[ Trapezoids (miscellaneous) [Auxiliary similar triangles

The bases $BC$ and $AD$ of trapezoid $ABCD$ are equal to $a$ and $b$. Four lines parallel to the bases are drawn. The first passes through the midpoints of the lateral sides, the second through the intersection point of the diagonals of the trapezoid, the third divides the trapezoid into two similar ones, and the fourth into two equal-area ones. Find the segments of these lines contained within the trapezoid, and arrange the found values in ascending order.
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For definiteness, let's assume that $B C=a<b=A D$.

Let $M$ and $N$ be the midpoints of the lateral sides of the trapezoid. Then $M N$ is its midline, hence $M N=1 / 2$ ( $B C$ $+A D)=1 / 2(a+b)$.

Let a line parallel to the bases of the trapezoid pass through the point $O$ of intersection of the diagonals and intersect the lateral sides $A B$ and $C D$ at points $P$ and $Q$ respectively. Triangle $A O D$ is similar to triangle $C O B$ with a similarity ratio of $A D / B C=b / a$, and triangle $A O P$ is similar to triangle $A C B$ with a similarity ratio of $A O / A C=$ $b / a+b$, thus, $O P=b / a+b \cdot B C=a b / a+b$.

Similarly, $O Q=a b / a+b$. Therefore, $P Q=2 a b / a+b$.

Let a line parallel to the bases intersect the lateral sides $A B$ and $C D$ of the trapezoid at points $E$ and $F$ respectively, and in this case, trapezoid $B E F C$ is similar to trapezoid $E A D F$. Then $B C: E F=E F: A D$, hence, $E F^{2}=$ $B C \cdot A D=a b$. Therefore, $E F=\sqrt{a b}$.

Let a line parallel to the bases intersect the lateral sides $A B$ and $C D$ of the trapezoid at points $G$ and $H$ respectively, and in this case, trapezoids $B G H C$ and $A G H D$ are equal in area.

Denote $G H=x$. If lines $A B$ and $C D$ intersect at point $T$, then triangle $T G H$ is similar to triangle $T B C$ with a similarity ratio of $G H / B C=x / a$, triangle $T A D$ is similar to triangle $T B C$ with a similarity ratio of $A D / B C=b / a$. Denote $S_{T B C}=S$. Then $S_{T G H}=x / a \cdot 2 S, S_{T A D}=b / a \cdot 2 S, S_{B G H C}=S_{T G H}-S_{T B C}=x / a \cdot 2 S-S, S_{A G H D}=S_{T A D}-$ $S_{T G H}=b / a \cdot 2 S-x / a \cdot 2 S$

From the equation $x / a \cdot 2 S-S=b / a \cdot 2 S-x / a \cdot 2 S$, we find that $x=\sqrt{\frac{a^{2}+b^{2}}{2}}$.

From the diagram, it is clear that the segment closest to the smaller base is $P Q$, followed by $E F, M N$, and $G H$.

## Answer

$\frac{2 a b}{a+b}<\sqrt{a b}<\frac{a+b}{2}<\sqrt{\frac{a^{2}+b^{2}}{2}}$

Author: $\underline{\text { Markelov S.V. }}$

A polygon can be cut into two equal parts in three different ways.

Is it true that it must have a center or axis of symmetry?

## Solution

See, for example, the figure:
![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-29.jpg?height=358&width=1972&top_left_y=482&top_left_x=63)

## Answer

False.