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Jokes/Interweb Things Thread, For Your Mathematicians: World's Hardest Easy Geometry Problem [Answer Not Supplied] in Fun Stuff; Link: World's Hardest Easy Geometry Problem This could keep your maths students busy on the run up to Christmas. I ...
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    DaveP's Avatar
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    For Your Mathematicians: World's Hardest Easy Geometry Problem [Answer Not Supplied]

    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.

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    PICNIC's Avatar
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    My wife's (maths teacher) solution. She says she hasn't checked her arithmetic...

    https://dl.dropbox.com/u/3944825/solution.jpg

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    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...

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    tmcd35's Avatar
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    Quote Originally Posted by Garacesh View Post
    I got x=90 (highlight)
    Er, it doesn't look like a right angle to me!

    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.

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    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.

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    nephilim's Avatar
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    I've got x=80 and x=75....though will double check it later
    Last edited by nephilim; 17th December 2012 at 05:31 PM.

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    aerospacemango's Avatar
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    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!!

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    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.

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    nephilim's Avatar
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    Well I'll be...can't read my own writing...lol...shows how tired I am

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    tmcd35's Avatar
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    Quote Originally Posted by LeightonJames View Post
    I have the answer:

    X = 20
    interesting! we both made EFD the same value, and my x is 20deg greater than yours! If you are correct then BDE cant be the right angle I assumed it was!

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    LeightonJames's Avatar
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    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

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    TechMonkey's Avatar
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    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.
    Highlight below

    You have the angle for FAD (10) & ADF (40) so 180 - 50 =130 not 140

    Highlight above

    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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    LeightonJames's Avatar
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    Quote Originally Posted by TechMonkey View Post
    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.
    Highlight below

    You have the angle for FAD (10) & ADF (40) so 180 - 50 =130 not 140

    Highlight above

    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.
    Yeah, that is weird. I have just worked it through again but cannot see how i got my answer lol!! Stupid thing.



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