Let the vertex of a triangle ABC be A(1, 2), | vedclass

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(C) Given $A = (1, 2)$ and midpoint of $AB$ is $M = (5, -1)$.Since $M = (A+B)/2$, we have $(1+B_x)/2 = 5 \implies B_x = 9$ and $(2+B_y)/2 = -1 \implie

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$497$

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(C) Given $A = (1, 2)$ and midpoint of $AB$ is $M = (5, -1)$.Since $M = (A+B)/2$, we have $(1+B_x)/2 = 5 \implies B_x = 9$ and $(2+B_y)/2 = -1 \implies B_y = -4$. Thus, $B = (9, -4)$.Given centroid $G = (3, 4)$, we have $(A+B+C)/3 = G$, so $A+B+C = 3G = (9, 12)$.$C = (9, 12) - (1, 2) - (9, -4) = (-1, 14)$.Let circumcenter be $O = (\alpha, \beta)$. $O$ is equidistant from vertices $A, B, C$, so $OA^2 = OB^2 = OC^2$.$OA^2 = (\alpha-1)^2 + (\beta-2)^2$$OB^2 = (\alpha-9)^2 + (\beta+4)^2$$OC^2 = (\alpha+1)^2 + (\beta-14)^2$Equating $OA^2 = OB^2$: $(\alpha-1)^2 - (\alpha-9)^2 = (\beta+4)^2 - (\beta-2)^2$$16\alpha - 80 = 12\beta + 12 \implies 4\alpha - 3\beta = 23$.Equating $OB^2 = OC^2$: $(\alpha-9)^2 - (\alpha+1)^2 = (\beta-14)^2 - (\beta+4)^2$$-20\alpha + 80 = -36\beta + 180 \implies 5\alpha - 9\beta = -25$.Solving the system: $12\alpha - 9\beta = 69$ and $5\alpha - 9\beta = -25$.Subtracting gives $7\alpha = 94 \implies \alpha = 94/7$.Substituting $\alpha$: $4(94/7) - 3\beta = 23 \implies 376/7 - 23 = 3\beta \implies 215/7 = 3\beta \implies \beta = 215/21$.Then $\alpha + \beta = 94/7 + 215/21 = (282+215)/21 = 497/21$.$21(\alpha + \beta) = 497$.

Let the vertex of a triangle $ABC$ be $A(1, 2)$, and the mid-point of the side $AB$ be $M(5, -1)$. If the centroid of this triangle is $G(3, 4)$ and its circumcenter is $O(\alpha, \beta)$, then $21(\alpha + \beta)$ is equal to:

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