Let the mid points of the sides of a triangle ABC be (52,7), (52,3) and (4,5).

Question

Let the mid points of the sides of a triangle $ABC$ be $\left(\dfrac52,7\right),\ \left(\dfrac52,3\right)$ and $(4,5)$. If its incentre is $(h,k)$, then $3h+k$ is equal to:

Accepted Answer

Let $D\left(\frac52,7\right), E\left(\frac52,3\right), F(4,5)$ be the midpoints of $BC, CA, AB$. Then the vertices are
$$A = E+F-D = (4,1), \quad B = F+D-E = (4,9), \quad C = D+E-F = (1,5)$$

Side lengths: $a=BC = \sqrt{3^2+4^2}=5$, $b=CA=\sqrt{3^2+4^2}=5$, $c=AB=\sqrt{0^2+8^2}=8$.

Incentre:
$$h = \frac{5(4)+5(4)+8(1)}{5+5+8} = \frac{48}{18}=\frac83, \qquad k = \frac{5(1)+5(9)+8(5)}{18}=\frac{90}{18}=5$$

$$3h+k = 3\cdot\frac83+5 = 8+5=13$$

Solutions · 1