Maths problem - how do you solve it.
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, surly the same principle works though ..
comeon arcade fire,,, you're from Cambridge!!! show 'em what were made of!!Arcade Fire said:I'm pretty trashed now, but bookmarking this for tomorrow because I couldn't work it out in the couple of minutes that I stared at it, which probably means that it's hard - and therefore interesting.
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I don't think it is; or else I've not seen something that you've seen. Your second and third lines are irrelevant - they evaluate to 0=0. I don't see where you've got the fourth line from. I think you've made a mistake and assumed that an angle which is equal to 10+x is the same as one which is equal to 130-x (which gives you the answer of x=60).Angus-Higgins said:
Edit: I see you edited your answer away
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Hey Howard - haven't seen you for a good while! How's tricks?Drawoh Tesremos said:
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no nm..divine_madness said:I do have a solution if anyone wants to see it
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Pfft. The true geniuses I met at uni were better at maths when trashed. You fail sir.Arcade Fire said:I'm pretty trashed now, but bookmarking this for tomorrow because I couldn't work it out in the couple of minutes that I stared at it, which probably means that it's hard - and therefore interesting.
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Arcade Fire said:I think you've made a mistake and assumed that an angle which is equal to 10+x is the same as one which is equal to 130-x (which gives you the answer of x=60).
Edit: I see you edited your answer away
Yeah, I spotted that. I am too tired for this I think.
(But it is fun).Angus Higgins
divine_madness said:I do have a solution if anyone wants to see it
It's quite complicated, if you're tired I don't think you'll get it anytime soon :/
edit - on second inspection, the solution i found is for a subtley different problem, not sure if it still applies, I can post it anyway if people want?
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divine_madness said:
put me out of my bloody misery (please and thank you
)SOLUTION - SPOILER, SKIP PAST IF YOU DON'T WANT TO KNOW.
1- Make a parallel to AB from D. DF parallel to AB
2- AF and DB intersects at the point O, the parallel logic makes that BAF = 60 = ABD = AFD = FDB, and also CFD = CBA = 80
3- Take a look at the triangle ABE. If we try to find the point G, which is the centre of the circle in which this triangle inserts, we get: GE = GA = GB.
By simple mathematics calculate: GÂB = GBA = a
GAE = GEA = b
GEB = GBE = c
a+a+b+b+c+c (the sum of the angles of the triangle ABE) = 180
a+b+c = 90
a+b = BAE = 70
b+c = AEB =30
c+a = EBA = 80
a+b+c = 70+c=90
so, c = 20, a = 60 , and b = 10
(If you draw the lines and see the points it'll be very easy…)
So, the centre of this triangle G forms GAB = GBA = 60. We can see that this point coincides with the point O, in fact G =O. so we have now, O , previously drawn , is the centre of this triangle AEB, so OE = OA = OB
4- We had already, BEO = 20 (the c angle) = FEO
The angle AFC = CFD + DFA = 80 + 60 =140
So in the triangle FEO, we have, 140, 20 and the last angle to find, FOE = 20
So FOE is an isosceles triangle, so, FO = FE
5- We have the triangle ODF, equilateral one (all 60 degrees) so OF =OD =DF
6- From 4 and 5 we can deduce that, FE = FD, isosceles triangle with EFD = 80, so each of FED = FDE = 50 degrees.
7- FED = FEA + (our famous angle) so, 50 = 30 + (our famous angle, I'll call it x, or AED)
8- x = 20 degrees.
END SOLUTION
1- Make a parallel to AB from D. DF parallel to AB
2- AF and DB intersects at the point O, the parallel logic makes that BAF = 60 = ABD = AFD = FDB, and also CFD = CBA = 80
3- Take a look at the triangle ABE. If we try to find the point G, which is the centre of the circle in which this triangle inserts, we get: GE = GA = GB.
By simple mathematics calculate: GÂB = GBA = a
GAE = GEA = b
GEB = GBE = c
a+a+b+b+c+c (the sum of the angles of the triangle ABE) = 180
a+b+c = 90
a+b = BAE = 70
b+c = AEB =30
c+a = EBA = 80
a+b+c = 70+c=90
so, c = 20, a = 60 , and b = 10
(If you draw the lines and see the points it'll be very easy…)
So, the centre of this triangle G forms GAB = GBA = 60. We can see that this point coincides with the point O, in fact G =O. so we have now, O , previously drawn , is the centre of this triangle AEB, so OE = OA = OB
4- We had already, BEO = 20 (the c angle) = FEO
The angle AFC = CFD + DFA = 80 + 60 =140
So in the triangle FEO, we have, 140, 20 and the last angle to find, FOE = 20
So FOE is an isosceles triangle, so, FO = FE
5- We have the triangle ODF, equilateral one (all 60 degrees) so OF =OD =DF
6- From 4 and 5 we can deduce that, FE = FD, isosceles triangle with EFD = 80, so each of FED = FDE = 50 degrees.
7- FED = FEA + (our famous angle) so, 50 = 30 + (our famous angle, I'll call it x, or AED)
8- x = 20 degrees.
END SOLUTION
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But x = 60 because its an equilateral triangle.
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No. 60 is an angle which x just cannot be.PanMaster said:
Simple application of the circle theorems demonstrates this.
Label the main triangle ABC, with A at the top, and B at bottom left,
and then put D between A and B, and E between A and C.
If x = 60, then angle CDE would equal angle CBE, and so DECB would be a cyclic quadrilateral.
However, angle ECD is not equal to angle EBD, and so DECB cannot be a cyclic quadrilateral.
Hence x cannot be 60.
Doing well thanks, Arcade Fire. Good to see you again. Not here as much as I used to be. Spend most of the summer away from home. Just back here for a few days, and saw this thread which looked interesting.Arcade Fire said: