$[\quad$ Average values $\quad]$ [Pascal's Triangle and the Binomial Theorem] [ Products and Factorials $\quad]$

Vasya bought $n$ pairs of identical socks. For $n$ days, Vasya had no problems: every morning he took a new pair from the closet and wore it all day. After $n$ days, Vasya's mother washed all the socks in the washing machine and paired them up as they came out, since, to reiterate, the socks are identical. We will call a pair of socks successful if both socks in this pair were worn by Vasya on the same day.

a) Find the probability that all the resulting pairs are successful.

b) Prove that the expected number of successful pairs is greater than 0.5.
a) The probability that the first pair is a match is $1 / 2n-1$, since when the mother picked up the first sock, there were $2n-1$ socks left, and only one of them forms a matching pair with the first. Similarly, the probability that the second pair is a match (given that the first pair is a match) is $1 / 2n-3$, and so on. The probability that all pairs are matches is obtained as the product $\frac{1}{2n-1} \cdot \frac{1}{2n-3} \cdot \ldots \cdot \frac{1}{3}=\frac{1}{(2n-1)!!}=\frac{(2n)!!}{(2n)!}=\frac{2^n n!}{(2n)!}$.

To find an approximate expression, we apply Stirling's formula: $m! \approx \sqrt{2 \pi m} \cdot m^m e^{-m}$:

$\frac{2^n n!}{(2n)!} \approx \frac{2^n \sqrt{2 \pi n} \cdot n^n e^{-n}}{\sqrt{4 \pi n} \cdot (2n)^{2n} e^{-2n}} = \frac{1}{\sqrt{2}}\left(\frac{e}{2n}\right)^n$.

b) The probability that a specific pair is a match, found in part a), is $1 / 2n-1$. We number the pairs from 1 to $n$. We introduce indicators:

$\bar{\xi}_k=1$, if the $k$-th pair is a match, $\xi_k=0$, if the $k$-th pair is not a match. Then $E \xi_k = 1 / 2n-1$.

The random variable $X$ "number of matching pairs" is equal to the sum of the indicators: $X = \xi_1 + \xi_2 + \ldots + \xi_n$. Therefore, $\mathrm{E} X = n \mathrm{E} \xi_1 = n / 2n-1 > 0.5$.

## Answer

a) $\frac{2^n n!}{(2n)!} \approx \frac{1}{\sqrt{2}}\left(\frac{e}{2n}\right)^n$

$\left[\begin{array}{cc}\text { In } \\ \text { [ inscribed and circumscribed circles }\end{array}\right]$

Quadrilateral $ABCD$ is inscribed in a circle, $I$ is the center of the inscribed circle of triangle $ABD$. Find the minimum value of $BD$ if $AI = BC = CD = 2$.

## Solution

From the condition, it follows that the arcs $BC$ and $CD$ are equal, so the bisector $AI$ of angle $BAD$ intersects the circle at point $C$ (see figures). By the trident lemma,

$CI = CB = CD = 2$. We can reason in different ways from here.

First method. Let $K$ and $N$ be the feet of the perpendiculars dropped from points $C$ and $I$ to $BD$ and $AD$ respectively (left figure), then from the equality of right triangles $BKC$ and $ANI$ it follows that $CK = IN$. The perpendicular $IP$ to diagonal $BD$ is also equal to $IN$ (these are the radii of the inscribed circle of triangle $ABD$), so $BD$ intersects segment $CI$ at its midpoint $L$. Thus, $CL = IL = 1$.

By the theorem on the product of segments of chords $BL \cdot DL = AL \cdot CL$. Let $BL = x, DL = y$, then $xy = 3$. From the Cauchy inequality, it follows that the minimum value of the sum $x + y$ is achieved if $x = y = \sqrt{3}$. Therefore, the value of $BD$ $= 2\sqrt{3}$ is the smallest possible.
![](https://cdn.mathpix.com/cropped/2024_05_06_579637707648b9944b26g-21.jpg?height=784&width=1172&top_left_y=1002&top_left_x=457)

Second method. By Ptolemy's theorem (see problem $\underline{52468}$) $AD \cdot BC + AB \cdot CD = AC \cdot BD$, that is, $2AD + 2AB = 4BD$. Therefore, $BD = 1/2(AB + AD)$ (a triangle $ABD$ with this property is called a difference triangle). Without loss of generality, we can assume that $AB \geq AD$. Draw a circle with center $C$ and radius 2, which intersects side $AB$ at point $E$ (right figure). The bisector $AC$ of angle $BAD$ is its axis of symmetry and the axis of symmetry of the angle, so points $E$ and $D$ are symmetric with respect to $AC$. Therefore, $AE = AD$.

By the theorem on the product of segments of secants $AE \cdot AB = AI \cdot AF$ (IF is the diameter of the constructed circle). Therefore, $AD \cdot AB = 2 \cdot 6 = 12$.

By the Cauchy inequality $BD = 1/2(AB + AD) \geq \sqrt{AB \cdot AD} = 2\sqrt{3}$. Equality is achieved if $AB = AD$.

## Answer