aime26-11

Proposed solution

I need to find the value of r for the fourth sphere that's tangent to plane P at point D and also tangent to a second plane T.

First, let me establish coordinates. Since ∠BAC = 90°, I'll place:
- A at the origin (0, 0, 0)
- B on the positive x-axis at (6, 0, 0) 
- C on the positive y-axis at (0, 4, 0)

The centroid G of triangle ABC is at:
G = (1/3)(A + B + C) = (1/3)(0, 0, 0) + (1/3)(6, 0, 0) + (1/3)(0, 4, 0)) = (2, 4/3, 0)

The reflection D of G across line BC can be found as follows:
Line BC has direction vector (0, 4, 0) - (6, 0, 0) = (-6, 4, 0).
The plane containing BC and perpendicular to the plane P is the yz-plane.

Wait, let me reconsider. Line BC goes from B(6, 0, 0) to C(0, 4, 0). The vector from B to C is (-6, 4, 0).

To reflect point G = (2, 4/3, 0) across line BC:
The line BC can be parameterized as (6, 0, 0) + t(-6, 4, 0) = (6-6t, 4t, 0)

The projection of G onto line BC is found by minimizing the distance from G to points on the line.
Let P(t) = (6-6t, 4t, 0)
The vector from G to P(t) is: (6-6t-2, 4t-4/3, 0) = (4-6t, 4t-4/3, 0)

For the projection, this vector should be perpendicular to the direction vector (-6, 4, 0):
(4-6t)(-6) + (4t-4/3)(4) = 0
-24 + 36t + 16t - 16/3 = 0
52t = 24 + 16/3 = 72/3 + 16/3 = 88/3
t = 88/(3×52) = 88/156 = 22/39

The projection point is:
P(22/39) = (6-6×22/39, 4×22/39, 0) = (6-132/39, 88/39, 0) = (234/39-132/39, 88/39, 0) = (102/39, 88/39, 0)

The reflection D is:
D = 2×(102/39, 88/39, 0) - (2, 4/3, 0)
= (204/39, 176/39, 0) - (2, 4/3, 0)
= (204/39, 176/39, 0) - (78/39, 52/39, 0)
= (126/39, 124/39, 0)

Let me simplify: 126/39 = 42/13 and 124/39 = 124/39.

Actually, let me double-check this calculation more carefully.

The line BC goes from (6,0,0) to (0,4,0). A point on this line is:
(6,0,0) + s((0,4,0)-(6,0,0)) = (6,0,0) + s(-6,4,0) = (6-6s, 4s, 0)

The vector from G(2, 4/3, 0) to a point on the line is:
(6-6s-2, 4s-4/3, 0) = (4-6s, 4s-4/3, 0)

This should be perpendicular to (-6, 4, 0):
(4-6s)(-6) + (4s-4/3)(4) = 0
-24 + 36s + 16s - 16/3 = 0
52s = 24 + 16/3 = 72/3 + 16/3 = 88/3
s = 88/(3×52) = 88/156 = 22/39

Projection point: (6-6(22/39), 4(22/39), 0) = (6-132/39, 88/39, 0) = (234/39-132/39, 88/39, 0) = (102/39, 88/39, 0)

Reflection D: 2×(102/39, 88/39, 0) - (2, 4/3, 0)
= (204/39, 176/39, 0) - (78/39, 52/39, 0) = (126/39, 124/39, 0)

Let me simplify: 126/39 = 42/13 and 124/39.

Actually, let me work with a cleaner approach. Let's use the fact that for reflection across line BC:
If we have point G and want to reflect it across line through B,C, we can use:
D = G + 2×(projection of (G-B) onto direction vector of BC) - (G-B))

But let me just work with the coordinates I computed.

The points are:
A = (0, 0, 0)
B = (6, 0, 0)  
C = (0, 4, 0)
D = (126/39, 124/39, 0) = (42/13, 124/39, 0)

Now for the spheres:
- Sphere with radius 1 is tangent to plane P at A(0,0,0)
- Sphere with radius 2 is tangent to plane P at B(6,0,0)  
- Sphere with radius 3 is tangent to plane P at C(0,4,0)
- Sphere with radius r is tangent to plane P at D(42/13, 124/39, 0)

Let the second plane T be at distance h from plane P (both sides of P).
The spheres are all on the same side of T.

For a sphere with center (x,y,z) and radius k that's tangent to plane P (z=0) at point (a,b,0), the center is at (a,b,k).

So:
- Sphere 1: center at (0,0,1)
- Sphere 2: center at (6,0,2)  
- Sphere 3: center at (0,4,3)
- Sphere 4: center at (42/13, 124/39, r)

Let the second plane T be at z = h. All spheres are tangent to this plane and on the same side of it.

The distance from each sphere's center to plane T must equal the radius:
- Distance from (0,0,1) to plane z=h is |1-h| = 1
- Distance from (6,0,2) to plane z=h is |2-h| = 2  
- Distance from (0,4,3) to plane z=h is |3-h| = 3
- Distance from (42/13, 124/39, r) to plane z=h is |r-h| = r

From the first three equations:
|1-h| = 1 → h = 0 or h = 2
|2-h| = 2 → h = 0 or h = 4  
|3-h| = 3 → h = 0 or h = 6

The only common solution is h = 0, but that would put the spheres on oppos