TITLE: Find angle UFO in the picture attached
QUESTION [17 upvotes]: I sent this problem to Presh Talwalkar who suggested me to send it to this site.
I tried many things but was not able to find the correct solution.

I made various segments trying to get an equilateral triangle similar to the Russian triangle problem, but no success. 
I also tried to flip the triangle UFO over side NO but again no success.  
I tried to find like triangles, but not enough. 
Could you please give me a hint?

Thanks,
R. de Souza

REPLY [2 votes]: None of the other answers is purely synthetic. Let me post one.

First of all, we easily calculate $\angle FEU = 5^\circ$.
Let $P$ be a point on $EU$ such that $PE=PF$. Then $\angle PFE = \angle FEP = 5^\circ$ and so $\angle FPU = \angle PFE + \angle FEP = 10^\circ$. 
Let $Q$ be a point of $PU$ such that $PQ=QF$. Then $\angle QFP = \angle FPQ = 10^\circ$ so $\angle FQU = \angle QFP + \angle FPQ = 20^\circ$.
Build an equilateral triangle $PQR$. Also, let the circle centered at $P$ with radius $PE=PF$ intersect $OE$ at $S$ (so in particular $PS=PF$). Note that 
$$\angle RPS = \angle FPS - \angle FPQ - \angle QPR = 2\angle FES - 10^\circ - 60^\circ = 2\cdot 40^\circ - 70^\circ = 10^\circ = \angle FPQ.$$ 
This along with $PQ=PR$ and $PF=PS$ shows that triangles $FPQ$, $SPR$ are congruent. This shows that $RS=FQ$.
Now we can prove that $\angle FQR = 140^\circ = \angle QRS$. Perhaps the fastest way is to note that $F, P, R$ lie on a circle with center $Q$ so $\angle FQR = 2\angle FPR = 2 \cdot (\angle FPQ + \angle QPR) = 140^\circ$. Similarly, $\angle QRS = 140^\circ$. So $\angle SQR = 90^\circ - \frac 12\angle QRS = 20^\circ$ hence $\angle FQS = \angle FQR - \angle SQR = 140^\circ - 20^\circ = 120^\circ$. We prove similarly that $\angle FRS = 120^\circ$. Hence $FQRS$ is inscribed in a circle, marked on the picture in red. (It is in fact an isosceles trapezoid but we don't need that.)
Also, $\angle SOF + \angle FQS = 60^\circ + 120^\circ = 180^\circ$, so $OFQS$ is cyclic. Hence $O$ also lies on the red circle. 
Since $\angle FOU = 20^\circ = \angle FQU$, it follows that $OUFQ$ is cyclic. This means that $U$ also lies on the red circle.
Phew! Now we can easily calculate $\angle UFO$. Note that $\angle QUF = \angle QRF = 20^\circ$ and $\angle OUE = 180^\circ - \angle EOU - \angle UEO = 180^\circ - 80^\circ - 35^\circ = 65^\circ$. Hence
$$\angle UFO = 180^\circ - \angle FOU - \angle OUF = 180^\circ - 20^\circ - (65^\circ + 20^\circ) = 75^\circ.$$