\section*{Problem 6B - 111246B}

As the "distance" between two functions \(f\) and \(g\), which are defined in the same interval, denote the largest of all values \(|f(x)-g(x)|\) that occur in this interval, provided such a largest value exists.

Let the functions \(f\) and \(g\) be given in the interval \(-2 \leq x \leq 2\) by \(f(x)=2-|x|\) and \(g(x)=-a x^{2}+2\) (where \(a\) is a positive real number), respectively.

Investigate whether there is a value of \(a\) for which the "distance" between \(f\) and \(g\) is as small as possible. If such an \(a\) exists, give all such values of \(a\).
}

Since for all \(x \in(-2,2)\) : \(f(x)=f(-x)\) and \(g(x)=g(-x)\) are true, one only needs to consider the function \(k\)

\[
k(x, a)=|h(x, a)| \quad \text { with } \quad h(x, a)=f(x)-g(8 x)
\]

(\(a\) is a parameter that can take any real number greater than zero) in the interval \((0,2)\). It is

\[
h(x, a)=a x^{2}-x=a\left(x-\frac{1}{2 a}\right)^{2}-\frac{1}{4 a}
\]

The image of \(h\) is, for every allowed \(a\), a parabola that opens upwards, intersects the x-axis at 0 and \(\frac{1}{a}\), and has the vertex \(\left(\frac{1}{2 a},-\frac{1}{4 a}\right)\).

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1926.jpg?height=368&width=734&top_left_y=1775&top_left_x=661)

To investigate the function \(l\)

\[
l(a)=\max _{x \in(0,2)} k(x, a) \quad, \quad a>0
\]

for an absolute minimum, three cases are distinguished:

1. Case: \(2 \leq \frac{1}{2 a} \leftrightarrow a \leq \frac{1}{4}\) (figure)

\[
l(a)=k(2, a)=-h(2, a)=-4 a+2
\]

2. Case: \(\frac{1}{2 a} \leq 2 \leq \frac{1}{a} \leftrightarrow \frac{1}{4} \leq a \leq \frac{1}{2}\)

\[
l(a)=k\left(\frac{1}{2 a}, a\right)=-h\left(\frac{1}{2 a}, a\right)=\frac{1}{4 a}
\]

3. Case: \(\frac{1}{a} \leq 2 \leftrightarrow \frac{1}{2} \leq a\); the interval \(\left[\frac{1}{2}, \infty\right)\) is further divided due to the possible cases here

\[
\left(3^{\prime}\right): \quad l(a)=k\left(\frac{1}{2 a}, a\right)=\frac{1}{4 a} \quad ; \quad\left(3^{\prime \prime}\right): \quad l(a)=k(2, a)=4 a-2
\]

The inequality \(\frac{1}{4 a} \geq 4 a-2\) corresponds to case 3. It is

\[
\frac{1}{4 a} \geq 4 a-2 \leftrightarrow 0 \geq 16 a^{2}-8 a-1 \leftrightarrow a \in\left[\frac{1}{4}(1-\sqrt{2}), \frac{1}{4}(1+\sqrt{2})\right]
\]

Since \(a \in\left[\frac{1}{2}, \infty\right)\) and \(\left[\frac{1}{2}, \infty\right) \cap\left[\frac{1}{4}(1-\sqrt{2}), \frac{1}{4}(1+\sqrt{2})\right]=\left[\frac{1}{2}, \frac{1}{4}(1+\sqrt{2})\right]\), it follows that \(l(a)=\frac{1}{4 a}\) for \(a \in\left[\frac{1}{2}, \frac{1}{4}(1+\sqrt{2})\right]\) and \(l(a)=4 a-2\) for \(a \in\left[\frac{1}{4}(1+\sqrt{2}), \infty\right)\).

Thus, overall,

\[
l(a)= \begin{cases}-4 a+2 & \text { for } a \in\left(0, \frac{1}{4}\right] \\ \frac{1}{4 a} & \text { for } a \in\left[\frac{1}{4}, \frac{1}{4}(1+\sqrt{2})\right] \\ \frac{1}{4 a-2} & \text { for } a \in\left[\frac{1}{4}(1+\sqrt{2}), \infty\right)\end{cases}
\]

Since \(l\) is strictly monotonically decreasing in the interval \(\left(0, \frac{1}{4}(1+\sqrt{2})\right]\) and strictly monotonically increasing in \(\left[\frac{1}{4}(1+\sqrt{2}), \infty\right)\), \(l\) has an absolute minimum at the point \(a=\frac{1}{4}(1+\sqrt{2})\) and only at this point.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1927.jpg?height=531&width=623&top_left_y=1082&top_left_x=722)

Adapted from \([5]\)

\subsection*{9.14 XII. Olympiad 1972}

\subsection*{9.14.1 First Round 1972, Class 12}