Find segment length $x$ from two identical overlapping squares

Question

I have the following picture:

$\hspace{4.5cm}$

---

Considering only lengths AN = 30 and LM = 10 are given, x is needed to be found (RM on the picture).

I started the problem based on triangle similarities (3 angles the same, so the proportion of the sides should be the same) and if I note the length of a square as len then I figured:

(len - 30) : (len - NP) = NP : 30

where len - 30 is NB.

So 30*(len - 30) = NP*(len-NP)

And from that I get that NP is equal to 30 and that is equal to len which cannot be from the picture. Am I doing something wrong and could x be found?

Answer 1

Let the unknown variable be the angle $\theta$. Then, all the segment lengths can be computed successively. They are all given in the diagram. Matching the left and right side lengths of the upright square to get

$$30+30\cos\theta = 30(1-\sin\theta) + 10 + 30(1+\cos\theta-\sin\theta)\tan\theta$$

which simplifies to

$$3(1-\sin\theta)=\cos\theta$$

Solve to get $\cos\theta=\frac35$ and $\sin\theta =\frac45$. Thus,

$$x= 30(1+\cos\theta-\sin\theta)\sec\theta=40$$

Answer 2

HINT.

From the similitude of triangles $APN$ and $RBN$ one gets: $$ {AN\over RN}={PN\over BN}=r, \quad\text{that is:}\quad AN=r\cdot RN,\quad PN=r\cdot BN. $$ Substituting that into $PN+RN=AN+BN$ one obtains: $$ RN(1-r)=BN(1-r). $$ As $RN>BN$, it follows that $r=1$, that is: $$ AN=RN,\quad PN=BN. $$