Segment Fault

(Thanks to @Marta Reece for pointing out the error I made when I first came up with a problem similar to this one.)

Suppose such a segment $AB$ does exist. Then, we can draw a cuboid with $AB$ as one of its long diagonals. Let this cuboid have side lengths $a, b, c,$ where the side of length $a$ is parallel to the $x$ -axis, the side of length $b$ is parallel to the $y$ -axis, and the side of length $c$ is parallel to the $z$ -axis. The length of the projection of $AB$ onto the $xy$ plane can be written as $\sqrt{a^2 + b^2}$ by the Pythagorean Theorem, and we can get similar expressions for the other two projections:

$\begin{aligned} \sqrt{a^2 + b^2} &= 10 \\ \sqrt{b^2 + c^2} &= 20 \\ \sqrt{c^2 + a^2} &= 30. \end{aligned}$

Squaring both sides of each equation gives us the system of equations

$\begin{aligned} a^2 + b^2 &= 10^2 \\ b^2 + c^2 &= 20^2 \\ c^2 + a^2 &= 30^2. \end{aligned}$

The first instinct would be to add all these equations up, divide both sides by 2, and square root the resulting expression to get $AB = \sqrt{\dfrac{10^2 + 20^2 + 30^2}{2}} \approx 26.46.$ However, we must be careful to make sure that we have solutions to the system of equations. Note that

$a^2 + b^2 + c^2 = \dfrac{10^2 + 20^2 + 30^2}{2} = 700,$

which is actually less that $900 = c^2 + a^2.$ That means that $b^2 = -200,$ but the square of a real number is always nonnegative, so in fact, there's no solution to the system.

We can also deduce the impossibility of the scenario by using the fact that a segment is at least as long as its projection onto a plane. So, we expect that $AB \geq 30.$ However, by the work we have done, we actually got $AB < 30,$ so we have reached a contradiction.

In either case, we conclude that $AB$ cannot exist with these projected segment lengths, and this situation is impossible .