[Inscribed Quadrilaterals (Miscellaneous)]

Quadrilateral $ABCD$ is inscribed in a circle. The bisectors of angles $B$ and $C$ intersect at a point lying on the segment $AD$.

Find $AD$, if $AB=5, CD=3$.
Let $M$ be the point of intersection of the bisectors of angles $B$ and $C$.

First method. Denote: $\angle A B C=2 \alpha, \angle B C D=2 \beta$, then $\angle A D C=180^{\circ}-2 \alpha, \angle B A D=180^{\circ}-2 \beta$ (see fig.).

![](https://cdn.mathpix.com/cropped/2024_05_06_6b47f7415227382cfc43g-41.jpg?height=445&width=523&top_left_y=192&top_left_x=779)

If $\alpha \beta$, a similar reasoning can be applied by marking a segment $A K$ on ray $A D$ equal to $A B$. The case $\alpha=\beta$ is impossible, otherwise $A B C D$ would be an isosceles trapezoid ($A B=D C$), which contradicts the condition.

Second method. Consider two cases.

1) $A B \| D C$. Then $A B C D$ is an isosceles trapezoid ($B C=A D$, see fig.). Draw $M N \| A B$. From the equality of alternate interior angles, we get:

$B N=N M=N C$, so $M N$ is the midline of the trapezoid, that is, $M N=1 / 2(A B+D C)$. Therefore, $A D=B C=$ $2 M N=A B+C D=8$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6b47f7415227382cfc43g-41.jpg?height=357&width=372&top_left_y=1262&top_left_x=846)
2) $A B$ intersects $D C$ at point $T$. Then $M$ is either the center of the inscribed circle of triangle $B T C$, or the center of the excircle of this triangle (see fig.). Since the reasoning in both cases is completely analogous, we will consider only the second one.

![](https://cdn.mathpix.com/cropped/2024_05_06_6b47f7415227382cfc43g-41.jpg?height=660&width=643&top_left_y=1785&top_left_x=730)

Let segment $A^{\prime} D^{\prime}$ be symmetric to side $A D$ with respect to the bisector $T M$ of angle $B T C$. Then $\angle D A^{\prime} M=\angle D^{\prime} A M$ $=\angle T C B$: since $A B C D$ is a cyclic quadrilateral, $A^{\prime} D^{\prime} \| B C$. From the equality of alternate interior angles, we get that $A^{\prime} M^{\prime}=A^{\prime} C$. Similarly, $D^{\prime} M=D^{\prime} B$, and also $A D^{\prime}=A^{\prime} D$ (from symmetry). Thus, $A D$ $=A^{\prime} D^{\prime}=A^{\prime} M+D^{\prime} M=A^{\prime} C+D^{\prime} B=A B+C D=8$.

## Answer

$A D=8$.

## Author: Lopatnikov A.A.

Can all cells of an infinite grid be colored in white and black so that each vertical line and each horizontal line intersect a finite number of white cells, and each slanted line intersects a finite number of black cells?

## Solution

Introduce a coordinate system $Oxy$ such that the vertical and horizontal grid lines have equations $x=n$ and $y=m$ ($n$ and $m$ are integers). Color black those and only those cells, all points of which satisfy one of the inequalities $|y| \geq x^{2}$ or $|x| \geq y^{2}$ (see fig.), and color the remaining cells white.

![](https://cdn.mathpix.com/cropped/2024_05_06_6b47f7415227382cfc43g-42.jpg?height=1212&width=1234&top_left_y=586&top_left_x=412)

Every vertical line will intersect a finite number of white cells between the parabolas $y= \pm x^{2}$, and every horizontal line will intersect a finite number of white cells between the parabolas $y= \pm x^{2}$. Every slanted line will intersect only a finite number of black cells, as its intersection with each of the four "black" regions can be either empty, a point, or a segment.

## Answer

Yes, it is possible.

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