The vertices of a $3 - 4 - 5$ right triangle are the centers of three mutually externally tangent circles, as shown.   What is the sum of the areas of these circles?

[asy]unitsize(1cm);
draw(Circle((1.8,2.4),1),linewidth(0.7));
draw(Circle((0,0),2),linewidth(0.7));
draw(Circle((5,0),3),linewidth(0.7));
draw((0,0)--(5,0)--(1.8,2.4)--cycle,linewidth(0.7));
label("$A$",(1.8,2.4),N);
label("$B$",(0,0),SW);
label("$C$",(5,0),E);
label("5",(2.5,0),S);
label("4",(3.4,1.2),NE);
label("3",(0.9,1.2),NW);
[/asy]
Let $r$, $s$, and $t$ be the radii of the circles centered at $A$, $B$, and $C$, respectively. Then $r+s=3$, $r+t=4$, and $s+t=5$.  Adding these three equations gives $2(r+s+t) = 12$, so $r+s+t = 6$.  Combining this with the original three equations, we get $r=1$, $s=2$, and $t=3$. Thus the sum of the areas of the circles is \[
\pi(1^2+2^2+3^2)=\boxed{14\pi}.
\]
14\pi
Level 2
Geometry
Problem: The vertices of a $3 - 4 - 5$ right triangle are the centers of three mutually externally tangent circles, as shown.   What is the sum of the areas of these circles?

[asy]unitsize(1cm);
draw(Circle((1.8,2.4),1),linewidth(0.7));
draw(Circle((0,0),2),linewidth(0.7));
draw(Circle((5,0),3),linewidth(0.7));
draw((0,0)--(5,0)--(1.8,2.4)--cycle,linewidth(0.7));
label("$A$",(1.8,2.4),N);
label("$B$",(0,0),SW);
label("$C$",(5,0),E);
label("5",(2.5,0),S);
label("4",(3.4,1.2),NE);
label("3",(0.9,1.2),NW);
[/asy]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the sum of the areas of the three circles, we first denote the radii of the circles centered at points A, B, and C as r_A, r_B, and r_C, respectively. The circles are mutually externally tangent, which means the distance between the centers of any two circles equals the sum of their radii.

From the triangle, we know the lengths of the sides: AB = 4, BC = 5, and AC = 3. We can set up the following equations based on the distances between the centers:

1. The distance between A and B: r_A + r_B = 4
2. The distance between B and C: r_B + r_C = 5
3. The distance between A and C: r_A + r_C = 3

We now have a system of three equations:

1. r_A + r_B = 4
2. r_B + r_C = 5
3. r_A + r_C = 3

To solve for the radii, we can express r_B in terms of r_A from the first equation:

r_B = 4 - r_A

Substituting r_B into the second equation gives:

(4 - r_A) + r_C = 5
r_C = 5 - 4 + r_A
r_C = 1 + r_A

Now we substitute r_C into the third equation:

r_A + (1 + r_A) = 3
2r_A + 1 = 3
2r_A = 2
r_A = 1

Now that we have r_A, we can find r_B and r_C:

r_B = 4 - r_A = 4 - 1 = 3
r_C = 1 + r_A = 1 + 1 = 2

Now we have the radii of the circles: r_A = 1, r_B = 3, and r_C = 2. 

Next, we calculate the areas of the circles using the formula for the area of a circle, A = πr^2:

Area of circle A: A_A = π(1^2) = π
Area of circle B: A_B = π(3^2) = 9π
Area of circle C: A_C = π(2^2) = 4π

Now, we sum the areas:

Total area = A_A + A_B + A_C = π + 9π + 4π = 14π.

Thus, the final answer is 14π.<|END_OF_LATENT|> The final answer is $14\pi$.