## Everyone loves an 'impossible maths exam' story

(85 Posts)The latest country to have an exam that has left students in tears is New Zealand. www.theguardian.com/world/2017/nov/22/impossible-new-zealand-maths-exam-even-flummoxes-teachers

Here's one of the questions if anyone fancies a go. I'm pretty sure there must be a more elegant solution than the one I came up with!

I must be missing something but if DG=GB=EG then wouldn't DE be a flat line?

So, 180 degrees?

ignore me

No idea but I guess 150!? (Do you get a point if you accidentally get it right?)

Can't furtle on my phone put what happens is you draw a line from G to B to create two triangles. Is it easier to calculate then?

**if** you draw a line, sorry.**

Oh dear DN (aged 15) will have taken that exam.

My solution:

DG=BG and BG bisects the kite so tan (0.5 x) = 1

so 0.5 x = 45 degrees and so angle x = 90 degrees.

**Gertrude** why did you change your mind? I agree with you...

It's definitely obtuse, **catslife**. I'm not sure where your right-angled triangle for trig comes from?

**Janek** DEG is an isosceles triangle, if you drop a line from G to the line DE, then that line will be shorter than DG and GE, not the same length as them, so x can't be 180.

Ah yes, i was starting to realise that myself...

So can you do it **noble**?

Triangle EGH has one side of length 1 unit and the hypotenuse length 2 units. So you can get the angles in that and then work out the rest from there.

I've done it, but I don't think my solution will be the one that the kids would be expected to come up with. I did a bit of trig, but it didn't feel very nice.

That's basically what I did, **parietal** (except base 1/2 and hyp 1). Maybe that is the way you're meant to do it then!

Angle BDG looks as if it could be a right angle, but does trigonometry count as a valid method?

I did it like this:

BG = 2GH

Call angle EGH y

cos y = 1/2, therefore y = 60 degrees

Internal angle of the kite DGE = 60 degrees as well, due to the sum of angles on the straight line being 180.

If you bisect x to give two isosceles triangles, then looking just at BGE, sum of its internal angles = x (ie 2 x 1/2x) + 30 = 180 degrees.

Therefore x = 150 degrees.

I would say it is well in line with last year's GCSE practice papers, in terms of reasoning. In fact, as it only requires minimal trigonometry and otherwise only sums of angles within a triangle and on a straight line, it is easier than some of those, which required circle theorems and all sorts.

I think **crisscross** is right, though my trig is rusty.

reasoning follow below so if you still want to do it look away now:

call the 3 lengths that are the same 2y. B and G bisect the side of the square. So the bottom triangle FG=y and DG=2y.

because the sides are the same length there are 2 isosceles triangles.

base angles = so each x/2 so apex angle BGD = 180-x

so DGF is 90-(180-x) = x-90.

oh to sin and ah becos so cos(DGF)=y/2y=1/2

cos(x-90)=1/2

sin(x)=0.5 x=30/150 and its obtuse so 150.

OK, I'll bite. What's actually the problem here? To do this question, students have to know what a kite is, and that the sine of 60 degrees is 1/2 (to be picky, the converse, that if an acute angle has a sine of 1/2, it must be 60 degrees), that the angles at the base of an isosceles are equal, and that the angles in a triangle add to 180 degrees. They have to use each of these facts once. Is any of those facts supposed to be beyond them, or was it just that there were more steps than they were used to?

As for the idea that their teachers were flummoxed... well.

Are we allowed a calculator? If so, fairly straightforward. If not, I'll have to go away and think again!

Looks like we agree.

I hate trying to write answers out when I can't see the diagram!

I got cos x/4 = 1/2.

It isn’t that hard if you spot from the key that GB = EG.

I’m very surprised the teachers were flummoxed.

D’oh.

I got 144°

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