Maximize this weird fraction!

Algebra Level 5

Let a , b , c > 0 a,b,c>0 be positive real numbers such that b c c a a b = 1 bc-ca-ab=1 . Suppose that the maximum of 4024 1 + a 2 4024 1 + b 2 2555 1 + c 2 \frac{4024}{1+a^2}-\frac{4024}{1+b^2}-\frac{2555}{1+c^2} is of the form d 2 e \frac{d^2}{e} , where d , e d,e are relatively prime integers. Find d + e d+e .

This problem was inspired by problem from Vietnamese Mathematical Olympiads 1999.


The answer is 4567.

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1 solution

From b c c a a b = 1 bc-ca-ab=1 , we can write it as a 1 b + 1 b 1 c + 1 c a = 1. a \cdot \frac{1}{b} + \frac{1}{b} \cdot \frac{1}{c} + \frac{1}{c} \cdot a = 1. The fact is that x y + y z + z x = 1 xy+yz+zx=1 if and only if there exist 0 < A , B , C < π 0< A,B,C < \pi such that A + B + C = π A+B+C=\pi and x = tan A 2 , y = tan B 2 , z = tan C 2 . x=\tan\frac{A}{2}, y=\tan\frac{B}{2}, z=\tan\frac{C}{2}. Here, we take x = a , y = 1 b , z = 1 c x=a, y=\frac{1}{b}, z=\frac{1}{c} . So, we have that a = tan A 2 , b = cot B 2 , c = cot C 2 . a=\tan\frac{A}{2}, b=\cot\frac{B}{2}, c=\cot\frac{C}{2}. Let P P be the expression we want to maximize. We can alternatively write it as P = 4024 cos 2 A 2 4024 sin 2 B 2 2555 sin 2 C 2 = 2012 ( 1 + cos A ) 2012 ( 1 cos B ) 2555 sin 2 C 2 = 2012 ( cos A + cos B ) 2555 sin 2 C 2 = 4024 cos A + B 2 cos A B 2 2555 sin 2 C 2 = 4024 cos π C 2 cos A B 2 2555 sin 2 C 2 = 4024 sin C 2 cos A B 2 2555 sin 2 C 2 . \begin{aligned} P &= 4024 \cos^2\frac{A}{2} - 4024 \sin^2\frac{B}{2} - 2555 \sin^2\frac{C}{2}\\[10pt] &= 2012(1+\cos A)-2012(1-\cos B)-2555\sin^2\frac{C}{2}\\[10pt] &= 2012(\cos A + \cos B) - 2555\sin^2\frac{C}{2}\\[10pt] &= 4024 \cos \frac{A+B}{2} \cos \frac{A-B}{2} - 2555\sin^2\frac{C}{2}\\[10pt] &= 4024 \cos \frac{\pi-C}{2} \cos \frac{A-B}{2} - 2555\sin^2\frac{C}{2}\\[10pt] &= 4024 \sin \frac{C}{2} \cos \frac{A-B}{2} - 2555\sin^2\frac{C}{2}. \end{aligned} Let u = sin C 2 u=\sin\frac{C}{2} and v = cos A B 2 v=\cos\frac{A-B}{2} . We then have that P = 4024 u v 2555 u 2 = 2555 ( u 2012 2555 v ) 2 + 201 2 2 2555 v 2 , \begin{aligned} P &= 4024uv-2555u^2\\[10pt] &= -2555 \left( u-\frac{2012}{2555}v \right)^2 + \frac{2012^2}{2555}v^2, \end{aligned} whose maximum is P max = 201 2 2 2015 P_{\max}=\frac{2012^2}{2015} , which occurs when u 2012 2015 v = 0 u-\frac{2012}{2015}v=0 , and v = 1 v=1 . Working on variables, we'll finally find out that the maximum occurs when ( a , b , c ) = ( a 0 , b 0 , c 0 ) (a,b,c)=(a_0,b_0,c_0) , where a 0 = 543 4567 , b 0 = 4567 543 , c 0 = 543 4567 2012 . a_0=\sqrt{\frac{543}{4567}}, b_0=\sqrt{\frac{4567}{543}}, c_0=\frac{\sqrt{543 \cdot 4567}}{2012}. This is to confirm that such maximum exists. Now, our task is to find the P max P_{\max} , which is 201 2 2 2555 \frac{2012^2}{2555} . One can check that gcd ( 2012 , 2555 ) = 1 \gcd(2012,2555)=1 , so that the answer is 2012 + 2555 = 4567 2012+2555=\boxed{4567} .

i did the same thing but i used x=tan(A). why i am not getting this solution in wolfram alpha.

Srikanth Tupurani - 11 months ago

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