Zig-Zag Midpoint Wag

We have the recurrence relation $\mathbf{m}_n \; = \; \tfrac12(\mathbf{m}_{n-2} + \mathbf{m}_{n-3})$ with the initial conditions $\mathbf{m}_0 = \mathbf{c}$ , $\mathbf{m}_1 = \tfrac12(\mathbf{a} + \mathbf{b})$ and $\mathbf{m}_2 = \tfrac12(\mathbf{b}+\mathbf{c})$ , where $\mathbf{a}$ , $\mathbf{b}$ , $\mathbf{c}$ are the position vectors of $A,B,C$ respectively. This recurrence relation leads to the indicial equation $0 \; = \; 2\lambda^3 - \lambda - 1 \; = \; (\lambda - 1)(2\lambda^2 + 2\lambda + 1)$ so the possible values of $\lambda$ are $-1$ , $-\tfrac12(1+i)$ , $-\tfrac12(1-i)$ , and so $\mathbf{m}_n \; = \; \mathbf{p} + \big(-\tfrac12(1+i)\big)^n\mathbf{q} + \big(-\tfrac12(1-i)\big)^n\mathbf{r}$ for some vectors $\mathbf{p}$ , $\mathbf{q}$ , $\mathbf{r}$ . Of course, $\mathbf{q}$ and $\mathbf{r}$ will be complex conjugates, so that $\mathbf{m}_n$ will turn out real.

Putting $n=0,1,2$ gives us three equations for $\mathbf{p}$ , $\mathbf{q}$ , $\mathbf{r}$ . $\begin{aligned} \mathbf{p} + \mathbf{q} + \mathbf{r} & = \; \mathbf{c} \\ \mathbf{p} - \tfrac12(1+i)\mathbf{q} - \tfrac12(1-i)\mathbf{r} & = \; \tfrac12(\mathbf{a} + \mathbf{b}) \\ \mathbf{p} + \tfrac12i\mathbf{q} - \tfrac12i\mathbf{r} & = \; \tfrac12(\mathbf{b} + \mathbf{c}) \end{aligned}$ We only need to determine $\mathbf{p} \; = \; \lim_{n \to \infty} \mathbf{m}_n$ since that will be the position vector of the desired point $P$ . Elementary algebra gives us $\mathbf{p} = \tfrac15\mathbf{a} + \tfrac25\mathbf{b} + \tfrac25\mathbf{c}$ .

Applying a coordinate system we have $\mathbf{a} \; = \; \binom{\tfrac12}{\tfrac12\sqrt{3}} \;, \; \mathbf{b} \; = \; \binom{0}{0} \;,\; \mathbf{c} \; = \; \binom{1}{0}$ and hence $\mathbf{p} \; = \; \binom{\tfrac12}{\tfrac{1}{10}\sqrt{3}}$ Thus $AP = \tfrac12\sqrt{3} - \tfrac{1}{10}\sqrt{3} = \tfrac25\sqrt{3}$ , while $BP = CP = \sqrt{\tfrac14 + \tfrac{3}{100}} = \tfrac15\sqrt{7}$ , and hence $AP + BP + CP \; = \; \frac25(\sqrt{3} + \sqrt{7})$ making the answer $2 + 5 + 1 + 3 + 1 + 7 = \boxed{19}$ .

Since $m_1$ and $m_2$ are midpoints of $\overline{AB}$ and $\overline{BC}$ , $\overline{m_1 m_2}$ is parallel to $\overline{AC}$

Similarly, $\overline{m_2 m_3}$ is parallel to $\overline{AB}$

$\overline{m_3 m_4}$ is parallel to $\overline{BC}$

$\overline{m_5 m_6}$ or $\overline{m_6 m_8}$ is parallel to $\overline{AC}$

$\overline{m_6 m_7}$ is parallel to $\overline{AB}$

$\overline{m_7 m_8}$ is parallel to $\overline{BC}$

So $\Delta m_6 m_7 m_8$ is similar and has the same orientation as $\Delta ABC$

$P$ in $\Delta m_6 m_7 m_8$ is still the same so $\overline{m_6 P}$ of $\Delta m_6 m_7 m_8$ corresponds to $\overline{AP}$ of $\Delta ABC$

Therefore, $A$ , $m_6$ , and $P$ must be on one line

Similarly, we can get $B$ , $m_7$ , and $P$ are collinear and $C$ , $m_8$ , and $P$ are collinear

Now using coordinates with $A(\frac{1}{2},\frac{\sqrt{3}}{2})$ , $B(0,0)$ , and $C(1,0)$ ,

We get $m_6(\frac{1}{2},\frac{\sqrt{3}}{8})$ , $m_7(\frac{15}{32},\frac{3\sqrt{3}}{32})$ , and $m_8(\frac{17}{32},\frac{3\sqrt{3}}{32})$

Using these we can compute the lines

$\overleftrightarrow{A m_6}$ is $x=\frac{1}{2}$

$\overleftrightarrow{B m_7}$ is $y=\frac{\sqrt{3}}{5}x$

and $\overleftrightarrow{C m_8}$ is $y=-\frac{\sqrt{3}}{5}(x-1)$

They all intersect at $P(\frac{1}{2},\frac{\sqrt{3}}{10})$

So $AP=\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{10}=\frac{2\sqrt{3}}{5}$

And $BP=CP=\sqrt{(\frac{1}{2})^2+(\frac{\sqrt{3}}{10})^2}=\sqrt{\frac{7}{25}}=\frac{\sqrt{7}}{5}$ ( $BP=CP$ since $P$ is on the perpendicular bisector of $BC$ which is $x=\frac{1}{2}$ )

Therefore, $AP+BP+CP=\frac{2\sqrt{3}}{5}+\frac{2\sqrt{7}}{5}=\frac{2}{5}(\sqrt{3}+\sqrt{7})$

And $a+b+c+d+e+f=2+5+1+3+1+7=\boxed{19}$

Relevant Wiki: Barycentric Coordinates

Using barycentric coordinates, found that

$m_1 (0,0,1)$ , $m_2 (\frac{1}{2},\frac{1}{2},0)$ , $m_3 (0,\frac{1}{2},\frac{1}{2})$ .

So generalize to

$m_n=\frac{1}{2} (m_{n-2} + m_{n-3})$

Using this, I want to do cancellation,

$2m_4=m_2+m_1$

$2m_5=m_3+m_2$

$2m_6=m_4+m_3$

$2m_7=m_5+m_4$

$2m_8=m_6+m_5$

$...$

$2m_{n-3}=m_{n-5}+m_{n-6}$

$2m_{n-2}=m_{n-4}+m_{n-5}$

$2m_{n-1}=m_{n-3}+m_{n-4}$

$2m_{n}=m_{n-2}+m_{n-3}$

Summing all of this, and do some cancellation, we get

$2m_n+2m_{n-1}+m_{n-2}=m_1+2m_2+2m_3$

Here is the main point: When n approaches infinity, it includes $m_n=m_{n-1}=m_{n-2}$

So

$5m_n=(\frac{1}{5},\frac{2}{5},\frac{2}{5})$

Return to normal xy-plane, let three verticle be $(0,0)$ , $(1,0)$ , $(\frac{1}{2},\frac{\sqrt{3}}{2})$

According to the barycenter coordinate of point P, we know that P is below upper verticle $\frac{4}{5}$ of the height of triangle.

So the ordinary coordinate of P is $(\frac{1}{2},\frac{\sqrt{3}}{10})$ .

Using simple formula we found that

$AP+BP+CP=\frac{2}{5} (\sqrt{7}+\sqrt{3})$

So $a+b+c+d+e+f=\boxed{19}$ .