Putnam 2022, Problem A-1

Determine all ordered pairs of real numbers $(a,b)$ such that the line $y = ax + b$ intersects the curve $y = \ln(1 + x^2)$ in exactly one point.

Solution:
Write \( f(x) = \ln(1 + x^2) \). We show that \( y = ax + b \) intersects \( y = f(x) \) in exactly one point if and only if \( (a, b) \) lies in one of the following groups:  
- \( a = b = 0 \)  
- \( |a| \geq 1 \), arbitrary \( b \)  
- \( 0 < |a| < 1 \), and \( b < \ln(1 - r_-)^2 - |a|r_- \) or \( b > \ln(1 - r_+)^2 - |a|r_+ \), where  
  \[
  r_{\pm} = \frac{1 \pm \sqrt{1 - a^2}}{a}.
  \]  

Since the graph of \( y = f(x) \) is symmetric under reflection in the \( y \)-axis, it suffices to consider the case \( a \geq 0 \): \( y = ax + b \) and \( y = -ax + b \) intersect \( y = f(x) \) the same number of times. For \( a = 0 \), by the symmetry of \( y = f(x) \) and the fact that \( f(x) > 0 \) for all \( x \neq 0 \) implies that the only line \( y = b \) that intersects \( y = f(x) \) exactly once is the line \( y = 0 \).  

We next observe that on \( [0, \infty) \), \( f'(x) = \frac{2x}{1 + x^2} \) increases on \( [0, 1] \) from \( f'(0) = 0 \) to a maximum at \( f'(1) = 1 \), and then decreases on \( [1, \infty) \) with \( \lim_{x \to \infty} f'(x) = 0 \). In particular, \( f'(x) \leq 1 \) for all \( x \) (including \( x < 0 \) since then \( f'(x) < 0 \)) and \( f'(x) \) achieves each value in \( (0, 1) \) exactly twice on \( [0, \infty) \).  

For \( a \geq 1 \), we claim that any line \( y = ax + b \) intersects \( y = f(x) \) exactly once. They must intersect at least once by the intermediate value theorem: for \( x \ll 0 \), \( ax + b < 0 < f(x) \), while for \( x \gg 0 \), \( ax + b > f(x) \) since \( \lim_{x \to \infty} \frac{\ln(1 + x^2)}{x} = 0 \). On the other hand, they cannot intersect more than once: for \( a > 1 \), this follows from the mean value theorem, since \( f'(x) < a \) for all \( x \). For \( a = 1 \), suppose that they intersect at two points \( (x_0, y_0) \) and \( (x_1, y_1) \). Then  
\[
1 = \frac{y_1 - y_0}{x_1 - x_0} = \frac{\int_{x_0}^{x_1} f'(x) dx}{x_1 - x_0} < 1
\]  
since \( f'(x) \) is continuous and \( f'(x) \leq 1 \) with equality only at one point.  

Finally we consider \( 0 < a < 1 \). The equation \( f'(x) = a \) has exactly two solutions, at \( x = r_- \) and \( x = r_+ \) for \( r_{\pm} \) as defined above. If we define \( g(x) = f(x) - ax \), then \( g'(r_{\pm}) = 0 \); \( g' \) is strictly decreasing on \( (-\infty, r_-) \), strictly increasing on \( (r_-, r_+) \), and strictly decreasing on \( (r_+, \infty) \); and \( \lim_{x \to -\infty} g(x) = \infty \) while \( \lim_{x \to \infty} g(x) = -\infty \). It follows that \( g(x) = b \) has exactly one solution for \( b < g(r_-) \) or \( b > g(r_+) \), exactly three solutions for \( g(r_-) < b < g(r_+) \), and exactly two solutions for \( b = g(r_{\pm}) \). That is, \( y = ax + b \) intersects \( y = f(x) \) in exactly one point if and only if \( b < g(r_-) \) or \( b > g(r_+) \).