Not a Typo

Algebra Level 5

Let $a,b$ and $c$ be real numbers satisfying $abc =a+b+c$ .

The maximum value of $\dfrac1{\sqrt{1+a^2}} + \dfrac1{\sqrt{1+b^2}} + \dfrac c{\sqrt{1+c^2}}$ can be expressed as $\dfrac {p \sqrt q}r$ , where $p,q,r$ are all positive integers with $p,r$ coprime and $q$ square-free.

Find $p+q+r$ .

Clarification: The third term in the addition is indeed $\dfrac c{\sqrt{1+c^2}}$ , not $\dfrac 1{\sqrt{1+c^2}}.$

The answer is 8.

1 solution

Reynan Henry
Dec 24, 2016

because $abc=a+b+c$ we can substitute $a=\tan{A},b=\tan{B},c=\tan{C}$ for some $\triangle ABC$ . Now

$\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{c}{\sqrt{1+c^2}}=\cos{A}+\cos{B}+\sin{C} \\ =2\sin{\frac{C}{2}}\cos{\frac{A-B}{2}}+2\sin{\frac{C}{2}}\cos{\frac{C}{2}}\le 2\sin{\frac{C}{2}}\left( 1+\cos{\frac{C}{2}} \right) \\ \le \frac{3\sqrt{3}}{2}$

(I found the maximum by differentiating.)

So $a+b+c=8$

Moderator note:

This solution is incomplete because the substitution only applies to a subset of possible cases.

As an explicit counter example, we could have $a = - \frac{1}{\sqrt{3} } , b = - \frac{1}{\sqrt{3} } , c = \sqrt{3}$ , but not find corresponding angles $A, B, C$ in a triangle.

If we only wanted $A+B+C = 180 ^ \circ$ , then we still run into issues like $\sin \frac{ C}{2} < 0$ .

Um it has to be $\frac{m\sqrt{n}}{p}$

Reynan Henry - 4 years, 5 months ago

This solution is flawed. If $a$ , $b$ , $c$ are real numbers such that $abc = a + b + c$ , it is not necessarily true that $a = \tan A$ , $b = \tan B$ , $c = \tan C$ , where $A$ , $B$ , $C$ are the angles of a triangle. This error is particularly egregious here, because the maximum occurs for $a = b = -\frac{1}{\sqrt{3}}$ and $c = \sqrt{3}$ , and this is a case where this representation does not work.

Jon Haussmann - 4 years, 5 months ago

ah it should be positive real sorry

Reynan Henry - 4 years, 5 months ago

@Jon Haussmann and @Reynan Henry ..I couldn't get what you are trying to say...please help me out...!

Ankit Kumar Jain - 4 years, 4 months ago

@Ankit Kumar Jain – i am using trigonometric substitution

Reynan Henry - 4 years, 4 months ago

@Reynan Henry – As Jon sir said that there is a flaw ...I could neither understand the flaw nor its rectification.

Ankit Kumar Jain - 4 years, 4 months ago

@Ankit Kumar Jain – the flaw is i forgot to mention that $a,b,c$ have to be positive and it is corrected now

Reynan Henry - 4 years, 4 months ago

@Reynan Henry – The issue is that the trigonometric substitution of "angles of a triangle" only works for positive real numbers (Do you know why?). I was hinting at this in my comment of "Be aware of what the conditions are in which the substitution is valid", but I didn't follow up on it as you didn't reply.

I have added the condition that $a, b, c$ are positive.

Calvin Lin Staff - 4 years, 4 months ago

@Calvin Lin – Sir ..Then what would be the equality case?

Ankit Kumar Jain - 4 years, 4 months ago

@Ankit Kumar Jain – Ahh, ic. I was too quick to make the edit.

Let me think about this further. I've reported the problem.

Calvin Lin Staff - 4 years, 4 months ago

@Ankit Kumar Jain – it is when $a=b=\tan{30},c=\tan{120}$

Reynan Henry - 4 years, 4 months ago

@Reynan Henry – Is it really giving $\frac{3\sqrt{3}}{2}$ when substituted in the expression..?

Ankit Kumar Jain - 4 years, 4 months ago

@Ankit Kumar Jain – you can try it

Reynan Henry - 4 years, 4 months ago

@Reynan Henry – Note that $\tan 120 ^ \circ < 0$ , so it doesn't satisfy the condition of "positive reals".

Calvin Lin Staff - 4 years, 4 months ago

Very nice substitution used! As always, be aware of what the conditions are in which the substitution is valid.

I used // to create line breaks in your Latex, which makes the long expression easier to read.

Calvin Lin Staff - 4 years, 5 months ago

Not a Typo

The answer is 8.

1 solution

Moderator note:

0 pending reports