A Triangle Evolves Into A Hexagon

Let $R$ be the radius of the circumcircle. The quadrilateral $BCB'C'$ is a rectangle, and $BC' \; = \; \sqrt{4R^2 - a^2} \; = \; \sqrt{a^2\mathrm{cosec}^2\,A - a^2} \; = \; a\cot A$ using the extended Sine Rule, and so $BCB'C'$ has area $a^2\cot A$ . Similarly the rectangle $ACA'C'$ has area $b^2\cot B$ , while the rectangle $ABA'B'$ has area $c^2 \cot C$ . The sum of these three rectangle areas, namely $a^2\cot A + b^2 \cot B + c^2 \cot C$ is equal to the area of the hexagon plus twice the area of the triangle $ABC$ .

On the other hand, the triangle $OBC$ has area $\tfrac12R^2 \sin 2A \; = \; \tfrac18a^2 \mathrm{cosec}^2\,A \sin2A \; = \; \tfrac14a^2 \cot A\;,$ and hence the area of the triangle $ABC$ can be written as $\tfrac14(a^2 \cot A + b^2\cot B + c^2 \cot C) \;.$ Thus four times the area of the triangle is the area of the hexagon plus twice the area of the triangle. Thus the area of the hexagon is twice the area of the triangle, namely $36$ square units.

Let $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$ be the three vertices.

Points diametrically opposite to these would be their reflections about the origin: $A'(-x_1, -y_1), B'(-x_2, -y_2), C'(-x_3, -y_3)$

Area of $\Delta ABC$ = $\frac {1}{2} \left| \begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\x_3 & y_3 & 1 \\\end{matrix} \right|$ this expands into six terms.

Area expressions of the three peripheral triangles A'BC, AB'C, ABC' will be identical except for the sign of four terms (which would cyclically vary for the three triangles).

In the sum of the peripheral areas, each term would appear thrice - once with one sign and twice with opposite signs. The pair with opposite sign would cancel out, leaving behind just one instance of each term in the sum. Which will be nothing but the expression for the area of $\Delta ABC$

Method II (This is probably not what Karthik has in mind, but found the cyclic nature of these areas interesting)

The areas can be numbered as shown (have used even for original triangle, odd for the peripherals) If the diameters AA', BB', CC' are taken as bases, then the other radii become the medians. Then the areas 1, 3, 5, 7, 9, 11 can be expressed as

1 = 4 + 6 - 2

3 = 2 + 12 - 4

5 = 8 + 10 - 6

7 = 4 + 6 - 8

9 = 2 + 12 - 10

11 = 8 + 10 - 12

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(1+3+5+7+9+11) = (2+4+6+8+10+12) as required

Is it right to consider it an equilateral triangle and then solve?

OAC' +OBC'+OA'B+OA'C+OB'C+OAB' is area of hexagon. This is OAC+OBC+OAB+OAC+OBC+OAB. Hence twice area of triangle

Trick solution: Consider $\bigtriangleup ABC$ as a equilateral triangle. It becomes immediately obvious that $AB'CA'BC'$ is a regular hexagon. Since $\bigtriangleup AOB \cong \bigtriangleup AC'B$ , $\bigtriangleup BOC \cong \bigtriangleup BA'C$ and $\bigtriangleup COA \cong \bigtriangleup CB'A$ , we can deem that the area of the polygon is twice the area of the equilateral triangle and hence the area is 36.

(This isn't a proof, it's just how I got the right answer.