Trigonometry Related Part III

Geometry Level 5

$\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{3c}{\sqrt{1+c^2}}$

Given that $a,b$ and $c$ are positive real numbers satisfying $ab+bc+ca=1$ .

If the maximum value of the expression above can be expressed as $\sqrt{M}$ for positive integer $M$ , determine $M$ .

The answer is 10.

4 solutions

Leah Smith
Dec 23, 2015

Any other solutions are welcome!

That's a crazy expression! Where did you come up with that?

Calvin Lin Staff - 5 years, 5 months ago

Well I came across a problem like: given a triangle with three angles $A,B,C$ , determine the maximum of $\frac{1}{2}(sinA+sinB) + 3sin\frac{C}{2}$ and developed that into a more algebraic problem.

Leah Smith - 5 years, 5 months ago

Interesting! Thanks for sharing :)

Calvin Lin Staff - 5 years, 5 months ago

Rajen Kapur
Dec 23, 2015

Let a, b and c equal cotangents of acute triangle with angles A, B, and C so that a = cot A, b = cot B and c = cot C, satisfying given equation ab + bc + ca = 1. Now the given expression has a value sin A cos A + sin B cos B + 3 cos C = cos (A - B) sin C + 3 cos C, which is maximum $\sqrt{10}$ when A = B.

good nice answer

Ankit Yadav - 5 years, 5 months ago

Atharva Sarage
Mar 5, 2016

a=tanA,b=tanB,c=tanC where A+B+C=pi/2 simplify and put A=B for max value of cos(A-B) WE GET sqrt 10

Gopal Narayanan
Dec 24, 2015

By symmetry we assume a=b. Then ab+bc+ca=1 becomes a^2+2ac-1=0 so we get a =-c+sqrt (a^2+1). Then put this value in the complex equation and differentiate. We get value of c=3. a=-3+sqrt (10). Substituting we get m=10

Trigonometry Related Part III

The answer is 10.

4 solutions

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