Question

Let the vertex A of a triangle ABC be (1, 2), and the mid-point of the side AB be (5, -1). If the centroid of this triangle is (3, 4) and its circumcenter is \((\alpha, \beta)\), then \(2(10\alpha + \beta)\) is equal to:

• A. 309
• B. 403
• C. 497
• D. 524

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We are given one vertex, the midpoint of a side, and the centroid. Using these, we can determine all three vertices of the triangle. Once the vertices are known, we can find the circumcenter by using the property that it is equidistant from all three vertices.
Step 2: Key Formula or Approach:
1. Midpoint Formula: \(M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
2. Centroid Formula: \(G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)
3. Circumcenter \(O(\alpha, \beta)\): \(OA^2 = OB^2 = OC^2\)
Step 3: Detailed Explanation:
Let vertex \(A = (1, 2)\) and vertex \(B = (x_B, y_B)\).
The midpoint of \(AB\) is \((5, -1)\):
\[ \frac{1 + x_B}{2} = 5 \implies x_B = 9 \] \[ \frac{2 + y_B}{2} = -1 \implies y_B = -4 \] So, \(B = (9, -4)\).
Let vertex \(C = (x_C, y_C)\). The centroid \(G = (3, 4)\):
\[ \frac{1 + 9 + x_C}{3} = 3 \implies 10 + x_C = 9 \implies x_C = -1 \] \[ \frac{2 - 4 + y_C}{3} = 4 \implies -2 + y_C = 12 \implies y_C = 14 \] So, \(C = (-1, 14)\).
Let the circumcenter be \(O(\alpha, \beta)\). Then \(OA^2 = OB^2 = OC^2\).
\[ OA^2 = (\alpha - 1)^2 + (\beta - 2)^2 = \alpha^2 + \beta^2 - 2\alpha - 4\beta + 5 \] \[ OB^2 = (\alpha - 9)^2 + (\beta + 4)^2 = \alpha^2 + \beta^2 - 18\alpha + 8\beta + 97 \] \[ OC^2 = (\alpha + 1)^2 + (\beta - 14)^2 = \alpha^2 + \beta^2 + 2\alpha - 28\beta + 197 \] Equating \(OA^2 = OB^2\):
\[ -2\alpha - 4\beta + 5 = -18\alpha + 8\beta + 97 \implies 16\alpha - 12\beta = 92 \implies 4\alpha - 3\beta = 23 \quad \text{--- (Eq 1)} \] Equating \(OA^2 = OC^2\):
\[ -2\alpha - 4\beta + 5 = 2\alpha - 28\beta + 197 \implies -4\alpha + 24\beta = 192 \implies -\alpha + 6\beta = 48 \quad \text{--- (Eq 2)} \] From Eq 2, \(\alpha = 6\beta - 48\). Substitute this into Eq 1:
\[ 4(6\beta - 48) - 3\beta = 23 \implies 24\beta - 192 - 3\beta = 23 \implies 21\beta = 215 \implies \beta = \frac{215}{21} \] Now, find \(\alpha\):
\[ \alpha = 6\left(\frac{215}{21}\right) - 48 = \frac{430}{7} - 48 = \frac{430 - 336}{7} = \frac{94}{7} = \frac{282}{21} \] Step 4: Final Answer:
We need to find the value of \(21(\alpha + \beta)\):
\[ 21(\alpha + \beta) = 21\left(\frac{282}{21} + \frac{215}{21}\right) = 282 + 215 = 497 \]