Multiple Triangle Problem Suppose there is a hypothetical acute triangle with an area of 1. The altitudes on this triangle are placed so they cut the triable into six smaller triangles. The areas of these triangles are $t_1$, $t_2$, $t_3$, $t_4$, $t_5$, and $t_6$. The triangles are labeled in a counterclockwise manner. Given that this is true, if the given equation $t_1t_3t_5+ t_2t_4t_6=5$, what would the solution for $t_1t_2t_3t_4t_5t_6$ be.
 A: The short answer is that it can be shown that $t_1t_3t_5=t_2t_4t_6.$ (A surprise, to me!) So if their sum is $5$ then each is $5/2$ and their product $t_1t_2t_3t_4t_5t_6=25/4.$
Without going into all the details, set up the acute triangle with vertices $$A=(-a,0),\ B=(b,0),\ C=(0,c).$$
Then the three altitudes meet at the point $X=(0,ab/c).$ Note that this is below the point $C$ in an acute triangle, i.e. $ab/c<c,$ since the addition law for tangent gives the tangent of angle $C$ as $(a/c+b/c)/(1-ab/c^2).$
The six triangles of the cut up triangle can then be numbered counterclockwise $t_k$ beginning with triangle $OBX$ for $t_1$ where $O=(0,0)$ is the origin. One can get expressions for $M,$ the foot of the perpendicular from $A$ to side $BC$, and for $N,$ the foot of the perpendicular from $B$ to side $AC.$ One gets
$$M_x=\frac{(c^2-ab)b}{b^2+c^2},\ \ M_y=\frac{(a+b)bc}{b^2+c^2},$$
and also 
$$N_x=-\frac{(c^2-ab)b}{b^2+c^2},\ \ N_y=\frac{(a+b)ac}{a^2+c^2}.$$
Now that the relevant coordinates have expressions, one can get the areas for each of the subtriangles $t_j$ by using a formula based on cross products. (This is the detail not put here, but can be supplied if desired.)
The strange thing (to me) is that when I then computed the two products $8t_1t_3t_5$ and $8t_2t_4t_6$ they each came out the same expression
$$\frac{a^3b^3(c^2-ab)^3(a+b)}{(a^2+c^2)(b^2+c^2)c^3}.$$
Thus the two products being the same makes the question of the post simple. I guess if one already knew that result (for acute triangles these area products are equal) the problem could arguably be said to be trivial.
Added note: The above calculation didn't use that the area of the larger triangle is $1$. But note that with this assumption, the data of the problem cannot be right, since if so each $t_j<1$ implying each product of three is at most $1$, so adding two such cannot be $5$. I would think the problem should be asked without even giving the area of the larger triangle, or else at least some particular case should be used, and then the problem should give the area of the larger triangle along with a compatible value for the sum of the two triple products. [In this last case it would still be trivial provided one knew the triple products had to be equal.]
