5. An arbitrary point $N$ on the line has coordinates $(t; -2-t), t \in R$. An arbitrary point $M$ on the parabola has coordinates $\left(x; x^{2}-4x+5\right), x \in R$. The square of the distance between points $M$ and $N$ is $\rho^{2}(x, t)=(x-t)^{2}+\left(x^{2}-4x+7+t\right)^{2}$. We need to find the coordinates of points $M$ and $N$ for which $\rho^{2}$ takes its minimum possible value.

For a fixed point $M$, $\rho^{2}(t)$ depends on $t$, and at the point of minimum, its derivative with respect to the variable $t$ is zero: $-2(x-t)+2\left(x^{2}-4x+7+t\right)=0$ (*). For a fixed point $N$, the function $\rho^{2}(x)$ depends on $x$, and at the point of minimum, its derivative with respect to $x$ is zero: $2(x-t)+2\left(x^{2}-4x+7+t\right)(2x-4)=0$ (**).

We solve the system (*), (**):

\[
\left\{\begin{array}{c}
2\left(x^{2}-4x+7+t\right)(2x-3)=0 \\
4(x-t)+2\left(x^{2}-4x+7+t\right)(2x-5)=0
\end{array}\right.
\]

Case 1. $x = \frac{3}{2}$

Then, substituting $x = \frac{3}{2}$ into the second equation, we get $t = -\frac{7}{8}$. The critical points are $N^{*}\left(-\frac{7}{8}; -\frac{9}{8}\right)$ and $M^{*}\left(\frac{3}{2}; \frac{5}{4}\right)$. The distance between points $M^{*}$ and $N^{*}$ is $\frac{19 \sqrt{2}}{8}$. If $2r \leq \frac{19 \sqrt{2}}{8}$, then the circle does not intersect the parabola (it touches the points $M^{*}$ and $N^{*}$).

![](https://cdn.mathpix.com/cropped/2024_05_06_0a9723897c1a776dd229g-07.jpg?height=425&width=346&top_left_y=2009&top_left_x=752)

Case 2. $x \neq \frac{3}{2}$

Then $\left(x^{2}-4x+7+t\right)=0$ and from the second equation $x=t$. Substituting $x=t$ into the equation $\left(x^{2}-4x+7+t\right)=0$, we get the condition on $t$, i.e., $t^{2}-3t+7=0$. Since the last equation has no solutions, Case 2 is not feasible. $r_{\max} = \frac{19 \sqrt{2}}{16}$
Answer: $r_{\max }=\frac{19 \sqrt{2}}{16}$

Problem 6 Answer: $l_{\max }=\frac{4 b^{2}+a^{2}}{4 b}=\frac{10}{3}$

Variant 0

In rectangle $ABCD$ with sides $AB=a, BC=b, a<b$, a semicircle $K$ with diameter $AB$ is placed. A line $L$ is tangent to the semicircle $K$. Find the maximum possible length of the segment of line $L$ located within the rectangle $ABCD$.

Answer: $l_{\max }=\frac{4 b^{2}+a^{2}}{4 b}$

Solution

![](https://cdn.mathpix.com/cropped/2024_05_06_0a9723897c1a776dd229g-08.jpg?height=394&width=525&top_left_y=1342&top_left_x=640)

Fig 1

$MN-$ segment of line $L, E-$ point of tangency, $\square BOE=\alpha, \square BOM=\square MOE=\frac{\alpha}{2}, BM=ME$, $\square ONA=\square ONE=\frac{\alpha}{2}, AN=NE. MN=ME+NE=BM+AN=\frac{a}{2}\left(\operatorname{tg} \frac{\alpha}{2}+\operatorname{ctg} \frac{\alpha}{2}\right)$.

If line $L$ passes through vertex $D$, then $AN=AD=b \rightarrow \frac{a}{2} \operatorname{ctg} \frac{\tilde{\alpha}}{2}=b \rightarrow \operatorname{tg} \frac{\tilde{\alpha}}{2}=\frac{a}{2 b}$.

Case 1. $\alpha \in[\tilde{\alpha} ; \pi / 2]$ (Fig 1)

Notation: $t=\operatorname{tg} \frac{\alpha}{2} \in\left[\frac{a}{2 b} ; 1\right]$

The function $l_{1}(t)=\frac{a}{2}\left(t+\frac{1}{t}\right)$ is decreasing on the interval $\left[\frac{a}{2 b} ; 1\right]$, so $l_{1, \max }=l\left(\frac{a}{2 b}\right)=\frac{a}{2}\left(\frac{a}{2 b}+\frac{2 b}{a}\right)=\frac{4 b^{2}+a^{2}}{4 b}$

![](https://cdn.mathpix.com/cropped/2024_05_06_0a9723897c1a776dd229g-09.jpg?height=415&width=602&top_left_y=158&top_left_x=584)

Fig 2

Case 2. $\alpha \in[0 ; \tilde{\alpha}]$ (Fig 2), point $N$ lies on side $CD$.

$\square CMN=\alpha, BM=\frac{a}{2} \operatorname{tg} \frac{\alpha}{2}, MC=b-BM=b-\frac{a}{2} \operatorname{tg} \frac{\alpha}{2}, MN=\frac{MC}{\cos \alpha}=\left(b-\frac{a}{2} \operatorname{tg} \frac{\alpha}{2}\right) / \cos \alpha$

Notation: $t=\operatorname{tg} \frac{\alpha}{2} \in\left[0 ; \frac{a}{2 b}\right], \cos \alpha=\frac{1-t^{2}}{1+t^{2}}$

The function

$l_{2}(t)=\left(b-\frac{a t}{2}\right) \cdot \frac{1+t^{2}}{1-t^{2}}=\frac{(2 b-a t)\left(1+t^{2}\right)}{2\left(1-t^{2}\right)}, t \in\left[0 ; \frac{a}{2 b}\right]$

A sketch of the graph of the function $l_{2}(t)$ on the interval $\left[0 ; \frac{a}{2 b}\right]$ is shown in Fig 3

![](https://cdn.mathpix.com/cropped/2024_05_06_0a9723897c1a776dd229g-09.jpg?height=405&width=466&top_left_y=1411&top_left_x=652)

It is increasing on the interval, so $l_{2, \max }=l_{2}\left(\frac{a}{2 b}\right)=\frac{\left(2 b-\frac{a^{2}}{2 b}\right)\left(1+\frac{a^{2}}{4 b^{2}}\right)}{2\left(1-\frac{a^{2}}{4 b^{2}}\right)}=\frac{4 b^{2}+a^{2}}{4 b}$.

## Answers and solutions