Problem 1129

The first solution. The area of triangle $ABC$ is equal to the sum of the areas of triangles $CNF$ and $FTB$ and quadrilateral $TANF$. We will prove that $S_{\triangle CNF} + S_{\triangle FTB} = \frac{1}{2} S_{\triangle ABC}$, from which it will immediately follow that the area of $TANF$ is the remaining half.

Let the side of triangle $ABC$ be $a$, and $NA$ be $b$. Then $CF = FB = \frac{1}{2} a$, and $CN = b - a$. Note immediately that $S_{\triangle ABC} = \frac{\sqrt{3}}{4} a^2$.

$S_{\triangle CNF} + S_{\triangle FTB} = \frac{1}{2} \cdot CN \cdot CF \cdot \sin 60^\circ + \frac{1}{2} \cdot BT \cdot BF \cdot \sin 60^\circ = \frac{\sqrt{3}}{8} (a - b) a + \frac{\sqrt{3}}{8} b a = \frac{\sqrt{3}}{8} a^2$, which is half the area of triangle $ABC$.

The second solution. Triangles $CFA$ and $BFA$ are equal (for example, by two sides and the angle between them: $CF = FB$ by the condition, $AC = AB$, and $\angle ACF = \angle ABF = 60^\circ$, since triangle $ABC$ is equilateral). From this, it follows that the area of triangle $ABF$ is half the area of triangle $ABC$. Also, from the equality $\triangle CFA = \triangle BFA$, it follows that the heights of triangles $CFA$ and $BFA$ from point $F$ are equal. But these heights are also heights of triangles $NAF$ and $FTB$, and since $AN = TB$, the areas of triangles $NAF$ and $FTB$ are equal. It remains to note that $S_{\text{TANF}} = S_{\triangle NAF} + S_{\triangle ATF} = S_{\triangle FTB} + S_{\triangle ATF} = S_{\triangle ABF} = \frac{1}{2} S_{\triangle ABC}$.

Remark. Points are not deducted if the following facts are used without proof. The median $AF$ is the bisector and altitude in triangle $ABC$. $AF = \frac{\sqrt{3}}{2} AB$, $S_{\triangle ABF} = \frac{1}{2} S_{\triangle ABC}$, $\triangle CFA = \triangle BFA$.

Comment. If the proof uses the equality of the heights of triangles $NAF$ and $FTB$, but the proof of this fact is not provided (and the rest of the proof is correct) - 5 points.