MHB Prove a triangle is equilateral

[anemone]

Let $a,\,b,\,c$ be the sides and $A,\,B,\,C$ be the opposite angles angles of a triangle.

Show that if $ab^2\cos A=bc^2\cos B=ca^2\cos C$ then the triangle is equilateral.

 

[Fallen Angel]

Let's try.

Spoiler

From the equalitys we got $$\left\{\begin{array}{c}ab cos A - c^2cos B = 0 \\ b^2 cos A - ac cos C =0 \\ cbcos B-a^2cos C=0 \end{array}\right.$$.

Given any constants, $$a, b, c>0$$ we can consider the matrix $$M=\left(\begin{array}{ccc}ab & -c^2 & 0\\
b^2 & 0 & -ac \\ 0 & cb & -a^2\end{array}\right)$$
Then $$rank(M)=2$$ and it's easy to see that $$kernel(M)=<(1,1,1)>$$ and any solution of $$Mx=0$$ should be in the kernel.
Hence, $$cos \ A= cos \ B= cos \ C$$ and the triangle is equilateral.

Wait, this is not correct, mixed two different things over the paper, i will try to fix it.

 

[anemone]

Let's try.

Spoiler

From the equalitys we got $$\left\{\begin{array}{c}ab cos A - c^2cos B = 0 \\ b^2 cos A - ac cos C =0 \\ cbcos B-a^2cos C=0 \end{array}\right.$$.

Given any constants, $$a, b, c>0$$ we can consider the matrix $$M=\left(\begin{array}{ccc}ab & -c^2 & 0\\
b^2 & 0 & -ac \\ 0 & cb & -a^2\end{array}\right)$$
Then $$rank(M)=2$$ and it's easy to see that $$kernel(M)=<(1,1,1)>$$ and any solution of $$Mx=0$$ should be in the kernel.
Hence, $$cos \ A= cos \ B= cos \ C$$ and the triangle is equilateral.

Wait, this is not correct, mixed two different things over the paper, i will try to fix it.

Hi Fallen Angel,

Let me first thank you for participating in this challenge of mine!:)

But, I must tell you that I am out of my depth in linear algebra and hence, I might not be available to comment on your solution.:(

 

[anemone]

Let $a,\,b,\,c$ be the sides and $A,\,B,\,C$ be the opposite angles angles of a triangle.

Show that if $ab^2\cos A=bc^2\cos B=ca^2\cos C$ then the triangle is equilateral.

While we're still waiting for Fallen Angel's second submission to this challenge problem, I would like to reveal a hint to solve for it using elementary way:

Spoiler

Relate the given equality with the Law of Cosines.

 

[Fallen Angel]

Hi,

Sorry, at first I thought that working with linear algebra would give a pretty solution, but now I see it's essentially the same than doing it analitically but in a longer way.

 

[Greg]

Spoiler

$$\cos A=\frac{b^2+c^2-a^2}{2bc}$$$$\cos B=\frac{a^2+c^2-b^2}{2ac}$$$$\cos C=\frac{a^2+b^2-c^2}{2ab}$$$$ab^2\cos A=\frac{abc}{2}+\frac{ab^3}{2c}-\frac{a^3b}{2c}$$$$bc^2\cos B=\frac{abc}{2}+\frac{bc^3}{2a}-\frac{b^3c}{2a}$$$$a^2c\cos C=\frac{abc}{2}+\frac{a^3c}{2b}-\frac{ac^3}{2b}$$Hence we have$$\frac{ab}{2c}(b^2-a^2)=\frac{bc}{2a}(c^2-b^2)=\frac{ac}{2b}(a^2-c^2)$$For an equilateral triangle, equality holds.$$\text{ }$$For a triangle with $a>b=c$, the middle term is zero and the other two terms are non-zero, so equality does not hold.$$\text{ }$$For a triangle with $a>b>c$, the first term is negative and the third term is positive, so equality does not hold.$$\text{ }$$Thus, equality only holds if $a=b=c$, i.e. if triangle $ABC$ is equilateral.

 

[anemone]

Spoiler

$$\cos A=\frac{b^2+c^2-a^2}{2bc}$$$$\cos B=\frac{a^2+c^2-b^2}{2ac}$$$$\cos C=\frac{a^2+b^2-c^2}{2ab}$$$$ab^2\cos A=\frac{abc}{2}+\frac{ab^3}{2c}-\frac{a^3b}{2c}$$$$bc^2\cos B=\frac{abc}{2}+\frac{bc^3}{2a}-\frac{b^3c}{2a}$$$$a^2c\cos C=\frac{abc}{2}+\frac{a^3c}{2b}-\frac{ac^3}{2b}$$Hence we have$$\frac{ab}{2c}(b^2-a^2)=\frac{bc}{2a}(c^2-b^2)=\frac{ac}{2b}(a^2-c^2)$$For an equilateral triangle, equality holds.$$\text{ }$$For a triangle with $a>b=c$, the middle term is zero and the other two terms are non-zero, so equality does not hold.$$\text{ }$$For a triangle with $a>b>c$, the first term is negative and the third term is positive, so equality does not hold.$$\text{ }$$Thus, equality only holds if $a=b=c$, i.e. if triangle $ABC$ is equilateral.

Hi greg1313! Thanks for participating and your solution is correct! Well done!:)

Or we could also prove it using the method below:

Spoiler

$ab^2\cos A=bc^2\cos B=ca^2\cos C$

$\dfrac{ab(bc\cos A)}{c}=\dfrac{bc(ac\cos B)}{a}=\dfrac{ac(ba\cos C)}{b}$

$\dfrac{ab(b^2+c^2-a^2)}{2c}=\dfrac{bc(c^2+a^2-b^2)}{2a}=\dfrac{ac(a^2+b^2-c^2)}{2b}$

$\dfrac{b^2+c^2-a^2}{c^2}=\dfrac{c^2+a^2-b^2}{a^2}=\dfrac{a^2+b^2-c^2}{b^2}$

Therefore we get

$a^2b^2+a^2c^2-a^4=c^4+a^2c^2-b^2c^2\implies a^2b^2+b^2c^2=a^4+c^4$---(1)

$b^2c^2+a^2b^2-b^4=a^4+a^2b^2-a^2c^2\implies b^2c^2+a^2c^2=a^4+b^4$---(2)

$a^2c^2+b^2c^2-c^4=b^4+b^2c^2-a^2b^2\implies a^2c^2+a^2b^2=b^4+c^4$---(3)

Adding the equations (1), (2) and (3) up yield $2a^2b^2+2a^2c^2+2+b^2c^2=2a^4+2b^4+2c^4$

Equivalently $(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2=0$

So it must be $a^2=b^2=c^2$ and $a=b=c$.