To-may-to / tomato; potato / po-tah-to; impossible exam / underprepared students. This time it’s the hapless Kiwis who are making Downfall parody videos and complaining that their practice papers hadn’t prepared them for stuff on the syllabus.

Never mind; the formidable @solvemymaths has picked out the two most-complained-about questions, and it would only be fair to have a stab at them to show how possible they are.

Suppose the square has a side-length of two units. Then $FG=1$ and $GD=2$, making FGD a 60º ((Don’t tell the Mathematical Ninja)) angle; by symmetry, HGE is also 60º, and DGE must be 60º to make the angles on the line FGH sum to 180º.

### Approach 1: Circle theorems

Whenever I see several lines the same length emanating from a point, I think circle theorems! (I don’t think they belong in exams, but I do still like them.)

Because GD=GB=GE=2, we can draw a circle of radius two around point G that passes through D, B and E. Looking at the major arc DE, the angle subtended at the centre is 300º, so the angle subtended at the circumference, DBE, is half of that, 150º.

### Approach 2: Isosceles triangles

A more reasonable approach would be to spot that DGB is an isosceles triangle, with $GDB=GBD$. By symmetry, the apex angle has to be half of the 60º at DGE, so the base angles are both 75º. Again using symmetry, the angle $x$ must be double that, or $150º$.

### Approach 3: Regular polygons

An astute student might spot that the kite GDBE is a small piece of a regular dodecagon - the isosceles triangles each have a peak angle of 30º, so 12 of them fit around the centre point at G.

The external angle for a regular dodecagon is 30º, so the internal angle - $x$ - is 150º again.

Impossible? Hardly. It might be difficult if you haven’t learned the geometrical techniques required by a new syllabus, but that’s not the exam’s fault.