My wife's (maths teacher) solution. She says she hasn't checked her arithmetic...
Link: World's Hardest Easy Geometry Problem
This could keep your maths students busy on the run up to Christmas. I don't have the solution.
The hint links are live [and there is more information] when you go to the website.
I got x=90 (highlight)
Had to employ a teency bit of guesswork when I found no actual way of figuring out Higher D/Middle D and Higher E (Middle E being angle x), but the error checking came back that it was sound (all angles inside triangle x added up to 180, as did all angles from triangle C) so I think I got it..
Not sure if that's how it was meant to be done, though. The guess would have been much more difficult to get right if the angles weren't all multiples of 10...
Now, assuming that all angles inside a triangle add up to 180deg. That the angle on the opposite sides of two intersecting lines is always the same. And angle BDE is a right angle, then I make x=40 (highlight).
But then the second hint in particular put me off of that solution - "you need to draw additional lines and there are more steps than you might imagine." Also the angle pictured looks too shallow for my guess to be correct.
No, it doesn't look like a right angle to me, either, but hell, if exam papers have bogus labels, this could, too xD
After further delving into my methods, my error checking was flawed, and using the same method (or at least the same error checking) also allows for x=120 and x=100, which still allows for all angles inside triangle x and triangle C to add up to 180, so yeah, I'm doing it wrong.
Back to the drawing board!
(Also, I haven't looked at the hints. OR Mrs. PICNIC's solution)
Last edited by Garacesh; 17th December 2012 at 05:06 PM.
I've got x=80 and x=75....though will double check it later
Last edited by featured_spectre; 17th December 2012 at 05:31 PM.
Even though the drawings are not to scale, I cannot see how the angle x can be anything near to 80....
From a quick look and some guesswork, I would say it was closer to 30....
But, what do I know!!
I have the answer:
X = 20
I can walk you through it now:
We all know the following rules:
Angles inside a triangle = 180
Angles on a straight line = 180
Follow me as i guide you through it: (highlight the text to reveal the solution)
We'll call the center point F
Angle ACB = 20
Angle BDC = 140 (180-20-20=140 angles in a triangle)
Angle ADB = 40 (angles on a straight line)
Angle ADF = 140 (triangle)
Angle EFD = 50 (lines)
Angle EFB = 130 (lines)
Angle BEF = 30 (triangles)
Angle AEC = 130 (triangles)
We can now work out x using the angles on a straight line rule using 180 - (Angle BEF + Angle AEC)
Hope that helps you all
Email sent to creator of puzzle, just waiting for response
Last edited by LeightonJames; 17th December 2012 at 06:11 PM.
Well I'll be...can't read my own writing...lol...shows how tired I am
Well, I got a response back, turns out my answer was right just my proof wasn't (whatever the hell that means). plus i made a couple of typos in the steps above. Still pretty pleased with myself. yey me
If by ADF you actually mean DFA, because ADF is the same angle as ADB you have all ready found, I think that might be where your working is wrong.
You have the angle for FAD (10) & ADF (40) so 180 - 50 =130 not 140
Also I don't see how from the angles you have you can then work out x. I have all those and a few more I have created and can't see as you don't have enough angles on their own to figure it.
Last edited by TechMonkey; 18th December 2012 at 09:51 AM.
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