In triangle $XYZ,$ circles are drawn centered at $X$, $Y$, and $Z$, so that all pairs of circles are externally tangent. If $XY = 8,$ $XZ = 10,$ and $YZ = 12$, then find the sum of the areas of all three circles.

Asymptote code below
[asy]
unitsize(1 cm);

pair A, B, C;
real[] r;

A = (1,2);
B = (0,0);
C = (3,0);
r[1] = (abs(A - B) + abs(A - C) - abs(B - C))/2;
r[2] = (abs(A - B) + abs(B - C) - abs(A - C))/2;
r[3] = (abs(A - C) + abs(B - C) - abs(A - B))/2;

draw(A--B--C--cycle);
draw(Circle(A,r[1]));
draw(Circle(B,r[2]));
draw(Circle(C,r[3]));

label("$X$", A, N);
label("$Y$", B, SW);
label("$Z$", C, SE);
[/asy]

We are given a triangle $ XYZ $ with side lengths $ XY = 8 $, $ XZ = 10 $, and $ YZ = 12 $. Circles are centered at the vertices $ X $, $ Y $, and $ Z $, and all pairs of circles are externally tangent. We need to find the sum of the areas of these circles.

### Step 1: Denote the Radii of the Circles

Let:
- $ r_X $ be the radius of the circle centered at $ X $,
- $ r_Y $ be the radius of the circle centered at $ Y $,
- $ r_Z $ be the radius of the circle centered at $ Z $.

Since the circles are externally tangent to each other, the sum of the radii of two circles centered at any two vertices equals the distance between those two vertices. This gives us the following system of equations:
\[
r_X + r_Y = XY = 8,
\]
\[
r_X + r_Z = XZ = 10,
\]
\[
r_Y + r_Z = YZ = 12.
\]

### Step 2: Solve the System of Equations

We now solve the system of equations to find $ r_X $, $ r_Y $, and $ r_Z $.

1. From $ r_X + r_Y = 8 $, we can express $ r_Y $ as:
   \[
   r_Y = 8 - r_X.
   \]

2. Substitute this expression for $ r_Y $ into the equation $ r_Y + r_Z = 12 $:
   \[
   (8 - r_X) + r_Z = 12,
   \]
   which simplifies to:
   \[
   r_Z = r_X + 4.
   \]

3. Now substitute $ r_Z = r_X + 4 $ into the equation $ r_X + r_Z = 10 $:
   \[
   r_X + (r_X + 4) = 10,
   \]
   which simplifies to:
   \[
   2r_X + 4 = 10.
   \]
   Solving for $ r_X $, we get:
   \[
   2r_X = 6 \quad \Rightarrow \quad r_X = 3.
   \]

4. Substitute $ r_X = 3 $ into $ r_Y = 8 - r_X $ to find $ r_Y $:
   \[
   r_Y = 8 - 3 = 5.
   \]

5. Finally, substitute $ r_X = 3 $ into $ r_Z = r_X + 4 $ to find $ r_Z $:
   \[
   r_Z = 3 + 4 = 7.
   \]

Thus, the radii of the circles are $ r_X = 3 $, $ r_Y = 5 $, and $ r_Z = 7 $.

### Step 3: Calculate the Areas of the Circles

The area of a circle is given by the formula $ A = \pi r^2 $. Therefore, the areas of the three circles are:
- Area of the circle centered at $ X $: $ A_X = \pi r_X^2 = \pi \times 3^2 = 9\pi $,
- Area of the circle centered at $ Y $: $ A_Y = \pi r_Y^2 = \pi \times 5^2 = 25\pi $,
- Area of the circle centered at $ Z $: $ A_Z = \pi r_Z^2 = \pi \times 7^2 = 49\pi $.

### Step 4: Sum of the Areas

The sum of the areas of the three circles is:
\[
A_X + A_Y + A_Z = 9\pi + 25\pi + 49\pi = 83\pi.
\]

Thus, the sum of the areas of the three circles is $ \boxed{83\pi} $.

Do you want details on any specific part or have any questions? Here are some related questions:

1. How would the problem change if the circles were internally tangent instead of externally tangent?
2. What is the general method for solving such systems of equations in geometry?
3. Can we find the circumradius of triangle $ XYZ $ given its side lengths?
4. How would the solution differ if one of the side lengths was unknown?
5. What if the circles were overlapping instead of tangent—how would the areas be affected?

**Tip:** When solving systems of linear equations, substitution or elimination are powerful techniques depending on the number of variables and how the equations relate.

In triangle $XYZ,$ circles are drawn centered at $X$, $Y$, and $Z$, so that all pairs of circles are externally tangent. If $XY = 8,$ $XZ = 10,$ and $YZ = 12$, then find the sum of the areas of all three circles.

Find the sum of the areas of three circles centered at the vertices of triangle XYZ, where the circles are externally tangent. Given side lengths XY = 8, XZ = 10, and YZ = 12, this geometry problem covers solving systems of equations and applying circle properties. Ideal for high school students in Grades 9-12.

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