Let (5, a4) be the circumcenter of a triangle with vertices A(a,-2), B(a,6) and C( a4,-2).

Question

Let $\left(5,\dfrac a4\right)$ be the circumcenter of a triangle with vertices $A(a,-2),\ B(a,6)$ and $C\left(\dfrac a4,-2\right)$. Let $\alpha$ denote the circumradius, $\beta$ denote the area and $\gamma$ denote the perimeter of the triangle. Then $\alpha+\beta+\gamma$ is:

Accepted Answer

Since $A(a,-2)$ and $B(a,6)$ have the same $x$-coordinate, the circumcenter's $y$-coordinate equals the midpoint of $A,B$'s $y$-values: $\dfrac a4 = \dfrac{-2+6}2=2 \implies a=8$.

So $A=(8,-2),\ B=(8,6),\ C=(2,-2)$, and circumcenter $=(5,2)$.

$\alpha = $ circumradius $=$ distance from $(5,2)$ to $A(8,-2) = \sqrt{9+16}=5$.

Sides: $AB=8$, $AC=\sqrt{36+0}=6$, $BC=\sqrt{36+64}=10$. Right triangle (right angle at $A$, since $6^2+8^2=10^2$).

$\beta = $ area $=\dfrac12\times6\times8=24$.

$\gamma = $ perimeter $=6+8+10=24$.

$$\alpha+\beta+\gamma = 5+24+24=53$$

Solutions · 1