^2{\frac{W}{2}} = \frac{1 - \cos{W}}{2}$$

and the cosine law is

$$w^2 = u^2 + v^2 - 2 u v \cos{W}$$

Therefore,

$$-\frac{\cos{W}}{2} = \frac{w^2 - u^2 - v^2}{4 u v}$$

Therefore,

$$\sin^2{\frac{W}{2}} = \frac{1}{2} + \frac{w^2 - u^2 - v^2}{4 u v} = \frac{w^2 - (u - v)^2}{4uv}.$$

Now we can find that

$$\sin^2{\frac{A}{2}}=\frac{\left(a^2+2 a c-c^2\right) \left(a^2+c^2\right)}{8 a c^2 (a+c)}$$

$$\sin^2{\frac{B}{2}}=\frac{1}{4} \left(-\frac{a}{c}+\frac{4 a c}{(a+c)^2}-\frac{c}{a}+2\right)$$

$$\sin^2{\frac{C}{2}}=\frac{-a^4+2 a^3 c+2 a c^3+c^4}{8 a^2 c (a+c)}$$

We then just need to check the progression types.

Answer 2

This is not an answer but a hint and comment about your assertion you have already got that a,b,c are in H.P. Here a possible way to prove what you want with the squares of the sinus of middle angles.

►Considering the incenter you have $$\sin^2 (\frac{A}{2})= \frac {r^2}{d_1^2}\left(= \frac{1}{\frac{d_1^2}{r^2}}\right)\\ \sin^2 \bigl(\frac{B}{2}\bigr)=\frac{r^2}{d_2^2}\left(= \frac{1}{\frac{d_2^2}{r^2}}\right)\\\sin^2(\frac{C}{2})=\frac{r^2}{d_3^2}\left(= \frac{1}{\frac{d_3^2}{r^2}}\right)$$ where $r$ is the inradius and $d_1,d_2,d_3$ the distances from the incenter to the vertices $A,B,C$ respectively. Assuming convenient magnitudes of angles, we can deduce (supposing true the H.P. relation) $$\frac{d_2^2-d_1^2}{r^2}=\frac{d_3^2-d_2^2}{r^2}\iff 2d_2^2=d_1^2+d_3^2$$

Hence $$2d_2^2=d_1^2+d_3^2\iff \frac{2}{\sin^2 (\frac{B}{2})}=\frac{1}{\sin^2 (\frac{A}{2})}+\frac{1}{\sin^2 (\frac{C}{2})}\iff \text{the three squares of sinus are in H.P.} $$ It follows an alternative to try to prove your statement.

►REMARK.- In an equilateral triangle you can verify the asked property (since $2a^2=a^2+a^2$ and besides $\frac 12=\frac{2}{2+2}$) but in this case the given quadratic equation does not make sense.