Let the vertex A of a triangle ABC be (1,2), and the mid-point of the side AB be (5,-1).

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 $21(\alpha+\beta)$ is equal to :

Accepted Answer

Let the coordinates of vertex B be $({x}_{B},{y}_{B})$. We are given that vertex A is $(1,2)$ and the mid-point of side AB is $(5,-1)$.

Using the mid-point formula, the coordinates of the mid-point are given by:

$(\frac{{x}_{A}+{x}_{B}}{2},\frac{{y}_{A}+{y}_{B}}{2})=(5,-1)$

By equating the $x$ and $y$ coordinates, we get:

$\frac{1+{x}_{B}}{2}=5\, \implies \, 1+{x}_{B}=10\, \implies \, {x}_{B}=9$

$\frac{2+{y}_{B}}{2}=-1\, \implies \, 2+{y}_{B}=-2\, \implies \, {y}_{B}=-4$

So, the coordinates of vertex B are $(9,-4)$.

Let the coordinates of vertex C be $({x}_{C},{y}_{C})$. The centroid of a triangle with vertices $({x}_{1},{y}_{1})$, $({x}_{2},{y}_{2})$, and $({x}_{3},{y}_{3})$ is given by the formula:

$(\frac{{x}_{1}+{x}_{2}+{x}_{3}}{3},\frac{{y}_{1}+{y}_{2}+{y}_{3}}{3})$

We are given the centroid of triangle ABC is $(3,4)$. Using the coordinates of vertices A, B, and C:

$\frac{1+9+{x}_{C}}{3}=3\, \implies \, 10+{x}_{C}=9\, \implies \, {x}_{C}=-1$

$\frac{2+(-4)+{y}_{C}}{3}=4\, \implies \, -2+{y}_{C}=12\, \implies \, {y}_{C}=14$

So, the coordinates of vertex C are $(-1,14)$.

The circumcenter of a triangle is a point that is equidistant from all three of its vertices. Let the given circumcenter be marked as $O(\alpha ,\beta )$. Therefore, we have:

$OA=OB=OC\, \implies \, O{A}^{2}=O{B}^{2}=O{C}^{2}$

Using the distance formula, we will write the equations for the squares of the distances:

$O{A}^{2}=(\alpha -1{)}^{2}+(\beta -2{)}^{2}={\alpha }^{2}+{\beta }^{2}-2\alpha -4\beta +5$

$O{B}^{2}=(\alpha -9{)}^{2}+(\beta +4{)}^{2}={\alpha }^{2}+{\beta }^{2}-18\alpha +8\beta +97$

$O{C}^{2}=(\alpha +1{)}^{2}+(\beta -14{)}^{2}={\alpha }^{2}+{\beta }^{2}+2\alpha -28\beta +197$

Now, equating $O{A}^{2}=O{B}^{2}$:

${\alpha }^{2}+{\beta }^{2}-2\alpha -4\beta +5={\alpha }^{2}+{\beta }^{2}-18\alpha +8\beta +97$

$16\alpha -12\beta =92$

Dividing the entire equation by 4 gives our first linear equation:

$4\alpha -3\beta =23\,$

Next, equating $O{A}^{2}=O{C}^{2}$:

${\alpha }^{2}+{\beta }^{2}-2\alpha -4\beta +5={\alpha }^{2}+{\beta }^{2}+2\alpha -28\beta +197$

$-4\alpha +24\beta =192$

Dividing this equation by -4 gives our second linear equation:

$\alpha -6\beta =-48\, \implies \, \alpha =6\beta -48\,$

Substitute the expression for $\alpha $ from Equation 2 into Equation 1:

$4(6\beta -48)-3\beta =23$

$24\beta -192-3\beta =23$

$21\beta -192=23$

$21\beta =215\, \implies \, \beta =\frac{215}{21}$

Now, substitute the value of $\beta $ back into Equation 2 to find $\alpha $:

$\alpha =6(\frac{215}{21})-48$

$\alpha =\frac{1290}{21}-\frac{1008}{21}$

$\alpha =\frac{282}{21}$

The problem asks for the value of $21(\alpha +\beta )$. We can simply plug in our values:

$21(\alpha +\beta )=21(\frac{282}{21}+\frac{215}{21})$

$21(\alpha +\beta )=282+215=497$

Option C is the correct answer.

Solutions · 1