Let's construct a convex quadrilateral $A B C D$ in which the diagonals are perpendicular and of equal length, and $A B=$ 2, $B C=4, C D=6$ units. Let's also calculate the length of the diagonals.
I. Solution. a) Let's find a modification of the desired figure where we can utilize the equality and perpendicularity of the diagonals. Move one of the diagonals so that one of its endpoints coincides with one of the endpoints of the other diagonal. It is reasonable to choose $B-\mathrm{t}$ or $C$ as the coinciding endpoint, because from these points, the desired quadrilateral has 2-2 known length sides, while from $A$ and $D$, only 1-1.

Move the $A C$ diagonal by $\overrightarrow{A B}$ to the $B C^{\prime}$ position (1st figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_e50b143cdeb4253b6875g-1.jpg?height=653&width=916&top_left_y=400&top_left_x=580)

1st figure

Thus, the $D$ point can be obtained by rotating $C^{\prime}$ around $B$ by $90^{\circ}$, and we know the distances of both points from $B$ and $C$. Therefore, by fixing $C$ and $B$ at the given distance from each other, we can draw two geometric loci for the points $C^{\prime}$ and $D$: the circle $k_{1}$ with radius $A B$ and the circle $k_{2}$ with radius $C D$ centered at $C$, as well as their $90^{\circ}$ rotated versions around $B$, $k_{2}^{*}$ and $k_{1}^{*}$, in opposite directions. (In the figure, the center of $k_{1}^{*}$ is $C^{*}$, and $k_{2}^{*}$ can be omitted.) Then $D$ is one of the intersection points of $k_{2}$ and $k_{2}^{*}$, from which $C^{\prime}$ can be obtained by a $90^{\circ}$ reverse rotation on $k_{1}$, and finally $A$ by translating $B$ by $\overrightarrow{C^{\prime} C}$.

Given the required dimensions, $k_{2}$ and $k_{1}^{*}$ intersect at two points, one of which results in a convex quadrilateral (1st figure), which is the correct one; the other (2nd figure, $C^{*}, D_{2}$, and $A_{2}$) is concave, which does not meet the criteria.

![](https://cdn.mathpix.com/cropped/2024_05_02_e50b143cdeb4253b6875g-1.jpg?height=339&width=415&top_left_y=1554&top_left_x=844)

2nd figure

b) We calculate the lengths of the diagonals only in the case of the 1st figure, based on the completed construction. In the triangle $B D C^{*}$, $C^{*} D=2$, $C^{*} B=4$, and the angle between them is $45^{\circ}$ larger than the angle at $C^{*}$ in the triangle $C D C^{*}$. In the latter triangle, $C C^{*}=4 \sqrt{2}$, so

$$
\cos C C^{*} D \varangle=\frac{(4 \sqrt{2})^{2}+2^{2}-6^{2}}{2 \cdot 4 \sqrt{2} \cdot 2}=0,
$$

the angle is $90^{\circ}$, hence

$B D^{2}=B C^{* 2}+C^{*} D^{2}-2 B C^{*} \cdot C^{*} D \cos 135^{\circ}=20+8 \sqrt{2}$,

$$
B D=5.596 \text { units. }
$$

II. Solution for part a) of the problem. Let the midpoints of the sides $A B, B C, C D$ be $P, Q, R$, respectively. From the assumptions,

$$
P Q=\frac{1}{2} A C=\frac{1}{2} B D=Q R \quad \text { and } \quad P Q R \varangle=90^{\circ}
$$

because $P Q\|A C \perp B D\| Q R$. Based on these, we construct a quadrilateral similar to the desired one from any isosceles right triangle $P^{*} Q^{*} R^{*}$. We use the fact that in any scaled version of our figure, $B^{*} P^{*}: B^{*} Q^{*}=1: 2$ and $C^{*} Q^{*}: C^{*} R^{*}=2: 3$, so $B^{*}$ is on the Apollonian circle $k_{b}$ of the point pair $P^{*}, Q^{*}$ in the ratio $1: 2$, $C^{*}$ is on the Apollonian circle $k_{c}$ of the point pair $Q^{*}, R^{*}$ in the ratio $2: 3$, and $C^{*}$ is the reflection of $B^{*}$ over $Q^{*}$, so $C^{*}$ is also on the reflection $k_{b}^{\prime}$ of $k_{b}$ over $Q^{*}$ (3rd figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_e50b143cdeb4253b6875g-2.jpg?height=705&width=594&top_left_y=306&top_left_x=749)

3rd figure

Thus, the intersection point of $k_{c}$ and $k_{b}^{\prime}$ is $C^{*}$, its reflection over $R^{*}$ is $D^{*}$, the reflection of $Q^{*}$ is $B^{*}$, and the image of $B^{*}$ over $P^{*}$ is $A^{*}$. Finally, we scale the quadrilateral $A^{*} B^{*} C^{*} D^{*}$ appropriately.

The calculation of the diagonals can also be performed based on this construction, for example, in a coordinate system where the origin is $Q^{*}$, and the axes pass through $P^{*}$ and $R^{*}$.

III. Solution for part b) of the problem. Let the length of the diagonals be $d$, their intersection point be $M$, and one of their parts be $M C=x$, $M B=y$, where due to convexity $0<x<d$ and $0<y<d$. For these,

$$
\begin{aligned}
(d-x)^{2}+y^{2} & =4 \\
x^{2}+y^{2} & =16 \\
x^{2}+(d-y)^{2} & =36
\end{aligned}
$$

Subtracting (2) from (1) and (3), we get first-degree equations for $x$ and $y$, and from them,

$$
x=\frac{d^{2}+12}{2 d}, \quad y=\frac{d^{2}-20}{2 d}
$$

Substituting these into (2), we get a quadratic equation for $d^{2}$:

$$
d^{4}-40 d^{2}+272=0, \quad d^{2}=20 \pm 8 \sqrt{2}
$$

The smaller root of $d^{2}$ would give a negative $y$ according to (4), which we cannot use. The larger value of $d^{2}$ matches the result from the first solution.