Hexagon in Triangle

This question is most easily handled using trilinear coordinates. The triangle vertices $A,B,C$ have coordinates $A:\hspace{0.3cm} 1\,:\,0\,:\,0 \hspace{2cm} B:\hspace{0.3cm} 0\,:\,1\,:\,0 \hspace{2cm} C:\hspace{0.3cm} 0\,:\,0\,:\,1$ while the points $A_p,A_q.B_p,B_q,C_p,C_q$ where $BA_p =p\,BC$ and $BA_q = q\,BC$ (and similarly for the other four points) have coordinates $\begin{array}{rlcrlcrl} A_p: & 0\,:\, (1-p)c\,:\, pb & \hspace{2cm} & B_p: & pc \,:\, 0\,:\, (1-p)a & \hspace{2cm}& C_p: & (1-p)b\,:\, pa\,:\, 0 \\ A_q: & 0\,:\, (1-q)c\,:\, qb && B_q: & qc \,:\, 0\,:\, (1-q)a && C_q: & (1-q)b\,:\, qa\,:\, 0 \end{array}$ The points $A_{1,p},A_{1,q},B_{1,p},B_{1,q},C_{1,p},C_{1,q}$ can now be calculated. $A_{1,p}$ is the intersection of $BB_p$ and $CC_p$ , while $A_{1,q}$ is the intersection of $BB_q$ and $CC_q$ , with the other four points similarly defined. Solving the trilinear equations for the intersection of these lines, we obtain the coordinates $\begin{array}{rlcrl} A_{1,p}: & p(1-p)bc\,:\, p^2ac\,:\, (1-p)^2ab & \hspace{2cm}& A_{1,q}: & q(1-q)bc\,:\, q^2ac\,:\, (1-q)^2a \\ B_{1,p}: & (1-p)^2bc\,:\, p(1-p)bc\,:\, p^2ab && B_{1,q}: & (1-q)^2bc\,:\, q(1-q)bc\,:\, q^2ab\\ C_{1,p}: & p^2bc\,:\, (1-p)^2ac\,:\, p(1-p)ab && C_{1,q}: & q^2bc\,:\, (1-q)^2ac\,:\, q(1-q)ab \end{array}$ Similarly, the points $A_2,A_3,B_2,B_3,C_2,C_3$ can be determined, where $A_2$ is the intersection of $BB_p$ with $CC_q$ , while $A_3$ is the intersection of $BB_q$ with $CC_p$ (and the other four points are similarly defined). We obtain $\begin{array}{rlcrl} A_2: & p(1-q)bc\,:\, pqac\,:\,(1-p)(1-q)ab & \hspace{2cm} & A_3: & (1-p)qbc\,:\, pqac\,:\,(1-p)(1-q)ab \\ B_2: & (1-p)(1-q)bc\,:\, p(1-q)ac\,:\, pqab && B_3: & (1-p)(1-q)bc\,:\, (1-p)qac\,:\,pqab \\ C_2: & pqbc\,:\, (1-p)(1-q)ac\,:\, p(1-q)ab && C_3: & pqbc\,:\, (1-p)(1-q)ac\,:\, (1-p)qab \end{array}$

This construction will not result in a central hexagon unless $q$ is large enough. A hexagon will be formed so long as $A_{1,q}$ lies inside the triangle $ABA_p$ , $B_{1,q}$ lies inside the triangle $BCB_p$ and $C_{1,q}$ lies inside the triangle $CAC_p$ . Symmetry gives us that these three things either all occur at the same time, or none occur. Thus the critical case to consider is when $A_{1,q}$ lies on the line $AA_p$ , which occurs when $A_{1,q} = B_3 = C_2$ . This implies that $p(1-q) = \lambda pq$ , $pq = \lambda(1-p)(1-q)$ and $(1-p)(1-q) = \lambda(1-p)q$ for some $\lambda > 0$ , which tells us that $q \; = \; \frac{\sqrt{1-p}}{\sqrt{p} + \sqrt{1-p}}$ Equally well, for there to be an internal hexagon, the point $A_{1,p}$ has to lie in the triangle $AA_qC$ , the point $B_{1,p}$ has to lie in the triangle $BB_qA$ , and the point $C_{1,p}$ has to lie in the triangle $CC_qB$ . Again, these three things either all happen, or none of them happen; the critical case to consider is when $A_{1,p}$ lies on $AA_q$ , which occurs when $A_{1,p} = B_2 = C_3$ . The condition for this is $q \; = \; \frac{(1-p)^2}{p^2 + (1-p)^2}$ We only have an internal hexagon when $\frac{\sqrt{1-p}}{\sqrt{p}+\sqrt{1-p}} < q < \frac{(1-p)^2}{p^2 + (1-p)^2}$ . Note that this requires that $0 < p < \tfrac12$ . Then the area of the internal hexagon is equal to the area of the triangle $A_{1,p}B_{1,p}C_{1,p}$ minus the areas of the triangles $A_{1,p}B_2C_3$ , $B_{1,p}C_2A_3$ and $C_{1,p}A_2B_3$ . Standard formulae for the areas of triangles given by trilinear coordinates tell us that the last three triangles all have the same area, and that the area of the internal hexagon is $\frac{\Delta}{(1 - p + p^2)^3}\left|\begin{array}{ccc} p(1-p) & p^2 & (1-p)^2 \\ (1-p)^2 & p(1-p) & p^2 \\ p^2 & (1-p)^2 & p(1-p) \end{array}\right| - \frac{3\Delta}{(1-p+p^2)(1-p+pq)(1-q+pq)}\left|\begin{array}{ccc}p(1-p) & p^2 & (1-p)^2 \\ (1-p)(1-q) & p(1-q) & pq \\ pq & (1-p)(1-q) & (1-p)q \end{array}\right|$ which simplifies to give $\frac{ -2 + 5(p + q) - 3(p^2 + 4pq + q^2) + 8pq(p + q) - 8p^2q^2}{(1 - p + pq)(1 - q + pq)}\Delta$ for $\frac{\sqrt{1-p}}{\sqrt{p}+\sqrt{1-p}} < q < \frac{(1-p)^2}{p^2+(1-p)^2}$ , where $\Delta$ is the area of the original triangle $ABC$ .

If we consider the expression for the area of the internal hexagon with $p = 0.1$ and for $\frac{\sqrt{0.9}}{\sqrt{0.9} + \sqrt{0.1}} = \tfrac34 < q < \tfrac{81}{82} = \frac{0.9^2}{0.1^2+0.9^2}$ , we obtain, solving a quadratic equation for $q$ - thank you, @Atomsky Jahid ) a relative area of $0.7$ for the internal hexagon when $q = \tfrac{720}{739}$ , and hence $\lceil 10^5 q \rceil = \boxed{97429}$ .

This problem is extremely suitable for the use of Routh's theorem . Routh's theorem

The ratio of the area of a cevian triangle to the original triangle depicted above is $\frac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)}$

Now, let's go back to the triangle of our problem. We will first find the area of the red triangle and then subtract the areas of of the smaller triangles around its vertices to get the area of the hexagon. It should be noted that the three smaller triangles whose areas are to be subtracted have equal areas. Our problem

Here, $\frac{BA_{1}}{BC} = \frac{1}{10}$ $\implies \frac{A_{1}C}{BA_{1}} = 9$

Let's assume $\frac{A_{2}C}{BA_{2}} = x$ $\implies \frac{BA_{2}}{BC} = q = \frac{1}{1+x}$

Now, we get the following equation. $\frac{(9^3-1)^2}{(9^2+9+1)^3} - \frac{3(81x-1)^2}{(9x+10)(81+10)(10x+1)} = \frac{7}{10}$ This simplifies to $x = \frac{19}{720}$

Therefore, $q = \frac{1}{1+x} = \frac{720}{739}$

The triangle can be stretched and skewed to an equilateral triangle with unit sides and still preserve the ratio of areas. Placing $B$ at the origin and $C$ at $(1, 0)$ , then $A$ is $(\frac{1}{2}, \frac{\sqrt{3}}{2})$ , $O$ is $(\frac{1}{2}, \frac{\sqrt{3}}{6})$ , $A_1$ is $(\frac{1}{10}, 0)$ , $A_2$ is $(q, 0)$ , $B_1$ is $(\frac{19}{20}, \frac{\sqrt{3}}{20})$ , $C_1$ is $(\frac{9}{20}, \frac{9\sqrt{3}}{20})$ , and $C_2$ is $(\frac{1}{2}(1 - q), \frac{\sqrt{3}}{2}(1 - q))$ .

The equation of the line $BB_1$ is $y = \frac{\sqrt{3}}{2}x$ , the line $CC_2$ is $y = \frac{\sqrt{3}(q - 1)}{q + 1}(x - 1)$ , the line $AA_1$ is $y = \frac{5\sqrt{3}}{4}(x - \frac{1}{10})$ , and the line $AA_2$ is $y = \frac{\sqrt{3}}{1 - 2q}(x - q)$ .

$AA_1$ and $CC_1$ intersect at $D(\frac{9 - 7q}{2q + 18}, \frac{9\sqrt{3} - 9\sqrt{3}q}{2q + 18})$ , $AA_1$ and $BB_1$ intersect at $E(\frac{19q - 19}{18q - 20}, \frac{\sqrt{3}q - \sqrt{3}}{18q - 20})$ , and $AA_2$ and $BB_1$ intersect at $F(\frac{19q}{2q + 18}, \frac{\sqrt{3}q}{2q + 18})$ .

The area of $\triangle DOE$ is $A_{\triangle DOE} = \frac{1}{2}((D_x - O_x)(E_y - O_y) - (E_x - O_x)(D_y - O_y)) = \frac{164\sqrt{3}q^2 - 244\sqrt{3}q + 81\sqrt{3}}{108q^2 + 852q - 1080}$ and the area of $\triangle EOF$ is $A_{\triangle DOE} = \frac{1}{2}((E_x - O_x)(F_y - O_y) - (F_x - O_x)(E_y - O_y)) = \frac{16\sqrt{3}q^2 - 36\sqrt{3}q + 18\sqrt{3}}{27q^2 + 213q - 270}$ .

The area of $\triangle ABC$ is $A_{\triangle ABC} = \frac{\sqrt{3}}{4}$ and by symmetry the area of the hexagon is $3(A_{\triangle DOE} + A_{\triangle EOF})$ , and since $\frac{\text{area of hexagon}}{\text{area of △ABC}} = 0.7$ , we have $\frac{3(\frac{164\sqrt{3}q^2 - 244\sqrt{3}q + 81\sqrt{3}}{108q^2 + 852q - 1080} + \frac{16\sqrt{3}q^2 - 36\sqrt{3}q + 18\sqrt{3}}{27q^2 + 213q - 270})}{\frac{\sqrt{3}}{4}} = \frac{7}{10}$ , which rearranges to $739q^2 - 1459q + 720 = 0$ and solves to $q = \frac{720}{739}$ for $q < 1$ .

Therefore, $\lceil 100000q \rceil = \lceil 100000 \cdot \frac{720}{739} \rceil = \boxed{97429}$ .