Calculate Incentre Linear Combination From Midpoints of Triangle

To find the incentre $(h, k)$ of triangle $ABC$, we first determine the coordinates of its vertices $A, B,$ and $C$ using the given midpoints of its sides.

Let the midpoints of the sides $BC, CA,$ and $AB$ be: $D = \left(\frac{5}{2}, 7\right)$ $E = \left(\frac{5}{2}, 3\right)$ $F = (4, 5)$

The vertices of triangle $ABC$ can be expressed in terms of the midpoints as follows: $A = E + F - D = \left(\frac{5}{2} + 4 - \frac{5}{2}, 3 + 5 - 7\right) = (4, 1)$ $B = D + F - E = \left(\frac{5}{2} + 4 - \frac{5}{2}, 7 + 5 - 3\right) = (4, 9)$ $C = D + E - F = \left(\frac{5}{2} + \frac{5}{2} - 4, 7 + 3 - 5\right) = (1, 5)$

Next, we calculate the lengths of the sides of triangle $ABC$:

1. Length of side $a = BC$: $a = \sqrt{(4 - 1)^2 + (9 - 5)^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$

2. Length of side $b = CA$: $b = \sqrt{(4 - 1)^2 + (1 - 5)^2} = \sqrt{3^2 + (-4)^2} = \sqrt{25} = 5$

3. Length of side $c = AB$: $c = \sqrt{(4 - 4)^2 + (9 - 1)^2} = \sqrt{0 + 8^2} = 8$

The coordinates of the incentre $(h, k)$ are given by the formula: $h = \frac{a x_A + b x_B + c x_C}{a + b + c}$ $k = \frac{a y_A + b y_B + c y_C}{a + b + c}$

Substituting the known values into the formulas: $h = \frac{5(4) + 5(4) + 8(1)}{5 + 5 + 8} = \frac{20 + 20 + 8}{18} = \frac{48}{18} = \frac{8}{3}$

$k = \frac{5(1) + 5(9) + 8(5)}{5 + 5 + 8} = \frac{5 + 45 + 40}{18} = \frac{90}{18} = 5$

Now, we calculate $3h + k$: $3h + k = 3\left(\frac{8}{3}\right) + 5 = 8 + 5 = 13$

Thus, the value of $3h + k$ is equal to $13$.