Advanced Vectors - Part 5

If we write $\mathbf{a} = \overrightarrow{OA}$ , $\mathbf{b} = \overrightarrow{OB}$ , $\mathbf{c} = \overrightarrow{OC}$ , then $|\mathbf{a}| \; = \; |\mathbf{b}| \; = \; |\mathbf{c}| \; = \; R$ where $R$ is the outradius, so that $AB^2 \; = \; |\mathbf{a}-\mathbf{b}|^2 \; = \; 2R^2 - 2\mathbf{a}\cdot\mathbf{b} \hspace{2cm} AC^2 \; = \; |\mathbf{a}-\mathbf{c}|^2 \; = \; 2R^2 - 2\mathbf{a}\cdot\mathbf{c}$ and hence $\overrightarrow{AO} \cdot \overrightarrow{BC} \; = \; -\mathbf{a} \cdot (\mathbf{c} - \mathbf{b}) \; = \; \mathbf{a}\cdot\mathbf{b} - \mathbf{a}\cdot\mathbf{c} \; = \; \big(R^2 - \tfrac12AB^2\big) - \big(R^2 - \tfrac12AC^2\big) \; = \; \tfrac12(AC^2 - AB^2)$ In this case, the answer is $\boxed{\tfrac{9}{2}}$ .

I chose my special case to be a right angled triangle, since it would then be easier to find the circumcenter of the triangle. I had point A as (0,0), B as (0,4) and C as (5,0). The circumcenter would then have the coordinates (2.5,2). The dot product then becomes 2.5 * 5 + (-4) * 2 = 4.5