Help pls (2 Viewers)

[Sy123]

LOL, there has to be a faster method...

I don't think its that long, there are only some core things that needs to be done
1. Establish the goal of proving BD is a tangent
2. Finding BD in terms of known pronumerals
3. Making a numerical equation for BD
4. Proving that the equation holds

2, 3 is quite fast though.

But I do agree there should be a faster solution present

 

[HSC2014]

My question is: How would/did you even know x = 20 in the first place to form your approach? (without people telling you)

 

[Sy123]

My question is: How would/did you even know x = 20 in the first place to form your approach? (without people telling you)

That is an issue, in an exam, if I had enough time left in a 3U exam (which usually if you are fast enough one can finish the 3U exam in under an hour) and you have a protractor and compass with you, since the diagram is really bad and that its a 'find' question, I would construct the diagram and find x manually, which would subsequently give me a hint as to what to do.

I do agree that this method is a little inefficient

 

[HSC2014]

Ah, good tip. Thanks

 

[braintic]

Has anyone found a way to do this without trig?
The fact that the answer is a nice round 20 degrees suggests to me that there is probably a purely Euclidean method.

 

[fionarykim]

Woahhh wrh thats crazyyy, id prob just skip

That is an issue, in an exam, if I had enough time left in a 3U exam (which usually if you are fast enough one can finish the 3U exam in under an hour) and you have a protractor and compass with you, since the diagram is really bad and that its a 'find' question, I would construct the diagram and find x manually, which would subsequently give me a hint as to what to do.

I do agree that this method is a little inefficient

Wow finish 3u in under an hr? Wow wish i could do that ><

 

[Riproot]

I thought it was 30

 

[braintic]

I thought it was 30

I think we've agreed that the answer is 20 degrees, unless you want to justify your answer.

 

[SpiralFlex]

Non trigonometric method for extension 1 may involve constructing lines. (You would be very hard press to get such an question like this, because it leans towards the competition side.)

Construct a line parallel to EM' parallel to DC & draw DM'. The point where DM' and CE crosses we will label O. Connect the upper vertice A to O. Then use triangles to deduce the 20 degrees. (This is not obvious - but rather constructing a few lines to get an answer)

A high school approach: (Rushed working, check details yourself - fill in the whys)

x = 20.

Edit

 

[HSC2014]

You must be using a different diagram SpiralFlex o_o
"Angle DFC = Angle EFD = Angle AEF = 40 degrees (why?)"
But Angle DFC = 50, Angle EFD = 130

 

[SpiralFlex]

You must be using a different diagram SpiralFlex o_o
"Angle DFC = Angle EFD = Angle AEF = 40 degrees (why?)"
But Angle DFC = 50, Angle EFD = 130

Whoop sorry yeah lol but you know what I mean

 

[SpiralFlex]

Turns out I didn't make a mistake it's how the forum aligns "<" so I had to use angle instead

 

[HSC2014]

Hmm disagree with last line "Angle AEB = 50 - x"
I was only able to get Angle AEB = 10+x which makes it still unsolvable

 

[SpiralFlex]

Sorry my mistake

 

[SpiralFlex]

I will finish the solution above later this afternoon

 

[SpiralFlex]

Rough sketch done on a shaky bus standing up

I'll try to put proper wording tonight got a meeting

 

[HAX0R]

1. Calculate some known angles:
ACB = 180-(10+70)-(60+20) = 20°
AEB = 180-70-(60+20) = 30°

2. Draw a line from point D parallel to AB, labeling the intersection with BC as a new point F and conclude:
DCF ACB
CFD = CBA = 60+20 = 80°
DFB = 180-80 = 100°
CDF = CAB = 70+10 = 80°
ADF = 180-80 = 100°
BDF = 180-100-20 = 60°

3. Draw a line FA labeling the intersection with DB as a new point G and conclude:
ADF BFD
AFD = BDF = 60°
DGF = 180-60-60 = 60° = AGB
GAB = 180-60-60 = 60°
DFG (with all angles 60°) is equilateral
AGB (with all angles 60°) is equilateral

4. CFA with two 20° angles is isosceles, so FC = FA

5. Draw a line CG, which bisects ACB and conclude:
ACG CAE
FC-CE = FA-AG = FE = FG
FG = FD, so FE = FD
6. With two equal sides, DFE is isosceles and conclude:
DEF = 30+x = (180-80)/2 = 50
Answer: x = 20°

 

[HAX0R]

x=20.

This is a well known problem, and has been described as the hardest easy geometry problem.

 

[HAX0R]

Divide by sin(70), convert sin 80 = cos 10, then use double angle formula



Divide by 2, convert 1 of the sin(20) into cos(70)



Use double angle formula again, use the fact that sin(180-x) = sin(x)



Use the sum to product trig identity, convert sin 70 = cos 20















(angle in alternate segment)



Difficult question.

Lol, Sy using some complex stuff here.
The idea of the problem was to use elementary geometry.

 

[Speed6]

Interesting geometry problem