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(C) Let the midpoints be $D(\frac{5}{2}, 7)$, $E(\frac{5}{2}, 3)$, and $F(4, 5)$.The vertices of $	riangle ABC$ are $A, B, C$. Using the property tha

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$13$

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(C) Let the midpoints be $D(\frac{5}{2}, 7)$, $E(\frac{5}{2}, 3)$, and $F(4, 5)$.The vertices of $	riangle ABC$ are $A, B, C$. Using the property that the vertices of the original triangle are $A = E+F-D$, $B = D+F-E$, and $C = D+E-F$:$A = (\frac{5}{2}+4-\frac{5}{2}, 3+5-7) = (4, 1)$$B = (\frac{5}{2}+4-\frac{5}{2}, 7+5-3) = (4, 9)$$C = (\frac{5}{2}+\frac{5}{2}-4, 7+3-5) = (1, 5)$The side lengths are $a = BC = \sqrt{(4-1)^2 + (9-5)^2} = \sqrt{3^2+4^2} = 5$, $b = AC = \sqrt{(4-1)^2 + (1-5)^2} = \sqrt{3^2+(-4)^2} = 5$, and $c = AB = \sqrt{(4-4)^2 + (9-1)^2} = 8$.Since $a=b=5$, the triangle is isosceles. The incentre $(h, k)$ is given by $(\frac{ax_A+bx_B+cx_C}{a+b+c}, \frac{ay_A+by_B+cy_C}{a+b+c})$.$h = \frac{5(4)+5(4)+8(1)}{5+5+8} = \frac{20+20+8}{18} = \frac{48}{18} = \frac{8}{3}$.$k = \frac{5(1)+5(9)+8(5)}{5+5+8} = \frac{5+45+40}{18} = \frac{90}{18} = 5$.Thus, $3h + k = 3(\frac{8}{3}) + 5 = 8 + 5 = 13$.

Let the midpoints of the sides of a triangle $ABC$ be $(\frac{5}{2}, 7)$, $(\frac{5}{2}, 3)$, and $(4, 5)$. If its incentre is $(h, k)$, then $3h + k$ is equal to:

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