aime26-11

Proposed solution

Given triangle \( \triangle ABC \) with \( AB = 6 \), \( AC = 4 \), and \( \angle BAC = 90^\circ \), we first find the coordinates of the vertices assuming \( A = (0,0) \), \( B = (6,0) \), and \( C = (0,4) \). The centroid \( G \) of \( \triangle ABC \) is given by:
\[
G = \left( \frac{0+6+0}{3}, \frac{0+0+4}{3} \right) = \left( 2, \frac{4}{3} \right)
\]

Next, we reflect \( G \) across \( BC \). The line \( BC \) has the equation \( y = -\frac{2}{3}x + 4 \). The reflection of a point \( (x_1, y_1) \) across a line \( ax + by + c = 0 \) is given by:
\[
(x', y') = \left( \frac{x_1(a^2 - b^2) - 2by_1 - 2ac}{a^2 + b^2}, \frac{y_1(b^2 - a^2) - 2ax_1 - 2bc}{a^2 + b^2} \right)
\]
For \( G = (2, \frac{4}{3}) \) and the line \( -\frac{2}{3}x + y - 4 = 0 \), we have \( a = -\frac{2}{3} \), \( b = 1 \), and \( c = -4 \). Substituting these values, we get:
\[
x' = \frac{2 \left( \left( -\frac{2}{3} \right)^2 - 1^2 \right) - 2 \left( -\frac{2}{3} \right) \left( \frac{4}{3} \right) - 2 \left( -\frac{2}{3} \right)(-4)}{ \left( -\frac{2}{3} \right)^2 + 1^2 } = \frac{2 \left( \frac{4}{9} - 1 \right) + \frac{16}{9} + \frac{32}{9}}{\frac{4}{9} + 1} = \frac{2 \left( \frac{4 + 16 + 32}{9} \right)}{\frac{13}{9}} = \frac{2 \cdot 52}{13} = \frac{104}{13} = 8
\]
\[
y' = \frac{\frac{4}{3} \left( 1^2 - \left( -\frac{2}{3} \right)^2 \right) - 2 \left( -\frac{2}{3} \right) \cdot 2 - 2 \cdot 1 \cdot (-4)}{ \left( -\frac{2}{3} \right)^2 + 1^2 } = \frac{\frac{4}{3} \left( 1 - \frac{4}{9} \right) + \frac{8}{3} + 8}{\frac{13}{9}} = \frac{\frac{4}{3} \cdot \frac{5}{9} + \frac{8}{3} + 8}{\frac{13}{9}} = \frac{\frac{20}{27} + \frac{72}{27} + \frac{216}{27}}{\frac{13}{9}} = \frac{208}{27} \cdot \frac{9}{13} = \frac{208}{27} \cdot \frac{3}{13} = \frac{208}{39} = \frac{208}{39}
\]
Thus, the reflection \( D \) of \( G \) across \( BC \) is \( D = (8, \frac{208}{39}) \).

We now consider the four spheres with radii 1, 2, 3, and \( r \) tangent to the plane \( \mathcal{P} \) at points \( A, B, C, \) and \( D \), respectively. Each sphere is also tangent to a second plane \( \mathcal{T} \). The distance from the center of each sphere to the plane \( \mathcal{P} \) is equal to its radius. Therefore, the centers of the spheres are \( (1, 0) \), \( (6, 0) \), \( (0, 4) \), and \( (8, \frac{208}{39}) \) respectively.

To find \( r \), we need to determine the distance from \( D \) to the plane \( \mathcal{P} \). The distance from a point \( (x_1, y_1) \) to a plane \( ax + by + cz + d = 0 \) is given by:
\[
\frac{|ax_1 + by_1 + cz + d|}{\sqrt{a^2 + b^2 + c^2}}
\]
Since the plane \( \mathcal{P} \) is \( z = 0 \), the distance from \( D = (8, \frac{208}{39}, 0) \) to \( \mathcal{P} \) is simply \( \frac{208}{39} \). Therefore, the radius \( r \) of the sphere tangent to \( D \) is:
\[
r = \frac{208}{39}
\]

We simplify \( \frac{208}{39} \) by finding the greatest common divisor (GCD) of 208 and 39. The prime factorization of 208 is \( 2^4 \cdot 13 \) and for 39 is \( 3 \cdot 13 \). The GCD is 13. Thus,
\[
\frac{208}{39} = \frac{208 \div 13}{39 \div 13} = \frac{16}{3}
\]

So, \( r = \frac{16}{3} \). The value of \( m + n \) where \( r = \frac{m}{n} \) is \( 16 + 3 = 19 \).

The final answer is:
\[
\boxed{19}
\]