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Step-by-Step Solution
Step 1: Identify the Given Elements
We have a triangle with vertices
A($a, -2$), B($a, 6$), and C($\frac{a}{4}, -2$).
The circumcenter of this triangle is given as
$O\bigl(5, \frac{a}{4}\bigr)$.
We denote:
$\alpha$ = Circumradius
$\beta$ = Area of the triangle
$\gamma$ = Perimeter of the triangle
We need to find $\alpha + \beta + \gamma$.
Step 2: Use the Circumcenter Property to Find $a$
Because $O\bigl(5, \tfrac{a}{4}\bigr)$ is the circumcenter,
its distance to A must equal its distance to B. Let
$O = \bigl(5, \tfrac{a}{4}\bigr)$,
$A = (a, -2)$, and $B = (a, 6)$.
We set $OA^2 = OB^2$.
First, compute:
$$
OA^2 = (a - 5)^2 + \Bigl(-2 - \frac{a}{4}\Bigr)^2,
$$
$$
OB^2 = (a - 5)^2 + \Bigl(6 - \frac{a}{4}\Bigr)^2.
$$
Since $(a - 5)^2$ appears in both, we only need to equate
$$
\Bigl(-2 - \frac{a}{4}\Bigr)^2 = \Bigl(6 - \frac{a}{4}\Bigr)^2.
$$
Solving this gives $a = 8.$
Therefore, the circumcenter simplifies to
$$
O = \Bigl(5, \frac{8}{4}\Bigr) = (5, 2).
$$
Step 3: Update the Vertices of the Triangle
Substituting $a = 8$:
A($8, -2$)
B($8, 6$)
C($2, -2$) because $\tfrac{8}{4} = 2$.
Step 4: Find the Circumradius $\alpha$
The circumradius is $OA$, the distance from $O(5, 2)$ to
any vertex (for instance, A($8, -2$)):
$$
OA = \sqrt{(8 - 5)^2 + (-2 - 2)^2}
= \sqrt{3^2 + (-4)^2}
= \sqrt{9 + 16}
= \sqrt{25}
= 5.
$$
Thus, $\alpha = 5$.
Step 5: Compute the Area of the Triangle $\beta$
Use the determinant formula for the area of the triangle with vertices
$(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$:
$$
\beta = \frac{1}{2}
\Bigl|
x_1(y_2 - y_3)
+ x_2(y_3 - y_1)
+ x_3(y_1 - y_2)
\Bigr|.
$$
Substituting A($8, -2$), B($8, 6$), and C($2, -2$):
$x_1 = 8, \; y_1 = -2$
$x_2 = 8, \; y_2 = 6$
$x_3 = 2, \; y_3 = -2$
$$
\beta = \frac{1}{2}
\bigl|
8(6 - (-2))
+ 8((-2) - (-2))
+ 2((-2) - 6)
\bigr|.
$$
Simplify inside:
$$
= \frac{1}{2}
\bigl|
8 \times 8 + 8 \times 0 + 2 \times (-8)
\bigr|
= \frac{1}{2} \bigl|64 - 16\bigr|
= \frac{1}{2} \times 48
= 24.
$$
Hence, the area $\beta = 24.$
Step 6: Determine the Perimeter $\gamma$
The perimeter is the sum of the lengths of sides $AB$, $BC$, and $CA$.
$$
AB = \sqrt{(8 - 8)^2 + (6 - (-2))^2}
= \sqrt{0 + 8^2}
= 8.
$$
$$
BC = \sqrt{(2 - 8)^2 + (-2 - 6)^2}
= \sqrt{(-6)^2 + (-8)^2}
= \sqrt{36 + 64}
= \sqrt{100}
= 10.
$$
$$
CA = \sqrt{(2 - 8)^2 + (-2 - (-2))^2}
= \sqrt{(-6)^2 + 0^2}
= \sqrt{36}
= 6.
$$
Therefore,
$$
\gamma = AB + BC + CA = 8 + 10 + 6 = 24.
$$
Step 7: Final Sum $\alpha + \beta + \gamma$
We have:
$$
\alpha = 5, \quad \beta = 24, \quad \gamma = 24.
$$
Thus,
$$
\alpha + \beta + \gamma = 5 + 24 + 24 = 53.
$$
Hence, the required value is $53$.
