Let A(1,1), B(1,-1), C(-1,1) be the vertices | vedclass

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(B) The vertices are $A(1,1), B(1,-1), C(-1,1)$.Since $AB$ is vertical (along $x=1$) and $AC$ is horizontal (along $y=1$),$	riangle ABC$ is a right-a

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$2$

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(B) The vertices are $A(1,1), B(1,-1), C(-1,1)$.Since $AB$ is vertical (along $x=1$) and $AC$ is horizontal (along $y=1$),$	riangle ABC$ is a right-angled triangle at $A(1,1)$.$1$. Circumcentre $S$: For a right-angled triangle,the circumcentre is the midpoint of the hypotenuse $BC$.$S = \left(\frac{1+(-1)}{2}, \frac{-1+1}{2}ight) = (0,0)$.$2$. Orthocentre $O$: For a right-angled triangle,the orthocentre is the vertex at the right angle.$O = A = (1,1)$.$3$. Incentre $I$: The side lengths are $c = AB = \sqrt{(1-1)^2 + (1-(-1))^2} = 2$,$b = AC = \sqrt{(1-(-1))^2 + (1-1)^2} = 2$,and $a = BC = \sqrt{(1-(-1))^2 + (-1-1)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$.The coordinates of the incentre $I$ are given by $\left(\frac{ax_A + bx_B + cx_C}{a+b+c}, \frac{ay_A + by_B + cy_C}{a+b+c}ight)$.$I = \left(\frac{2\sqrt{2}(1) + 2(1) + 2(-1)}{2\sqrt{2}+2+2}, \frac{2\sqrt{2}(1) + 2(-1) + 2(1)}{2\sqrt{2}+2+2}ight) = \left(\frac{2\sqrt{2}}{2\sqrt{2}+4}, \frac{2\sqrt{2}}{2\sqrt{2}+4}ight) = \left(\frac{\sqrt{2}}{\sqrt{2}+2}, \frac{\sqrt{2}}{\sqrt{2}+2}ight) = \left(\frac{1}{1+\sqrt{2}}, \frac{1}{1+\sqrt{2}}ight) = (\sqrt{2}-1, \sqrt{2}-1)$.Now,calculate $IS + OS$:$OS = \sqrt{(1-0)^2 + (1-0)^2} = \sqrt{1+1} = \sqrt{2}$.$IS = \sqrt{(\sqrt{2}-1-0)^2 + (\sqrt{2}-1-0)^2} = \sqrt{2(\sqrt{2}-1)^2} = \sqrt{2}(\sqrt{2}-1) = 2 - \sqrt{2}$.$IS + OS = (2 - \sqrt{2}) + \sqrt{2} = 2$.

Let $A(1,1), B(1,-1), C(-1,1)$ be the vertices of $\triangle ABC$. Let $S$ be the circumcentre,$O$ be the orthocentre,and $I$ be the incentre of the $\triangle ABC$. Then $IS + OS =$ ?

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