Trigonometry! #125

The given equation factorizes as follows:

$(x-4)(x-6)(x-2)=0\implies x=4,6,2$

Take the given cyclic sum as $S$ . Now, using the Law of Cosines, we have,

$S=\sum_{cyclic}\frac{\cos A}{a}=\sum_{cyclic} \frac{b^2+c^2-a^2}{2abc}$

It can clearly be seen that the denominator of the general term of the cyclic sum is symmetric since $2abc=2bca=2cab$ . The sum can hence be expanded out as:

$S=\frac{(b^2+c^2-a^2)+(c^2+a^2-b^2)+(a^2+b^2-c^2)}{2abc}=\frac{a^2+b^2+c^2}{2abc}$

The final result obtained is a symmetric expression in $a,b,c$ and so, the answer remains the same for all the three cases of sides that can be given by $\{a,b,c\}=\{4,5,2\}$ . Considering any one of the three cases (like $a=5,b=4,c=2$ ), we have,

$S=\frac{16+25+4}{80}=\boxed{\frac{9}{16}}$

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A diagram with the case taken:

Image

From the given equation on the sides of the triangle ABC, we get three equations: a+b+c=11, ab+bc+ca=38, abc=40 On solving these simultaneous equations we get a,b,c=2,4,5(in any order) Now, we are asked the cyclic summation of cosA/a. On, simplifying, we get that we are asked the value of (a^2 + b^2 + c^2)/ 2abc . So, putting the values we get 45/80 i.e. 9/16. So, p/q = 9/16. Hence, p+q = 9 + 16 = 25