Let the midpoint of segment $AB$ be $F$, and let $Z$ be any point on segment $AF$. We erect a perpendicular to $AB$ at $F$ and measure the distance $FX = FA$. We also erect a perpendicular to $AB$ at $B$ and measure the distance $BY = AZ$ such that $X$ and $Y$ are on the same side of line $AB$. What could be the measure of angle $XZY$?
Let's create a diagram in GeoGebra, perhaps we can make a conjecture. The online solution can be found on the following page, and a GeoGebra file can be downloaded where the $Z$ point can be moved along the $A F$ segment:

https://www.komal.hu/feladat? $\mathrm{a}=$ feladat $\& \mathrm{f}=\mathrm{C} 1549 \& \mathrm{l}=$ hu.

Based on this, the conjecture is that the value of the angle in question is $45^{\circ}$, and the $\triangle XYZ$ is an isosceles right triangle. Let's prove this.

If the $Z$ point coincides with the $A$ point, then the $Y$ point coincides with the $B$ point, and the $\triangle XZY$ is an isosceles right triangle, $\angle XZY=45^{\circ}$ (1st figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_0606ff366f7525abc9b5g-1.jpg?height=344&width=528&top_left_y=476&top_left_x=777)

If the $Z$ point coincides with the $F$ point, then the quadrilateral $XZBY$ is a square, and $\triangle XZY$ is also an isosceles right triangle, and $\angle XZY=45^{\circ}$ (2nd figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_0606ff366f7525abc9b5g-1.jpg?height=336&width=501&top_left_y=995&top_left_x=801)

If the $Z$ point is placed elsewhere, we provide three solutions for this case. Two different solutions from these can be read on the website.

I. Solution. Draw a perpendicular from the $X$ point to the line $BY$, and let the foot of the perpendicular be $P$. The quadrilateral $FBPX$ is a square because three of its angles are right angles and $FB=FX=a$. Therefore, $PB=a$, and thus $PY=a-b$ (3rd figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_0606ff366f7525abc9b5g-1.jpg?height=284&width=529&top_left_y=1609&top_left_x=782)

3rd figure

Since $PY=FZ=a-b$, $PX=FX=a$, and $\angle YPX=\angle ZFX=90^{\circ}$, the triangles $\triangle YPX$ and $\triangle ZFX$ are congruent. This implies that $XY=XZ$ and $\angle YXP=\angle ZXF$. Therefore,

$$
90^{\circ}=\angle FXB=\angle FXZ+\angle ZXP=\angle FXZ+\angle ZXF=\angle ZXP,
$$

and thus

$$
\angle XZY=\frac{180^{\circ}-90^{\circ}}{2}=45^{\circ}
$$

II. Solution. Place the diagram in a coordinate system such that the $F$ point is at the origin, the coordinates of the $A$ point are $(-a; 0)$, and the coordinates of the $Z$ point are $(-b; 0)$. Then the coordinates of the $B$ point are $(a; 0)$, and the coordinates of the $Y$ point are $(a; a-b)$ (4th figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_0606ff366f7525abc9b5g-2.jpg?height=509&width=786&top_left_y=125&top_left_x=648)

Write the dot product of the vectors $\overrightarrow{ZX}$ and $\overrightarrow{ZY}$ in two ways. Since $\overrightarrow{ZX}=(b; a)$ and $\overrightarrow{ZY}=(a+b; a-b)$, we have:

$$
\overrightarrow{ZX} \cdot \overrightarrow{ZY}=b(a+b)+a(a-b)=ab+b^2+a^2-ab=a^2+b^2
$$

On the other hand,

$$
\begin{aligned}
\overrightarrow{ZX} \cdot \overrightarrow{ZY} & =\sqrt{b^2+a^2} \cdot \sqrt{(a+b)^2+(a-b)^2} \cdot \cos \alpha= \\
& =\sqrt{a^2+b^2} \cdot \sqrt{2(a^2+b^2)} \cdot \cos \alpha=\sqrt{2}(a^2+b^2) \cos \alpha
\end{aligned}
$$

The two expressions are equal:

$$
\begin{aligned}
\sqrt{2}(a^2+b^2) \cos \alpha & =a^2+b^2 \\
\cos \alpha & =\frac{1}{\sqrt{2}} \\
\alpha & =45^{\circ}
\end{aligned}
$$

since $\alpha$ is between $0^{\circ}$ and $90^{\circ}$.

Gál Bence (Szolnok, Varga Katalin Gimn., 11th grade)

III. Solution. Since $FA=FX=FB$ and $\angle AFX=90^{\circ}$, the triangle $\triangle ABX$ is half of a square with a diagonal of $2a$, $XA=XB$ and $\angle AXB=90^{\circ}$ (5th figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_0606ff366f7525abc9b5g-2.jpg?height=393&width=613&top_left_y=1517&top_left_x=737)

5th figure

Rotate the line $e:=AB$ by $90^{\circ}$ around the point $X$. The resulting line $e'$ will be perpendicular to the line $e$, so it will coincide with the line perpendicular to $AB$ at $B$ (6th figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_0606ff366f7525abc9b5g-2.jpg?height=393&width=569&top_left_y=2102&top_left_x=759)

6th figure

Since $AX=BX$ and $\angle AXB=90^{\circ}$, the image of $A$ under the rotation is $B$. We know that $b=AZ=A'Z'=BY$, so the image of $Z$ is $Y=Z'$. Since the rotation preserves distances, $XZ=XZ'=XY$, and since we rotated by $90^{\circ}$, $\angle ZXZ'=\angle ZXY=90^{\circ}$. Therefore, $\triangle XZY$ is an isosceles right triangle, and the angle in question, $\angle XZY=45^{\circ}$.