A geometry problem by Md Zuhair

By Cosine Rule ,

$\cos(A) = \dfrac{b^2+c^2-a^2}{2bc} , \cos(B) = \dfrac{a^2+c^2-b^2}{2ac} , \cos(C) = \dfrac{b^2+a^2-c^2}{2ba}$

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Plugging in the values in the given equation.

$\left(\dfrac{b^2+c^2-a^2}{2bc}\right)^2 + \left(\dfrac{a^2+c^2-b^2}{2ac}\right)^2 + \left(\dfrac{b^2+a^2-c^2}{2ba}\right)^2=1$

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

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Call . $a^2+b^2-c^2=x , b^2+c^2-a^2=y , c^2+a^2-b^2=z$

$\Rightarrow \dfrac{x+y}2 = b^2,\dfrac{y+z}2=c^2,\dfrac{z+x}2=a^2$

$(1)$ becomes $x^2y+x^2z+y^2z+y^2x+z^2x+z^2y = (x+y)(y+z)(z+x)$

$\Rightarrow (x+y)(y+z)(z+x)-2xyz = (x+y)(y+z)(z+x)$

$\Rightarrow$ either of $x , y , z = 0$

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Either of them has to happen $\boxed{a^2+b^2=c^2 , c^2+b^2=a^2 , b^2+c^2=a^2}$

This shows that the triangle is right angled.

An Intuitive solution to this problem could be something like. Cos(A)^2 = 1 - Sin(A)^2, Cos(B)^2 = 1 - Sin(B)^2, Cos(C)^2 = 1 - Sin(C)^2

Substituting this in the given equation we get,

3 - (Sin(A)^2 + Sin(B)^2 + Sin(C)^2 = 1

Sin(A)^2 + Sin(B)^2 + Sin(C)^2 = 2 ...... (1)

the max value of Sin(A) is 1 so the max value of Sin^2(A) is also one.

Now for (1) to be true, Atleast one of Sin(A)^2 or Sin(B)^2 or Sin(C)^2 has to be 1, so as to make the total sum as 2. thus, one of the angles has to be a right angle.