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# Quadratic equation + Trigonometry problem

If the quadratic equation $$ax^2+bx+c=0$$ has equal roots where a, b and c denote the lengths of the sides opposite to vertices A, B and C of a triangle ABC respectively, then find the sum of integers in the range of

$\frac{\sin A}{\sin C}+\frac{\sin C}{\sin A}$

I am stuck on this problem from quite some time and finally decided to post this here.

Since the roots are equal, I get $$\sin^2B-4\sin A\sin C=0$$. I thought of replacing B with $$\pi-A-C$$ and plug that in Wolfram Alpha but I get no nice solutions. I honestly don't know how to proceed in this problem.

Any help is appreciated. Thanks!

Note by Pranav Arora
3 years, 9 months ago

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Let the circumradius of the triangle be $$R$$, so from the sine rule: $\frac{a}{\sin A}= \frac{b}{\sin B}= \frac{c}{\sin C}= 2R$ Then, note that $\frac{\sin A}{\sin C} + \frac{\sin C}{\sin A}= \frac{\frac{2R}{a}}{\frac{2R}{c}} + \frac{\frac{2R}{c}}{\frac{2R}{a}}= \frac{a}{c} + \frac{c}{a} = \frac{a^2+c^2}{ac}$
The roots of the given equation will be equal iff $b^2-4ac= 0$ From the cosine rule, we obtain $b^2= c^2 + a^2 - 2ac \cos (B)$ Hence, $c^2 + a^2 - 2ac \cos (B) - 4ac = 0$ $\implies c^2 + a^2 = ac(4 + 2 \cos (B) )$ $\implies \frac{c^2+a^2}{ac}= 4 + 2 \cos (B)$ From the inequalities (note that the inequalities are strict) $$-1 < \cos (B) < 1$$, we obtain: $4< \frac{c^2+a^2}{ac} < 6$ · 3 years, 9 months ago

Is the actual answer 12 ? I did nothing special, just denote given expression by E. and as we get from the condition of equal roots , replace your sin^2x, by (1-cos^2x) then write the expression for cosx for a triangle. you will obtain an eqn. like, [(E/2)-2]^2= 1-4ac which in turn gives the simple inequality, (E-2)(E-6)<0
since a and c both>0 i.e 2<E<6.(the lower limit is also obvious from A.M-G.M inequality, anyway,) Therefore the sum of the integers lying between this range is =(3+4+5)=12.(ans) SORRY FOR IM NOT SO GOOD IN USING LATEX. can you suggest me a better way to learn latex codes? the links in brilliant are not enough for learning. · 3 years, 9 months ago

You can try this. · 3 years, 9 months ago

Yes, the answer is 12 but how do you obtain $$((E/2)-2)^2=1-4ac$$ ? · 3 years, 9 months ago