Daughter Triangles

The three cevians through the points $T_A$ , $T_B$ , $T_C$ of tangency with the incircle are concurrent at the Gergonne point $Ge$ . Suppose that $AT_CGe$ is equilateral. We know that $AT_A^2 \; = \; c^2 + (s-b)^2 - 2(s-b)c \cos B \hspace{2cm} CT_C^2 \; = \; a^2 + (s-b)^2 - 2(s-b)a \cos B \hspace{2cm} \cos B \; = \; \frac{a^2 + c^2 - b^2}{2ac}$

at all times. Since the angles $\angle T_AAB = \angle AT_CC = 60^\circ$ , we also deduce that $(s-b)^2 \; = \; AT_A^2 + c^2 - c AT_A \hspace{2cm} b^2 \; = \; (s-a)^2 + CT_C^2 - (s-a)CT_C$ These equations are all homogeneous of degree $2$ , and so we can attempt to solve for $b,c$ in terms of $a$ . It turns out that there are only two real solutions of these equations:

• $b = a - c$ ,
• $b = \tfrac{19}{49}a$ , $c = \tfrac{40}{49}a$

The first solution only produces a trivial triangle, so the second is the only proper solution. If we put $a = 49$ , $b=19$ and $c=40$ , then $s = 54$ . Thus the equilateral triangle $AT_CGe$ has area $\tfrac{\sqrt{3}}{4}(s-a)^2 = \tfrac{25}{4}\sqrt{3}$ . On the other hand the original triangle has area $\sqrt{s(s-a)(s-b)(s-c)} = 210\sqrt{3}$ , and hence the ratio of areas is $\frac{210\sqrt{3} \times 4}{25\sqrt{3}} \; = \; \frac{168}{5} = \boxed{33.6}$