aime26-09

Proposed solution

I need to find the area of hexagon $AA'CC'BB'$ where $\triangle A'B'C'$ is a rotation of $\triangle ABC$ about its circumcenter.

First, let me find the circumradius of $\triangle ABC$ with sides $AB = 13$, $BC = 14$, and $CA = 15$.

Using the formula $R = \frac{abc}{4K}$ where $K$ is the area.

I'll find the area using Heron's formula:
$s = \frac{13 + 14 + 15}{2} = 21$

$K = \sqrt{s(s-a)(s-b)(s-c)}} = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{7056} = 84$

So $R = \frac{13 \cdot 14 \cdot 15}{4 \cdot 84} = \frac{2730}{336} = \frac{65}{8}$

Now I need to understand the rotation. The triangle is rotated about its circumcenter so that $\overline{AC}$ becomes perpendicular to $\overline{BC}$.

Let me place the triangle in a coordinate system with the circumcenter at the origin.

For a triangle inscribed in a circle of radius $R$, if we place the circumcenter at the origin, then each vertex is at distance $R$ from the origin.

Let me think about this more systematically. The rotation moves $\triangle ABC$ to $\triangle A'B'C'$ such that:
- $A'C' \perp BC$
- $A'$ and $B$ are on opposite sides of line $B'C'$

Since we're rotating about the circumcenter, all vertices maintain distance $R$ from the origin.

Let me consider the angle of rotation. If I place $C$ at some position, then after rotation, $C'$ will be at the same position (since it's the center of rotation).

Wait, let me reconsider what "rotating about its circumcenter" means.

Actually, since we're rotating $\triangle ABC$ about its circumcenter to get $\triangle A'B'C'$, and we want $A'C' \perp BC$, I need to find the angle of rotation.

Let me place the triangle in a coordinate system with circumcenter at origin.
Let's say $C$ is at some point on the circle. After rotation, $C'$ will be at the same point (since it's the center of rotation).

Actually, let me think about this more carefully. The problem says we rotate $\triangle ABC$ about its circumcenter so that after rotation:
- $\overline{AC}$ becomes perpendicular to $\overline{BC}$
- $A'$ and $B$ are not on the same side of line $B'C'$

This means we're rotating the triangle so that the image of $AC$ is perpendicular to the image of $BC$.

Let me think about the rotation angle. If I rotate by angle $\theta$, then:
- The vector from circumcenter to $A$ rotates by $\theta$
- The vector from circumcenter to $C$ rotates by $\theta$

But we want the new position of $AC$ (which is $A'C'$) to be perpendicular to the new position of $BC$ (which is $B'C'$).

Let me approach this differently. I'll find the area of the hexagon directly.

The hexagon $AA'CC'BB'$ has vertices that are alternating between original and rotated positions.

Since we're rotating about the circumcenter, all points are at distance $R = \frac{65}{8}$ from the origin.

The key insight is that the hexagon consists of two triangles sharing a common vertex at the circumcenter, plus some other regions.

Actually, let me reconsider what the problem means. 

If we rotate $\triangle ABC$ about its circumcenter to get $\triangle A'B'C'$, then:
- $A'$ is where $A$ ends up after rotation
- $B'$ is where $B$ ends up after rotation  
- $C'$ is where $C$ ends up after rotation

But since we're rotating about the circumcenter, $C' = C$, so $C = C'$.

Wait, that's not right. The problem says "Triangle $\triangle A'B'C'$ is obtained by rotating $\triangle ABC$ about its circumcenter". This means we rotate the entire triangle, so each vertex maps to a new position.

Let me re-read carefully: "$\overline{AC}$ is perpendicular $\overline{BC}$" - this should be "$\overline{AC}$ is perpendicular to $\overline{BC}$".

So after rotation, $A'C' \perp B'C'$.

The hexagon $AA'CC'BB'$ has vertices in that order. Since we're rotating about the circumcenter, and all points are distance $R$ from center:
- $|OA| = |OA'| = |OC| = |OC'| = |OB| = |OB'| = R$

Let me consider the rotation angle. If we rotate by angle $\alpha$, then:
- $A' = \text{rotation of } A \text{ by angle } \alpha$
- $