Quadratic equation + Trigonometry problem

If the quadratic equation ax2+bx+c=0ax^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

sinAsinC+sinCsinA\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 sin2B4sinAsinC=0 \sin^2B-4\sin A\sin C=0 . I thought of replacing B with πAC \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
6 years, 9 months ago

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6 votes

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Let the circumradius of the triangle be RR, so from the sine rule: asinA=bsinB=csinC=2R \frac{a}{\sin A}= \frac{b}{\sin B}= \frac{c}{\sin C}= 2R Then, note that sinAsinC+sinCsinA=2Ra2Rc+2Rc2Ra=ac+ca=a2+c2ac \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 b24ac=0b^2-4ac= 0 From the cosine rule, we obtain b2=c2+a22accos(B)b^2= c^2 + a^2 - 2ac \cos (B) Hence, c2+a22accos(B)4ac=0 c^2 + a^2 - 2ac \cos (B) - 4ac = 0     c2+a2=ac(4+2cos(B)) \implies c^2 + a^2 = ac(4 + 2 \cos (B) )     c2+a2ac=4+2cos(B) \implies \frac{c^2+a^2}{ac}= 4 + 2 \cos (B) From the inequalities (note that the inequalities are strict) 1<cos(B)<1-1 < \cos (B) < 1, we obtain: 4<c2+a2ac<64< \frac{c^2+a^2}{ac} < 6

Sreejato Bhattacharya - 6 years, 9 months ago

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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.

Shubhabrota Chakraborty - 6 years, 9 months ago

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Yes, the answer is 12 but how do you obtain ((E/2)2)2=14ac ((E/2)-2)^2=1-4ac ?

Pranav Arora - 6 years, 9 months ago

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You can try this.

Nishant Sharma - 6 years, 9 months ago

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