IMO_2018_1

Finally a solvable IMO problem! (that does not require things like inversion or complex bash)

Here’s the first diagram that I drew (same lengths in the same color):

Now to solve this problem I extended \overline{EG} to meet the circle at Y and \overline{FD} to meet the circle at X.

Now looking at \angle{FBD}, it subtends minor arc FA. And since \angle{DXA} also subtends minor arc FA, those two angles are congruent.
In addition to this, we know that \angle FDB = \angle ADX because of vertical angles.
Putting these two together we get that \triangle BFD is similar to \triangle XAD and in turn, AD=AX.

We can do a similar argument for \triangle YAE and get that it is also isosceles.
And now our updated diagram looks like this:


(Note that I can draw a circle at center A that goes through Y, D, E, and X since the lengths are all the same).

Finally, to prove that the two lines are parallel I must prove that \angle YED = \angle YGF.
\angle YED = \angle YXD (same arc with respect to the smaller circle)
\angle YXD = \angle YXF (same ray)
\angle YXF = \angle YGF (same arc with respect to the bigger circle).

Thus \overline{DE} and \overline{FG} are parallel (or are the same line) and we are done.

:smiley: :smile: :smiley:

2 Likes

:grinning: :smile: :grinning: :smile: :grinning: :smile: :grinning: :smile: :grinning: :smile: :grinning: :smile: :grinning: :smile: INDEED. Unbelievable, well done! Pinned.