We are given an isosceles right triangle $ABC$. On its sides, isosceles right triangles $BCO_1$, $ACO_2$, and $ABO_3$ are constructed. We are given $\vec{CB} = \vec{r}$, $\vec{CA} = \vec{p}$, and $\vec{O_2C} = \vec{q}$. We need to calculate the vectors $\vec{O_2O_3}$, $\vec{O_1O_3}$, and $\vec{O_2O_1}$.

1. Problem Description

We are given an isosceles right triangle $ABC$. On its sides, isosceles right triangles $BCO_1$, $ACO_2$, and $ABO_3$ are constructed. We are given $\vec{CB} = \vec{r}$, $\vec{CA} = \vec{p}$, and $\vec{O_2C} = \vec{q}$. We need to calculate the vectors $\vec{O_2O_3}$, $\vec{O_1O_3}$, and $\vec{O_2O_1}$.

2. Solution Steps

First, we can express the position vectors of $A$ and $B$ with respect to $C$ as $\vec{CA} = \vec{p}$ and $\vec{CB} = \vec{r}$.

Since $ACO_2$ is an isosceles right triangle and $\angle ACO_2 = 90^{\circ}$, we can express $\vec{CO_2}$ as a rotation of $\vec{CA}$ by $90^{\circ}$ counterclockwise and scaling by $\frac{1}{\sqrt{2}}$. Then, we can also say that $\vec{O_2C} = -\vec{CO_2}$.

Also, since $BCO_1$ is an isosceles right triangle and $\angle BCO_1 = 90^{\circ}$, we can express $\vec{CO_1}$ as a rotation of $\vec{CB}$ by $90^{\circ}$ clockwise and scaling by $\frac{1}{\sqrt{2}}$.

Since $ABO_3$ is an isosceles right triangle and $\angle AO_3B = 90^{\circ}$, we can express $\vec{O_3}$ as the midpoint of the hypotenuse $AB$.

Therefore, we can write $\vec{CO_3} = \frac{\vec{CA}+\vec{CB}}{2} = \frac{\vec{p}+\vec{r}}{2}$.

Now, we are given that $\vec{O_2C} = \vec{q}$. So, $\vec{CO_2} = -\vec{q}$.

Since $\vec{CO_2}$ is obtained by rotating $\vec{CA}=\vec{p}$ by $90^{\circ}$ counterclockwise and scaling by $\frac{1}{\sqrt{2}}$, we have:

$\vec{CO_2} = \frac{1}{\sqrt{2}}(-p_y, p_x)$.

So, $-\vec{q} = \frac{1}{\sqrt{2}}(-p_y, p_x)$.

Thus, $\vec{q} = \frac{1}{\sqrt{2}}(p_y, -p_x)$.

Similarly, $\vec{CO_1}$ is obtained by rotating $\vec{CB}=\vec{r}$ by $90^{\circ}$ clockwise and scaling by $\frac{1}{\sqrt{2}}$.

$\vec{CO_1} = \frac{1}{\sqrt{2}}(r_y, -r_x)$.

Now, we can express the vectors we need to calculate.

$\vec{O_2O_3} = \vec{O_2C} + \vec{CO_3} = \vec{q} + \frac{\vec{p}+\vec{r}}{2}$

$\vec{O_1O_3} = \vec{O_1C} + \vec{CO_3} = -\vec{CO_1} + \frac{\vec{p}+\vec{r}}{2} = -\frac{1}{\sqrt{2}}(r_y, -r_x) + \frac{\vec{p}+\vec{r}}{2}$

$\vec{O_2O_1} = \vec{O_2C} + \vec{CO_1} = \vec{q} + \frac{1}{\sqrt{2}}(r_y, -r_x)$

3. Final Answer

$\vec{O_2O_3} = \vec{q} + \frac{\vec{p}+\vec{r}}{2}$

$\vec{O_1O_3} = \frac{\vec{p}+\vec{r}}{2} - \frac{1}{\sqrt{2}}(r_y, -r_x)$

$\vec{O_2O_1} = \vec{q} + \frac{1}{\sqrt{2}}(r_y, -r_x)$