If two of these segments have lengths 8 and 6, what are all possible lengths of the third segment?

Question

Please help me for this question, I can't fully understand the problem and not sure where and how to start.

In a plane, two congruent squares share a common vertex but have no other points in common vertex but have no other points in common. Connect pairs of the remaining six vertices to get three different parallel segments. If two of these segments have lengths 8 and 6, what are all possible lengths of the third segment?

Answer 1

Using the labelling of JeanMarie:

trapezoid $AA^{\prime}B^{\prime}B$ is isosceles. Drop perpendiculars and notice that the shorter leg of the right triangle is 1. Then notice that each triangle is congruent to triangle $CDO$. Hence segment $CD=1$ and so $CC^{\prime}$ is 2.

Answer 2

The solution is 2. We prove it using complex numbers.

We take as real axis the axis of symmetry of the figure.

We denote the squares' vertices ABCD, and A'B'C'D', with primes attributed to symmetrical vertices as on figure below.

We take lowercase letters for the complex numbers associated with the corresponding uppercase letters.

Hypotheses: $ \ \ (H_1) \ AA'=6 \ \ and \ \ (H_2) \ BB'=8.$

We are going to show that $CC'=2.$

Let $s$ = side length of the square, and $\theta$ = polar angle of side $OA.$

We have: $a=se^{i\theta}, b=s(1+i)a.$

Using hypothesis $(H_1)$, we have: $a \sin \theta=1 \ \ (1).$

Besides, Consider the imaginary part of $s(1+i)(cos\theta + i\sin \theta).$

It is the half length of BB', i.e., we have, using hypothesis $(H_2)$:

$s(\cos\theta + \sin \theta)=4 \ \ (2).$

Substracting (1) from (2), on gets $\ a \cos \theta = 1 \ \ (3).$

Taking the quotient of (1) and (3): $ \ \tan \theta=3 \ \ (4).$

Using relationship $ \ 1+(\tan \theta)^2=1/(\cos \theta)^2$,

we get $\ \cos \theta=1/\sqrt{10} \ \ (5)$ and consequently $ \ \sin \theta=3/\sqrt{10} \ \ (6)$.

Using (5) in (3), we get $ \ s=\sqrt{10} \ \ (7)$.

Let $D$ be the projection of $C$ on the real-axis. Considering that angle O in triangle OCD is $\ \pi - (\theta +\pi/2) = \pi/2-\theta$, one has:

$\ CD=s \sin(\pi/2-\theta)=s \cos \theta=1 \ $ (using (5) and (7)).

Thus $ \ CC'=2CD=2$.

Answer 3

Thought you might be interested in more solutions (may be even applied to the above more solutions). I will leave you the computations of the various potential solutions (3?)