The teacher drew a coordinate plane on the board and marked some points on this plane. Unfortunately, Vasya's second-grader, who was on duty, erased almost the entire drawing, except for two points $A (1, 2)$ and $B (3,1)$. Will the excellent Andriyko be able to follow these two points to construct the beginning of the coordinate system point $O (0, 0)$? Point A on the board located above and to the left of point $B$.
1. **Draw the line \( l = AB \):**
   - The line \( l \) passes through points \( A(1, 2) \) and \( B(3, 1) \).
   - The slope of the line \( l \) is calculated as:
     \[
     m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 2}{3 - 1} = \frac{-1}{2}
     \]
   - The equation of the line \( l \) in point-slope form is:
     \[
     y - y_1 = m(x - x_1) \implies y - 2 = -\frac{1}{2}(x - 1)
     \]
     Simplifying, we get:
     \[
     y = -\frac{1}{2}x + \frac{1}{2} + 2 \implies y = -\frac{1}{2}x + \frac{5}{2}
     \]

2. **Draw the line \( k \) through \( A \) such that \( k \perp l \):**
   - The slope of line \( k \) is the negative reciprocal of the slope of line \( l \):
     \[
     m_k = -\frac{1}{m} = -\frac{1}{-\frac{1}{2}} = 2
     \]
   - The equation of line \( k \) passing through point \( A(1, 2) \) is:
     \[
     y - y_1 = m_k(x - x_1) \implies y - 2 = 2(x - 1)
     \]
     Simplifying, we get:
     \[
     y = 2x - 2 + 2 \implies y = 2x
     \]

3. **Draw the circle \( c \) with midpoint \( A \) and radius \( AB \):**
   - The distance \( AB \) is calculated using the distance formula:
     \[
     AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(3 - 1)^2 + (1 - 2)^2} = \sqrt{4 + 1} = \sqrt{5}
     \]
   - The equation of the circle \( c \) with center \( A(1, 2) \) and radius \( \sqrt{5} \) is:
     \[
     (x - 1)^2 + (y - 2)^2 = 5
     \]

4. **Find the intersection points of line \( k \) and circle \( c \):**
   - Substitute \( y = 2x \) into the circle's equation:
     \[
     (x - 1)^2 + (2x - 2)^2 = 5
     \]
     Simplifying, we get:
     \[
     (x - 1)^2 + 4(x - 1)^2 = 5 \implies 5(x - 1)^2 = 5 \implies (x - 1)^2 = 1 \implies x - 1 = \pm 1
     \]
     Therefore, \( x = 2 \) or \( x = 0 \).

   - For \( x = 2 \):
     \[
     y = 2(2) = 4 \implies (2, 4)
     \]
   - For \( x = 0 \):
     \[
     y = 2(0) = 0 \implies (0, 0)
     \]

5. **Verify the intersection points:**
   - The point \( (2, 4) \) does not lie on the line \( l \), so it is not the origin.
   - The point \( (0, 0) \) lies on the line \( k \) and satisfies the circle's equation, confirming it as the origin.

Therefore, Andriyko can construct the origin \( O(0, 0) \) using the given points \( A \) and \( B \).

The final answer is \( \boxed{ O(0, 0) } \).