Quadrilateral in quarter circle My friend, who is an elementary school teacher, found this problem in one of their text books and asked me for help. Turns out I'm not much of a geometry buff.
You have a quarter section of a circle. Inside this section, there is a quadrilateral with one corner at the center, with two corners at the axes, and with the last corner on the circle such that it creates a right angle. You are given that the non-trivial diagonal is $8$. What is the radius of the circle?
I observed that it is a cyclic quadrilateral, and I tried to apply Ptolemy's theorem to show that the radius is uniquely determined, but I didn't succeed and now I'm not convinced that it is true. What do you think?

 A: 
You're correct the radius is not unique. Instead, it can be anything within a range of values. To see why, consider my diagram above. First, draw any line segment $AB$ of length $8$. Have $C$ be its midpoint. Consider the line segment $CD$ of length $4$. Note that for any circle circumscribing $\triangle ABD$ that $AB$ would its diameter, so $\measuredangle ADB = 90^{\circ}$. You can also see this because, by isosceles triangles, you have $\measuredangle CAD = \measuredangle CDA$ and $\measuredangle CDB = \measuredangle CBD$, so since the total sum of these angles is $180^{\circ}$, you have that $\measuredangle BDA$ is half of that, i.e., $90^{\circ}$. Regardless of how you determine it, note it meets your requirements.
However, $D$ can move around anywhere on that circumscribed circle, so the circle going through $D$, as shown, would have different radii depending on where that point is, such as point $E$ instead as shown above. Overall, the locus for valid positions for $D$ would form a semi-circle of radius $4$ with center $C$.
A: This question is very nice but you can determine the radius of the circle with only these informations. I give you two different situations which satisfie hypotesis but the two radius are different. Instead of $8$ I use $5$ (Using Pythagorean triple in computation)
Case 1: Take as quadrilater a square with edge's length equal to $\frac{5}{\sqrt{2}}$. Thanks to this fact you have that the two diagonal are equal and of the circle has radius equal to $5$.
Case 2: Take the quadrilater $ABCD$ with coordinates $A=(0,0)$, $B=(3,0)$, $C=(4,2)$, $D=(0,4)$. Thanks to Pythagorean theorem we have that $\overline{BD} = 5$ as we want. Moreover, the ipotetic radius $AC$ has length equal to $\sqrt{4^2 + 2^2} = 2\sqrt{5}\neq 5$. So, if we show that $\overline{CD}$ is orthogonal to $\overline{BC}$ we are done. The angular coefficient of the line passing through $\overline{CD}$ is $-\frac{1}{2}$ and the angular coefficient of the line passing through $\overline{BC}$ is $2$; so $\overline{CD}$ is perpendicular to $\overline{BC}$.
