aime26-09

Proposed solution

To solve the problem, we start by finding the circumradius \( R \) of triangle \( \triangle ABC \) with side lengths \( AB = 13 \), \( BC = 14 \), and \( CA = 15 \). We use the formula for the circumradius of a triangle given its side lengths \( a, b, c \):

\[
R = \frac{abc}{4K}
\]

where \( K \) is the area of the triangle. First, we calculate the area \( K \) using Heron's formula. The semi-perimeter \( s \) is:

\[
s = \frac{a + b + c}{2} = \frac{13 + 14 + 15}{2} = 21
\]

Then, the area \( K \) is:

\[
K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{7056} = 84
\]

Now, substituting into the formula for \( R \):

\[
R = \frac{13 \cdot 14 \cdot 15}{4 \cdot 84} = \frac{2730}{336} = \frac{65}{8} = 8.125
\]

Next, we rotate \( \triangle ABC \) about its circumcenter such that \( AC \) becomes perpendicular to \( BC \). This rotation results in a new triangle \( \triangle A'B'C' \) where \( A' \) and \( B \) are not on the same side of \( B'C' \). We need to find the coordinates of the vertices after rotation and then calculate the area of hexagon \( AA'CC'BB' \).

We place \( B \) at the origin \((0,0)\), \( C \) at \((14,0)\), and \( A \) at coordinates \((x,y)\) such that:

\[
x^2 + y^2 = 13^2 \quad \text{and} \quad (x-14)^2 + y^2 = 15^2
\]

Solving these equations, we get:

\[
x^2 + y^2 = 169
\]
\[
(x-14)^2 + y^2 = 225
\]

Expanding and simplifying the second equation:

\[
x^2 - 28x + 196 + y^2 = 225
\]
\[
169 - 28x + 196 = 225
\]
\[
-28x + 196 = 56
\]
\[
-28x = -140
\]
\[
x = 5
\]

Substituting \( x = 5 \) back into \( x^2 + y^2 = 169 \):

\[
25 + y^2 = 169
\]
\[
y^2 = 144
\]
\[
y = 12 \quad \text{(since y is positive)}
\]

Thus, \( A \) is at \((5, 12)\). After rotation, \( A' \) will be at \((x', y')\) such that \( A'C \) is perpendicular to \( BC \). Since \( AC \) is rotated 90 degrees, \( A' \) will be at \((12, 5)\).

Now, we need to find the coordinates of \( B' \). Since \( B \) is at the origin and \( B' \) is the image of \( B \) after rotation, we need to find the new position of \( B \) after the rotation. The rotation matrix for 90 degrees counterclockwise is:

\[
\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}
\]

Applying this to \( B \):

\[
\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}
\begin{pmatrix}
0 \\
0
\end{pmatrix}
=
\begin{pmatrix}
0 \\
0
\end{pmatrix}
\]

Thus, \( B' \) remains at \((0,0)\).

The vertices of hexagon \( AA'CC'BB' \) are \( A(5,12) \), \( A'(12,5) \), \( C(14,0) \), \( B(0,0) \), and \( B'(0,0) \). We need to calculate the area of this hexagon. We can break it into simpler shapes, such as triangles and rectangles.

First, we calculate the area of \( \triangle ABC \):

\[
K = 84
\]

Next, we calculate the area of \( \triangle A'BC \):

\[
K = 84
\]

Finally, we calculate the area of \( \triangle A'CB' \):

\[
K = 84
\]

Summing these areas:

\[
84 + 84 + 84 = 252
\]

Thus, the area of hexagon \( AA'CC'BB' \) is approximately 252. The closest integer to this area is:

\[
\boxed{252}
\]