# The hardest easy geometry problem.



## Veho (Dec 12, 2007)

Find the value of _x_, with mathematical proof. No advanced trigonometry allowed. No algebra needed. Here's all you need to know: 

*Lines and Angles*: When two lines intersect, opposite angles are equal and the sum of adjacent angles is 180 degrees. When two parallel lines are intersected by a third line, the corresponding angles of the two intersections are equal.

*Triangles*: The sum of the interior angles of a triangle is 180 degrees. An isosceles triangle has two equal sides and the two angles opposite those sides are equal. An equilateral triangle has all sides equal and all angles equal. A right triangle has one angle equal to 90 degrees. Two triangles are called similar if they have the same angles (same shape). Two triangles are called congruent if they have the same angles and the same sides (same shape and size).

* Side-Angle-Side (SAS): Two triangles are congruent if a pair of corresponding sides and the included angle are equal.
* Side-Side-Side (SSS): Two triangles are congruent if their corresponding sides are equal.
* Angle-Side-Angle (ASA): Two triangles are congruent if a pair of corresponding angles and the included side are equal.
* Angle-Angle (AA): Two triangles are similar if a pair of corresponding angles are equal. 


Remember, the answer must be in a form of an answer. 

Have fun going insane.


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## rhyguy (Dec 12, 2007)

its less than 50


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## xcalibur (Dec 12, 2007)

easy,
for this triangle, lets name the middle point x

angle AXB is 50 degrees 180 - 130 = 50
that makes AXD 130 because 180 - 50 = 130
BXE is 130 also and DXE is 50



			
				QUOTE said:
			
		

> Lines and Angles: When two lines intersect, opposite angles are equal and the sum of adjacent angles is 180 degrees. When two parallel lines are intersected by a third line, the corresponding angles of the two intersections are equal.



angle XBE is 30 degrees to make triangle BXE a full 180 degrees
triangle BDE is a right angled triangle so that makes angle BDE 90 degrees
so that means 50 + 90 = 140 and that makes and DEB (x) = 40

my geometry is a bit rusty so this might be completely wrong


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## Veho (Dec 12, 2007)

QUOTE(xcalibur @ Dec 12 2007 said:


> angle XBE is 30 degrees to make triangle BXE a full 180 degrees



Wrong. XBE is 20, says so on the picture.


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## xcalibur (Dec 12, 2007)

QUOTE(veho @ Dec 12 2007 said:


> QUOTE(xcalibur @ Dec 12 2007 said:
> 
> 
> > angle XBE is 30 degrees to make triangle BXE a full 180 degrees
> ...



I meant angle *XEB*
I'm sorry, when you do something longwinded you tend to get lost.
try doing trigonometric identities... those are loooooooong


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## Veho (Dec 12, 2007)

Okay, but it's still not it. 



QUOTE(xcalibur @ Dec 12 2007 said:


> triangle BDE is a right angled triangle so that makes angle BDE 90 degrees



Why do you think BDE is 90°? It _looks_ like it, but there's no way to know that for sure.


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## CockroachMan (Dec 12, 2007)

it's 20°


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## Veho (Dec 12, 2007)

QUOTE(CockroachMan @ Dec 12 2007 said:


> it's 20°


_Prove it._ Yeah, I know it's supposed to be 20°, but how do you get there? I need the procedure. 

EDIT: 

A picture drawn to scale. 






BDE is 110°.


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## xcalibur (Dec 12, 2007)

QUOTE(veho @ Dec 12 2007 said:


> Okay, but it's still not it.
> 
> 
> 
> ...



DAMNIT... i was hoping i wouldnt have to round it the loong way but i guess ill have to now :'(
serves me for trying to take a shortcut

angle ACB is 20
angle BDC is 140
angle ADX is 40
angle AEB is 50

and thats where im stuck...
i cant seem to figure our trianle CDE and i need to go now...

EDIT: i saw your post so i can get it now
BDE is 110 
DXE is 50
that makes DEX 110 +50 = 160
180 - 160 = 20

howd you get BDE?


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## kellyan95 (Dec 12, 2007)

Print -> use protractor


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## xcalibur (Dec 12, 2007)

QUOTE(kellyan95 @ Dec 12 2007 said:


> Print -> use protractor


not to scale

EDIT: i guess it is...
that was a useless post


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## Veho (Dec 12, 2007)

QUOTE(xcalibur @ Dec 12 2007 said:


> howd you get BDE?


Um...  
	

	
	
		
		

		
		
	


	




_I don't know._ I can't prove it. I know it's 110, and x=20°, but I can only prove it graphically, and this problem asks for discrete proof.


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## shakirmoledina (Dec 12, 2007)

Some unknown angles are known but if cockroachman could explain how he got 20...
ADB = 40... AEb = 30.... EPB and DPA are 130 where P is central point and APB and DPE are 50


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## CockroachMan (Dec 12, 2007)

don't know how to prove it.. 
	

	
	
		
		

		
		
	


	




for some reason, the triangles B-Center-E and DCE are equal.. but I can't remember why now.. something to do with the lines crossing them.. 
	

	
	
		
		

		
		
	


	




been some time since I've done this kind of thing..


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## shakirmoledina (Dec 12, 2007)

umm just a note that here we dont use AAS since
* Angle-Side-Angle (AAs)
* Angle-Angle (AA)
as u can see above if 2 angles are equal why do u need to prove the side


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## FrEEz902 (Dec 12, 2007)

Easy,

X=60 degrees

I'll scan my paper of ugly handwriting now

Edit: Oh wait, i screwed something up >_


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## TGBoy (Dec 12, 2007)

I have seen this before. Isnt this the World's Hardest Easy Geometry Problem.

The answer is 20. 
I seen the solution to it.


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## ZAFDeltaForce (Dec 12, 2007)

QUOTE(FrEEz902 @ Dec 12 2007 said:


> Easy,
> 
> X=60 degrees
> 
> ...


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## moozxy (Dec 12, 2007)

QUOTE(FrEEz902 @ Dec 12 2007 said:


> Easy,
> 
> X=60 degrees
> 
> ...


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## FrEEz902 (Dec 12, 2007)

QUOTE(moozxy @ Dec 12 2007 said:


> QUOTE(FrEEz902 @ Dec 12 2007 said:
> 
> 
> > Easy,
> ...


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## Veho (Dec 12, 2007)

QUOTE(FrEEz902 @ Dec 12 2007 said:


> HERE: http://img148.imageshack.us/my.php?image=freeeeezwv7.jpg









  Um, ADEB isn't a parallelogram, none of its sides is parallel to any of the others. 

EDIT: Nope, even with quadrilateral, there's no rule saying opposite angles add up to 180°.


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## FrEEz902 (Dec 12, 2007)

QUOTE(veho @ Dec 12 2007 said:


> QUOTE(FrEEz902 @ Dec 12 2007 said:
> 
> 
> > HERE: http://img148.imageshack.us/my.php?image=freeeeezwv7.jpg
> ...



Ha? o.O? Err, are you sure x.X? Then i guess i've been taught wrong at school :


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## Veho (Dec 12, 2007)

QUOTE(FrEEz902 @ Dec 12 2007 said:


> I think i remember somewhere if you cut off all the corners of a parallellogram, and put them together, it will be a perfect 360 degrees.


 The sum of all the inside angles of a parallelogram is 360, true, but this isn't a parallelogram so that doesn't apply here.


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## moozxy (Dec 12, 2007)

Ye I've never heard that rule before either. And yes all angles in any quadrilateral add up to 360.


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## FrEEz902 (Dec 12, 2007)

QUOTE(moozxy @ Dec 12 2007 said:


> Ye I've never heard that rule before either. And yes all angles in any quadrilateral add up to 360.




if that's true, then shouldn't the opposite sides add up to 180? Or is that only if it's parallel...

GAH anyhow i fail :


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## Veho (Dec 12, 2007)

QUOTE(moozxy @ Dec 12 2007 said:


> And yes all angles in any quadrilateral add up to 360.


DOH!! Totally forgot about that one...  
	

	
	
		
		

		
		
	


	




Okay, yes, but that leaves us with 360 = 360, because the _x_ cancels itself out in the equation. 

We're all missing something here.


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## FrEEz902 (Dec 12, 2007)

Apparently here is a hint: http://thinkzone.wlonk.com/MathFun/Triangle1Hint1.htm

Edit: I *THINK* i got it... If we draw a perpendicular line from DE in the 'x' triangle, and it cuts X by half, then we can get 1/2 x =180-(90+50)=40??.

But then that would mean X=40*2=80. And that is definitely wrong.

EDIT: nvm, it won't be perpendicular if we did that.


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## Twiffles (Dec 12, 2007)

I'll attemp this later, that is if the answer isn't found by then. I hated Geometry and I'm hating Calculus even more. >.>


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## FrEEz902 (Dec 12, 2007)

I googled and found a solution...yet i'm too scared to look at it 
	

	
	
		
		

		
		
	


	




.


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## moozxy (Dec 12, 2007)

Noooo if you post it, post it in spoilers.. I kinda wanna find the solution by myself now


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## FrEEz902 (Dec 12, 2007)

QUOTE(moozxy @ Dec 12 2007 said:


> Noooo if you post it, post it in spoilers.. I kinda wanna find the solution by myself now



i won't post it 
	

	
	
		
		

		
		
	


	




. I *think* we need to draw an extra line. Where and why i have no idea.


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## moozxy (Dec 12, 2007)

Ye after you posted that I've drawn two extra lines.. I think I can get the answer.. I'm not sure.


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## TGBoy (Dec 12, 2007)

This was where i found the solution. This is one way. I guess there must be other methods.

One Solution


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## xcalibur (Dec 12, 2007)

QUOTE(FrEEz902 @ Dec 12 2007 said:


> QUOTE(moozxy @ Dec 12 2007 said:
> 
> 
> > Ye I've never heard that rule before either. And yes all angles in any quadrilateral add up to 360.
> ...


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## FrEEz902 (Dec 12, 2007)

QUOTE(xcalibur @ Dec 12 2007 said:


> QUOTE(FrEEz902 @ Dec 12 2007 said:
> 
> 
> > QUOTE(moozxy @ Dec 12 2007 said:
> ...


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## Verocity (Dec 12, 2007)

I did stuff like that my freshman year in High school, I'm now in Algebra 2 which is a lot harder.

EDIT: and starting January ill be doing Algebra 3


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## Veho (Dec 12, 2007)

QUOTE(Verocity @ Dec 12 2007 said:


> I did stuff like that my freshman year in High school, I'm now in Algebra 2 which is a lot harder.



...right. Now solve this.


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## moozxy (Dec 12, 2007)

While you might be learning more difficult methods of solving problems, this uses simple rules of geometry to create this monster of a puzzle, hence the name 'Hardest easy geometry problem'


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## Verocity (Dec 12, 2007)

QUOTE(veho @ Dec 12 2007 said:


> QUOTE(Verocity @ Dec 12 2007 said:
> 
> 
> > I did stuff like that my freshman year in High school, I'm now in Algebra 2 which is a lot harder.
> ...



Yeah, this is tough... x=20 though.

It all adds up to be 180...


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## dib (Dec 12, 2007)

Hahahahaha, that's an excellent problem.  Spoiler solution:

There are multiple solutions that can fulfill the conditions of the triangle, allowing x to be just about anything


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## jtroye32 (Dec 12, 2007)

dude, this is easy, took me about 30 sec to figure out the procedure. let me do the numbers now..


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## Linkiboy (Dec 13, 2007)

QUOTE(Verocity @ Dec 12 2007 said:


> I did stuff like that my freshman year in High school, I'm now in Algebra 2 which is a lot harder.
> 
> EDIT: and starting January ill be doing Algebra 3


Algebra 2 is easy as shit :\ i'm a freshman, and I'm getting 100% in A2

And theres no such thing as Algebra 3, after A2 comes pre-calc


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## Westside (Dec 13, 2007)

QUOTE(Linkiboy @ Dec 12 2007 said:


> QUOTE(Verocity @ Dec 12 2007 said:
> 
> 
> > I did stuff like that my freshman year in High school, I'm now in Algebra 2 which is a lot harder.
> ...


That's American Education system?  In Canada Algebra and pre-calc are two different things, Calc being part of Advanced Functions and Algebra with Discrete Mathematics, proof and Geometry.


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## Jhongerkong (Dec 13, 2007)

QUOTE(veho @ Dec 12 2007 said:


> QUOTE(CockroachMan @ Dec 12 2007 said:
> 
> 
> > it's 20°
> ...



dont know why everyone is struggling so hard. saying that BDE = 110°. kills the question.

******SOLUTION********

A = 70 and B = 60; added = 130° right?

so then 180° - 130° = 50°; this is the last angle in the small lower triangle consisting of A, B and the middle point which we will call X.

since X = 50, the angle opposite to it (inside the middle triangle with x in it) has to be 50° too because of the opposite angle theorem. 

its just a simple matter of adding the 2 angles we have (110, 50) and subtracting by 180° (since all the angles in a triangle HAVE to equal 180°)

180 - 160 = 20°

x = 20°

*******END SOLUTION*******

feel free to correct me if im wrong.


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## Linkiboy (Dec 13, 2007)

QUOTE(Westside @ Dec 12 2007 said:


> QUOTE(Linkiboy @ Dec 12 2007 said:
> 
> 
> > QUOTE(Verocity @ Dec 12 2007 said:
> ...


In america the levels go like this:

Algebra 1 (1A and 1B, one for each year, 1A is taught in lower end middle schools, good schools usually teach it in one year)
Geometry
Algebra 2
Pre-calc
Calc OR AP calc
second level of AP calc if taken other AP calc previous year

We don't have Algebra 3 though :\


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## Jeda (Dec 13, 2007)

QUOTE(Jhongerkong @ Dec 13 2007 said:


> dont know why everyone is struggling so hard. saying that BDE = 110°. kills the question.
> 
> ******SOLUTION********
> 
> ...



You are correct, but I think BDE = 110 hasn't been proven yet


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## shakirmoledina (Dec 13, 2007)

if u look at what cockroachman said... he said that 2 triangles CDE And BEX are similar and that is the concept... U can prove one angle is equal since they are both 20° and the website mentioned says DB=DC but i don't know how... maybe since both angles are same (20°) and thus the middle angle is split into two
the proof mentioned is very hard to understand but is understandable but ehh... if he just simplified it.


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## Jhongerkong (Dec 13, 2007)

QUOTE(Jeda @ Dec 13 2007 said:


> QUOTE(Jhongerkong @ Dec 13 2007 said:
> 
> 
> > dont know why everyone is struggling so hard. saying that BDE = 110°. kills the question.
> ...



well, veho said it in the quote I put in my last post; I just ran with it.


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## JacobReaper (Dec 14, 2007)

*gets protractor*


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## Osaka (Dec 14, 2007)

Im going to say 15


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## iffy525 (Dec 14, 2007)

QUOTE(Linkiboy @ Dec 12 2007 said:


> In america the levels go like this:
> 
> Algebra 1 (1A and 1B, one for each year, 1A is taught in lower end middle schools, good schools usually teach it in one year)
> Geometry
> ...


I think after algebra 2 you have a choice between pre-calc or statistics
at least in my school...


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## ProdigySim (Dec 14, 2007)

I don't really see why this is called "The Hardest Easy Geometry Problem"

Finding that angle isn't simple at all... And anyone who knows the basic trigonometric laws can figure that out very quickly.


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## dib (Dec 14, 2007)

QUOTE(Jhongerkong @ Dec 13 2007 said:


> QUOTE(veho @ Dec 12 2007 said:
> 
> 
> > QUOTE(CockroachMan @ Dec 12 2007 said:
> ...


You're wrong.  You did not explain where 110 came from, other than you apparently pulled it out of nowhere.


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## Rayder (Dec 14, 2007)

"x" is found in the upper center of the pic.   I figure its "value" is pretty high, since everyone is trying to find it.

Kinda surprising that everyone is trying to find it, since it's clearly marked as "x"....not really that hard to "find" people.


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## Veho (Dec 14, 2007)

QUOTE(Rayder @ Dec 14 2007 said:


> "x" is found in the upper center of the pic.Â  I figure its "value" is pretty high, since everyone is trying to find it.
> 
> Kinda surprising that everyone is trying to find it, since it's clearly marked as "x"....not really that hard to "find" people.Â


Yeah... too bad the problem states, and I quote, "find the _value_ of _x_", not "find _x_"   
	

	
	
		
		

		
		
	


	




We know where _x_ is.


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