For a unit circle, construct a square such that two of its adjacent vertices lie on the circle, and the side connecting the other two vertices is tangent to the circle. Calculate the side length of the square!
I. solution. Due to the axial symmetry of the circle and the square, the perpendicular line to the tangent at the point of tangency is the axis of symmetry for both figures (Fig. 1). The center $O$ of the circle lies on this axis of symmetry. The side $AB$ of the square is a chord of the circle. Let $AB = 2d$, and the distance from the side $AB$ to the center of the circle be $x$. From the right triangle $OFB$: $d = \sqrt{1 - x^2}$, and since $AB = BC$, $2d = 1 + x$.

From this, we get $5x^2 + 2x - 3 = 0$ and $x = \frac{3}{5}$ (only the positive root is considered). The side of the square is $\frac{8}{5}$ units. The problem has a unique solution, which can be easily constructed.

Kovács Péter (Tatabánya, Árpád Gymn., 9th grade)

II. solution. The desired square can also be constructed without calculation. Take an arbitrary point $E$ on the circle and draw the tangent to the circle at point $E$.

Draw a line through the center $O$ of the circle and the point $E$ (Fig. 2). Then draw a square whose axis of symmetry is the line $EO$ and whose two adjacent vertices lie on the tangent: let this be the square $ABCD$. Enlarge (or reduce) this square from the center $E$ so that the vertices $A$ and $B$ lie on the circle. The point $A'$ is the intersection of the line $EA$ and the circle, and the point $B'$ is the intersection of the line $EB$ and the circle. $A'B'$ will be the side of the desired square. From the similarity of the figures $ABCD$ and $A'B'C'D'$, it follows that $A'B'C'D'$ is also a square and satisfies the condition. The problem has only one solution (a line can intersect a circle in at most 2 points, one of which is $E$, the other is $A'$, or $B'$).

We still need to determine the length of the side of the square. Let $x$ be half the side of the square, for example, $A'F' = x$. Then $OF' = 2x - 1$, because $OE = 1$. $A'O = 1$, because it is a radius. In the triangle $A'F'O$, applying the Pythagorean theorem, we get $1 = x^2 + (2x - 1)^2$, which can be simplified to $5x^2 - 4x = 0$. From this, $x = \frac{4}{5}$, which is half the side of the square, so the length of the side of the square is $\frac{8}{5}$.

Based on the solution by Csató György (Hajdúszoboszló, Hőgyes E. Gymn., 10th grade)

Note. Using the notation from Fig. 1, the construction part of the problem can also be solved by starting from the square and constructing a circle that passes through points $A$, $B$, and $E$ (basic construction). The completed diagram is then scaled so that the radius of the circle is one unit in length. The side of the square can then be calculated using the method from the second solution.

Fried Katalin

![](https://cdn.mathpix.com/cropped/2024_05_02_b5efffe39469f51c7656g-1.jpg?height=753&width=383&top_left_y=1345&top_left_x=860)