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We are tasked with finding \( p + q \) based on the given conditions:

### Problem Breakdown:
- \( f(x) = \frac{1}{2}x^2 \)
- \( g(x) = x^3 \)
- A tangent line \( L \) touches the curve \( y = f(x) \) at the point \( (a, b) \), and it also touches the curve \( y = g(x) \) at the point \( (c, d) \).
- We are given that \( b + d + \frac{ac}{3} = \frac{p}{q} \), where \( a, b, c, d \in \mathbb{R} \), and \( c \neq 0 \).

### Step 1: Find the tangent line at \( (a, b) \) for \( y = f(x) \)
Given \( f(x) = \frac{1}{2}x^2 \), the derivative \( f'(x) \) gives the slope of the tangent line:
\[
f'(x) = x
\]
At \( x = a \), the slope is:
\[
\text{slope of } L = a
\]
The point of tangency is \( (a, b) = (a, f(a)) = (a, \frac{1}{2}a^2) \). Hence, the equation of the tangent line at \( (a, \frac{1}{2}a^2) \) is:
\[
y - \frac{1}{2}a^2 = a(x - a)
\]
Simplifying this:
\[
y = ax - a^2 + \frac{1}{2}a^2 = ax - \frac{1}{2}a^2
\]
So the equation of the tangent line is:
\[
y = ax - \frac{1}{2}a^2
\]

### Step 2: Find the point \( (c, d) \) where this tangent line also contacts the curve \( g(x) \)
We are given that \( g(x) = x^3 \). For the tangent line to contact \( g(x) \) at \( (c, d) \), we know:
\[
d = g(c) = c^3
\]
Also, the slope of the tangent line at \( (c, c^3) \) is given by the derivative of \( g(x) \), which is:
\[
g'(x) = 3x^2
\]
At \( x = c \), the slope is \( 3c^2 \). Since the tangent line has slope \( a \), we equate the slopes:
\[
a = 3c^2
\]
Thus, \( a = 3c^2 \).

### Step 3: Use the condition \( b + d + \frac{ac}{3} = \frac{p}{q} \)
We have:
- \( b = \frac{1}{2}a^2 \)
- \( d = c^3 \)
- \( a = 3c^2 \)

Substitute \( a = 3c^2 \) into the expression for \( b \):
\[
b = \frac{1}{2}(3c^2)^2 = \frac{1}{2} \times 9c^4 = \frac{9}{2}c^4
\]
Now, using \( b + d + \frac{ac}{3} = \frac{p}{q} \):
\[
\frac{9}{2}c^4 + c^3 + \frac{(3c^2)c}{3} = \frac{p}{q}
\]
Simplify the third term on the left-hand side:
\[
\frac{(3c^2)c}{3} = c^3
\]
So the equation becomes:
\[
\frac{9}{2}c^4 + c^3 + c^3 = \frac{p}{q}
\]
\[
\frac{9}{2}c^4 + 2c^3 = \frac{p}{q}
\]

### Step 4: Set \( c = 1 \) to simplify calculations
To find \( p \) and \( q \), we can choose \( c = 1 \), since \( c \neq 0 \). Substituting \( c = 1 \) into the equation:
\[
\frac{9}{2}(1)^4 + 2(1)^3 = \frac{p}{q}
\]
\[
\frac{9}{2} + 2 = \frac{p}{q}
\]
\[
\frac{9}{2} + \frac{4}{2} = \frac{p}{q}
\]
\[
\frac{13}{2} = \frac{p}{q}
\]
Thus, \( p = 13 \) and \( q = 2 \).

### Step 5: Compute \( p + q \)
Finally, \( p + q = 13 + 2 = 15 \).

The value of \( p + q \) is \( \boxed{15} \).

---

Would you like more details on any step or have any questions?  
Here are five related questions you can explore:
1. How do you find the tangent line of a curve at a point?
2. How do you compute the intersection of two curves?
3. What are the conditions for two curves to share a tangent line?
4. How does changing the value of \( c \) affect the solution for \( p + q \)?
5. What is the geometric significance of \( p + q \) in this problem?

**Tip:** When working with tangents, always match both the slope and the point of tangency for consistency across different curves.

Let f(x) = 1/2x^2 and g(x) = x^3. Let L be a tangent line that contacts a curve y = f(x) at (a, b). L also contacts a curve of y = g(x) at (c, d). Let b + d + ac/3 = p/q where a, b, c, d ∈ R and c ≠ 0. Compute p + q.

Learn how to compute p + q for a tangent line problem involving f(x) = 1/2x^2 and g(x) = x^3. This problem covers key calculus concepts such as derivatives, tangent lines, and solving equations with real numbers.

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