Not a Typo 2

Algebra Level 5

Real numbers a , b , c a,b,c are such that a b + b c + c a = 1 ab+bc+ca=1 and the maximum value of 1 1 + a 2 + 1 1 + b 2 + 6 c 1 + c 2 \frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{6c}{1+c^2} is m + p q r m+\frac{p\sqrt{q}}{r} , where m , p , q , r m,p,q,r are all positive integers with gcd ⁡ ( p , r ) = 1 \gcd(p,r)=1 and q q square free. Find the value of m + p + q + r m+p+q+r .


The answer is 14.

Reynan Henry by Reynan Henry , 22, Egypt

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Reynan Henry
Dec 31, 2016

Like the first problem we substitute a = tan ⁡ A 2 , b = tan ⁡ B 2 , c = tan ⁡ C 2 a=\tan{\frac{A}{2}},b=\tan{\frac{B}{2}},c=\tan{\frac{C}{2}} 1 1 + a 2 + 1 1 + b 2 + 6 c 1 + c 2 = cos ⁡ 2 A 2 + cos ⁡ 2 B 2 + 6 sin ⁡ C 2 cos ⁡ C 2 = cos ⁡ A + 1 2 + cos ⁡ B + 1 2 + 6 sin ⁡ C 2 cos ⁡ C 2 = 1 + cos ⁡ A + cos ⁡ B 2 + 6 sin ⁡ C 2 cos ⁡ C 2 = 1 + cos ⁡ A + B 2 cos ⁡ A − B 2 + 6 sin ⁡ C 2 cos ⁡ C 2 ≤ 1 + sin ⁡ C 2 + 6 sin ⁡ C 2 cos ⁡ C 2 = 1 + sin ⁡ C 2 ( 1 + 6 cos ⁡ C 2 ) ≤ 1 + 5 5 3 \begin{aligned} \frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{6c}{1+c^2}&=\cos^2{\frac{A}{2}}+\cos^2{\frac{B}{2}}+6\sin{\frac{C}{2}}\cos{\frac{C}{2}}\\ &=\frac{\cos{A}+1}{2}+\frac{\cos{B}+1}{2}+6\sin{\frac{C}{2}}\cos{\frac{C}{2}}\\&= 1 +\frac{\cos{A}+\cos{B}}{2}+6\sin{\frac{C}{2}}\cos{\frac{C}{2}}\\&= 1 +\cos{\frac{A+B}{2}}\cos{\frac{A-B}{2}}+6\sin{\frac{C}{2}}\cos{\frac{C}{2}}\\&\le 1+\sin{\frac{C}{2}}+6\sin{\frac{C}{2}}\cos{\frac{C}{2}}\\&= 1+\sin{\frac{C}{2}}\left(1+6\cos{\frac{C}{2}}\right)\\&\le 1 +\frac{5\sqrt{5}}{3}\end{aligned}

so m + p + q + r = 14 m+p+q+r=14

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Very nicely done!

Calvin Lin Staff - 4 years, 5 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...