Triangle Tangled

By Heron's formula, the area of the bigger triangle is $A = \sqrt{21(21-13)(21-14)(21-15)} = 84$ . Let the unknown sides of the three smaller triangles be $a$ , $b$ , and $c$ as shown in the figure, and the common perimeter be $2s$ . Then we have:

$\begin{cases} a+b+13 = 2s & ...(1) \\ b+c+14 = 2s & ...(2) \\ c+a+15 = 2s & ...(3) \end{cases} \implies \begin{cases} (1) - (2): & c = a - 1 \\ (2) - (3): & b = a+ 1 \\ \frac {(1)-(2)+(3)}2: & s = a + 7 \end{cases}$

Applying Heron's formula to the three smaller triangle:

$\begin{cases} A_1 = \sqrt{s(s-a)(s-b)(s-13)} = \sqrt{42(a+7)(a-6)} \\ A_2 = \sqrt{s(s-b)(s-c)(s-14)} = \sqrt{48(a+7)(a-7)} \\ A_3 = \sqrt{s(s-c)(s-a)(s-15)} = \sqrt{56(a+7)(a-8)} \end{cases}$

Then we have $A_1+A_2+A_3 = \sqrt{42(a+7)(a-6)} + \sqrt{48(a+7)(a-7)} + \sqrt{56(a+7)(a-8)} = 84$ . Solving for $a$ (I did it using numerical method), we have $a= \frac {91}{11}$ . Then,

$\begin{cases} A_1 = \dfrac {\sqrt{42(168)(25)}}{11} = \dfrac {420}{11} \\ A_2 = \dfrac {\sqrt{48(168)(14)}}{11} = \dfrac {336}{11} \\ A_3 = \dfrac {\sqrt{56(168)(3)}}{11} = \dfrac {168}{11} \end{cases} \implies A_3 : A_2 : A_1 = \boxed{2:4:5}$

Take origin of coordinates at the point where the sides 13 and 14 meet, and put the x-axis along the 14 side. Let the point where the three triangles meet be P(x,y). Use the cosine rule to show that the cosine of the angle at the origin is 5/13. Hence the coordinates of the vertices of the big triangle are O(0,0), B(14,0), and C(5,12).

Use the equal perimeters fact to show that if PB = d, then PC= d + 1, and PO = d + 2.

Then for OP, $x^{2}+y^{2} = (d+2)^{2}$

For BP, $(x-14)^{2}+ y^{2} = d^{2}$

And for CP, $(x-5)^{2} + (y-12)^{2} = (d+1)^{2}$

The first pair of equations gives $d = 7x - 50$ , And the second pair gives $y = x/6+ 3$ .

Substituting these results into the first equation gives a quadratic in x, and we find that P is the point (90/11, 48/11).

The areas of the two triangles with vertex at O are then found from $(x_{1} y_{2} - x_{2} y_{1})/2$ . The big triangle OBC has base 14 and perpendicular height 12, and area 84. The three small triangle areas then prove to be 168, 336, and 420, each divided by 11, giving the ratio 2:4:5.