Find the number of 10-tuples $(x_1, x_2, \dots, x_{10})$ of real numbers such that
\[(1 - x_1)^2 + (x_1 - x_2)^2 + (x_2 - x_3)^2 + \dots + (x_9 - x_{10})^2 + x_{10}^2 = \frac{1}{11}.\]
By the Cauchy-Schwarz inequality,
\begin{align*}
&[(1^2 + 1^2 + 1^2 + \dots + 1^2 + 1^2)][(1 - x_1)^2 + (x_1 - x_2)^2 + (x_2 - x_3)^2 + \dots + (x_9 - x_{10})^2 + x_{10}^2] \\
&\ge [(1 - x_1) + (x_1 - x_2) + (x_2 - x_3) + \dots + (x_9 - x_{10}) + x_{10}]^2 = 1.
\end{align*}From the given condition, we have equality, so by the equality condition for Cauchy-Schwarz,
\[\frac{1 - x_1}{1} = \frac{x_1 - x_2}{1} = \frac{x_2 - x_3}{1} = \dots = \frac{x_9 - x_{10}}{1} = \frac{x_{10}}{1}.\]Let
\[d = 1 - x_1 = x_1 - x_2 = x_2 - x_3 = \dots = x_9 - x_{10} = x_{10}.\]Then
\[(1 - x_1) + (x_1 - x_2) + \dots + (x_9 - x_{10}) + x_{10} = 11d,\]so $11d = 1.$  Then $d = \frac{1}{11},$ so
\[(x_1, x_2, x_3, \dots, x_{10}) = \left( \frac{10}{11}, \frac{9}{11}, \frac{8}{11}, \dots, \frac{1}{11} \right).\]In particular, there is only $\boxed{1}$ solution.
1
Level 4
Intermediate Algebra
Problem: Find the number of 10-tuples $(x_1, x_2, \dots, x_{10})$ of real numbers such that
\[(1 - x_1)^2 + (x_1 - x_2)^2 + (x_2 - x_3)^2 + \dots + (x_9 - x_{10})^2 + x_{10}^2 = \frac{1}{11}.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, we need to analyze the given equation, which is a sum of squares. Each term in the sum represents a squared distance between consecutive variables and the boundary conditions. The expression can be interpreted as a measure of how far the variables deviate from a certain configuration.

The equation can be rewritten as:
\[(1 - x_1)^2 + \sum_{i=1}^{9} (x_i - x_{i+1})^2 + x_{10}^2 = \frac{1}{11}.\]

This indicates that the total "energy" or "distance" represented by the left-hand side must equal \(\frac{1}{11}\). The sum of squares is minimized when each individual term is minimized. The minimum value of a squared term occurs when the term itself is zero.

To minimize the left-hand side, we set each term to zero:
1. \(1 - x_1 = 0 \Rightarrow x_1 = 1\)
2. \(x_i - x_{i+1} = 0 \Rightarrow x_i = x_{i+1}\) for \(i = 1, 2, \ldots, 9\)
3. \(x_{10} = 0\)

From the second condition, all \(x_i\) must be equal. Let \(x_1 = x_2 = \ldots = x_{10} = c\). Substituting this into the equation gives:
\[(1 - c)^2 + 9(c - c)^2 + c^2 = \frac{1}{11}.\]
This simplifies to:
\[(1 - c)^2 + c^2 = \frac{1}{11}.\]

Expanding this, we have:
\[1 - 2c + c^2 + c^2 = \frac{1}{11} \Rightarrow 2c^2 - 2c + 1 = \frac{1}{11}.\]
Multiplying through by 11 to eliminate the fraction:
\[22c^2 - 22c + 11 = 1 \Rightarrow 22c^2 - 22c + 10 = 0.\]

Using the quadratic formula \(c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 22\), \(b = -22\), and \(c = 10\):
\[c = \frac{22 \pm \sqrt{(-22)^2 - 4 \cdot 22 \cdot 10}}{2 \cdot 22} = \frac{22 \pm \sqrt{484 - 880}}{44} = \frac{22 \pm \sqrt{-396}}{44}.\]

Since the discriminant is negative, there are no real solutions for \(c\). This indicates that the only way to satisfy the original equation is to have all terms equal to zero, which leads to a unique configuration.

Thus, the only valid 10-tuple that satisfies the equation is when all variables are set to a specific value that maintains the equality, leading to the conclusion that there is exactly one configuration that meets the criteria.

Therefore, the final answer is \(1\).<|END_OF_LATENT|> The final answer is $1$.