Calculate Expression Involving Triangle Circumcenter Coordinates

To find the value of $21(\alpha + \beta)$, we first determine the coordinates of the vertices of the triangle $ABC$.

Let the vertices of $\triangle ABC$ be $A(1, 2)$, $B(x_B, y_B)$, and $C(x_C, y_C)$.

Step 1: Find the coordinates of vertex $B$. We are given that the midpoint of side $AB$ is $D(5, -1)$. Using the midpoint formula: $\frac{1 + x_B}{2} = 5 \implies x_B = 10 - 1 = 9$ $\frac{2 + y_B}{2} = -1 \implies y_B = -2 - 2 = -4$ Thus, vertex $B = (9, -4)$.

Step 2: Find the coordinates of vertex $C$. We are given that the centroid $G$ of $\triangle ABC$ is $(3, 4)$. Using the centroid formula: $\frac{x_A + x_B + x_C}{3} = 3 \implies \frac{1 + 9 + x_C}{3} = 3 \implies 10 + x_C = 9 \implies x_C = -1$ $\frac{y_A + y_B + y_C}{3} = 4 \implies \frac{2 - 4 + y_C}{3} = 4 \implies -2 + y_C = 12 \implies y_C = 14$ Thus, vertex $C = (-1, 14)$.

Step 3: Find the equations of the perpendicular bisectors to determine the circumcenter $(\alpha, \beta)$.

1. Perpendicular bisector of $AB$:

   • Midpoint of $AB$ is $D(5, -1)$.
   • Slope of line $AB = \frac{-4 - 2}{9 - 1} = \frac{-6}{8} = -\frac{3}{4}$.
   • Slope of the perpendicular bisector of $AB = \frac{4}{3}$.
   • Equation of the perpendicular bisector of $AB$: $y - (-1) = \frac{4}{3}(x - 5) \implies 3y + 3 = 4x - 20 \implies 4x - 3y = 23$ Since the circumcenter $(\alpha, \beta)$ lies on this bisector: $4\alpha - 3\beta = 23 \quad \text{--- (Equation 1)}$

2. Perpendicular bisector of $AC$:

   • Midpoint of $AC$ is $E\left(\frac{1 - 1}{2}, \frac{2 + 14}{2}\right) = (0, 8)$.
   • Slope of line $AC = \frac{14 - 2}{-1 - 1} = \frac{12}{-2} = -6$.
   • Slope of the perpendicular bisector of $AC = \frac{1}{6}$.
   • Equation of the perpendicular bisector of $AC$: $y - 8 = \frac{1}{6}(x - 0) \implies 6y - 48 = x \implies x - 6y = -48$ Since the circumcenter $(\alpha, \beta)$ lies on this bisector: $\alpha - 6\beta = -48 \quad \text{--- (Equation 2)}$

Step 4: Solve for $\alpha$ and $\beta$. From Equation 2, we have: $\alpha = 6\beta - 48$

Substituting $\alpha$ into Equation 1: $4(6\beta - 48) - 3\beta = 23$ $24\beta - 192 - 3\beta = 23$ $21\beta = 215 \implies \beta = \frac{215}{21}$

Now, substitute $\beta$ back into the expression for $\alpha$: $\alpha = 6\left(\frac{215}{21}\right) - 48 = \frac{430}{7} - 48 = \frac{430 - 336}{7} = \frac{94}{7} = \frac{282}{21}$

Step 5: Calculate $21(\alpha + \beta)$. $\alpha + \beta = \frac{282}{21} + \frac{215}{21} = \frac{497}{21}$

Thus, $21(\alpha + \beta) = 21 \times \frac{497}{21} = 497$