# Triangle and logarithm problem

#### Pranav

##### Well-known member
Problem:
Given that $a,b$ and $c$ are the sides of $\Delta ABC$ such that $$z=\log_{2^a+2^{-a}} \left(4(ab+bc+ca)-(a+b+c)^2\right)$$ then $z$ has a real value if and only if

A)a=b=2c
B)3a=2b=c
C)a-b=3c
D)None of these

Attempt:
I am not sure where to start with this kind of problem. I wrote the expression inside the logarithm as follows:
$$-(a^2+b^2+c^2-2(ab+bc+ca))$$
Since, anything inside the logarithm must be greater than zero, I have
$$a^2+b^2+c^2-2(ab+bc+ca)<0 \Rightarrow a^2+b^2+c^2<2(ab+bc+ca)$$
But $a^2+b^2+c^2 >ab+bc+ca$, I get $ab+bc+ca>0$. I doubt this is going to help.

Any help is appreciated. Thanks!

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Problem:
Given that $a,b$ and $c$ are the sides of $\Delta ABC$ such that $$z=\log_{2^a+2^{-a}} \left(4(ab+bc+ca)-(a+b+c)^2\right)$$ then $z$ has a real value if and only if

A)a=b=2c
B)3a=2b=c
C)a-b=3c
D)None of these

Attempt:
I am not sure where to start with this kind of problem. I wrote the expression inside the logarithm as follows:
$$-(a^2+b^2+c^2-2(ab+bc+ca))$$
Since, anything inside the logarithm must be greater than zero,
Hey Pranav! I agree that the argument to the logarithm should be greater than zero.
From that point on the problem is symmetric in $a,b,c$.
That is, they can be swapped around giving the same expression.
In other words, none of the answers A, B, or C fit.
Therefore the answer must be D.

I have
$$a^2+b^2+c^2-2(ab+bc+ca)<0 \Rightarrow a^2+b^2+c^2<2(ab+bc+ca)$$
But $a^2+b^2+c^2 >ab+bc+ca$, I get $ab+bc+ca>0$. I doubt this is going to help.

Any help is appreciated. Thanks!
I believe that should be a $\ge$ sign.

Anyway, since $a,b,c$ are the sides of a triangle (which you did not use yet), you can conclude that this is always the case.

#### Opalg

##### MHB Oldtimer
Staff member
If $a,b,c$ are the sides of a triangle then $b+c > a$, $c+a>b$ and $a+b>c$. Multiply the first of those inequalities by $a$, the second by $b$ and the third by $c$, and add. That gives $2(bc+ca+ab) > a^2+b^2+c^2.$ Therefore $4(bc+ca+ab) > a^2+b^2+c^2 + 2(bc+ca+ab) = (a+b+c)^2.$ So $4(bc+ca+ab) - (a+b+c)^2$ is always positive and therefore always has a real logarithm (to any base).

#### Pranav

##### Well-known member
That is, they can be swapped around giving the same expression.
Hi ILS! I understand the above statement but how do you conclude the following:
In other words, none of the answers A, B, or C fit. If $a,b,c$ are the sides of a triangle then $b+c > a$, $c+a>b$ and $a+b>c$. Multiply the first of those inequalities by $a$, the second by $b$ and the third by $c$, and add. That gives $2(bc+ca+ab) > a^2+b^2+c^2.$ Therefore $4(bc+ca+ab) > a^2+b^2+c^2 + 2(bc+ca+ab) = (a+b+c)^2.$ So $4(bc+ca+ab) - (a+b+c)^2$ is always positive and therefore always has a real logarithm (to any base).
Thanks a lot Opalg, very nicely done. #### Klaas van Aarsen

##### MHB Seeker
Staff member
Hi ILS! I understand the above statement but how do you conclude the following:
In other words, none of the answers A, B, or C fit.
Since the problem is symmetric in $a, b, c$ the correct answer will also have to be symmetric in $a, b, c$. Neither of the answers A, B, or C are.

#### Pranav

##### Well-known member
Since the problem is symmetric in $a, b, c$ the correct answer will also have to be symmetric in $a, b, c$. Neither of the answers A, B, or C are.
I see it now, thanks ILS! 