Can anybody solve this math problem?

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poc91nc

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Ok all you math wizzards....can anybody crack this problem? Very simple looking and innocent....but can't do it.

Given a triangle ABC as described below: Side AB = Side AC. Draw a line from C to side AB. call that line CD. Now draw a line from B to side AC. Call that line BE. Let angle EBC = 60 degrees, angle BCD equal 70 degrees, angle ABE equal 20 degrees, and angle DCE equal 10 degrees. Now draw line DE. The question--find what angle EDC is. (Do not do this trigonometrically ; do it geometrically to get an exact answer).
 

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Maybe it's cause I'm tired as hell, but it seems that there are a few possible answers. Right off the bat 50 and 60 are both acceptable. I'm sure there are others.

Is there anything that you didn't tell us about the problem? Any parallel lines, etc?
 
No, everything you need is in the diagram. There are officially six ways to solve the problem. I'll just say that some very very math savy people have tried to solve this problem with no luck. There six official solutions and I know none of them. Like the problem says....you have to do it geometrically. Draw something extra to extrapolate more information.

For those of you thinking of algebraic solutions....you're better off just looking for a geometric solution.
 
This is very weird... I keep doing the 180 degrees for each triangle and I keep getting 130 degrees for some of the angles (not the one we're looking for), when they look like they're 90 degrees... not drawn to scale I think... but then again I have never seen anything so out of scale.
 
Angle AED = Y
Angle CDE = X

20+Y=30+X
20+(150-X)=Y+40

Solve that and I think you will get the answer.

p.s.I think I am wrong.. let me get back to yall..
 
Angle AED = Y
Angle CDE = X

20+Y=30+X
20+(150-X)=Y+40

Solve that and I think you will get the answer.

p.s.I think I am wrong.. let me get back to yall..

I see how you derived both equations, but you made an assumption to get the second expression. The first one is legit, however....you assumed that the angle you denote as Y = BED....its not a problem if the deriviation was based off of some geometrical axiom. What was your reasoning?
 
Maybe it's cause I'm tired as hell, but it seems that there are a few possible answers. Right off the bat 50 and 60 are both acceptable. I'm sure there are others.

Is there anything that you didn't tell us about the problem? Any parallel lines, etc?

Yes....if you call both angles within the triangle in question ? = x and the second one within the same triangle = y....x + y = 130. Just looking at the statement alone there are a number of solutions that will work. But isn't that counterintuitive? X and Y are both fixed in space (the scale has nothing to do with it)....nothing is changing. How can one angle be a function of the other? Intuitively, a second independent equation must exist and it should be possible to extrapolate that second equation to give an exact solution for both angles.
 
This is a funny problem where 50, 60 and 70 works.. I bet there are other answers.

I figured out I made an false assumption in the previous post thus not correct.

This problem as given can have more than one answer.
 
No there is only one right answer. An angle is an angle haha, it can't be more than 1 measure. Try drawing it to scale.

There is apparently a longass solution that requires lots of extra lines. I will tackle it later.
 
No there is only one right answer. An angle is an angle haha, it can't be more than 1 measure. Try drawing it to scale.

There is apparently a longass solution that requires lots of extra lines. I will tackle it later.

Come on...draw it scale? We have MATLAB for an easy numerical solution....
Most likely the solution is very simple. The more you look at this problem...the more it becomes like quick sand. The more lines you draw...the more variables you will generate and hence the number of independent equations you will have to derive. The solution has to be something very simple that is being overlooked, but can't figure it out myself.
 
Come on...draw it scale? We have MATLAB for an easy numerical solution....
Most likely the solution is very simple. The more you look at this problem...the more it becomes like quick sand. The more lines you draw...the more variables you will generate and hence the number of independent equations you will have to derive. The solution has to be something very simple that is being overlooked, but can't figure it out myself.

No, you have to draw more lines and use more properties of geometry and triangles and whatnot to figure out some of the unknown angles. I don't know exactly how it's done yet. Once I have more time tonight I will try to figure it out.
 
No, you have to draw more lines and use more properties of geometry and triangles and whatnot to figure out some of the unknown angles. I don't know exactly how it's done yet. Once I have more time tonight I will try to figure it out.

Yes. I have drawn out many many patterns myself. Some very complicated. The point is NOT to create new variables (but doing what you said...draw more to extrapolate more information and hence expressions independent of the original). Thus, your diagram must be carefully done...if not you just have to insert new variables that are "derivatives" of the initial unknowns. Whenever I think I have a square matrix and stick it into the TI-85 to get a solution to the set of equations....I get a zero solution to the set (I forgot the term for it...been awhile since I've taken linear algebra).

PS: I know this problem is off topic and probably doesn't belong in the DAT forum. I am just taking a couple days off from the DAT (I've hit a plateau) and want to get my mind off the DAT till Monday. This is just a problem that has haunted me for years. Just wanted to share the problem with fellow predents who enjoy a good math puzzle.
 
Yes. I have drawn out many many patterns myself. Some very complicated. The point is NOT to create new variables (but doing what you said...draw more to extrapolate more information and hence expressions independent of the original). Thus, your diagram must be carefully done...if not you just have to insert new variables that are "derivatives" of the initial unknowns. Whenever I think I have a square matrix and stick it into the TI-85 to get a solution to the set of equations....I get a zero solution to the set (I forgot the term for it...been awhile since I've taken linear algebra).

PS: I know this problem is off topic and probably doesn't belong in the DAT forum. I am just taking a couple days off from the DAT (I've hit a plateau) and want to get my mind off the DAT till Monday. This is just a problem that has haunted me for years. Just wanted to share the problem with fellow predents who enjoy a good math puzzle.

Yea, it is a complicated problem, do you have hard bio questions!!!
 
feh. not sure if this is correct.

Angle X = CDE
Angle X = 20 degrees

Angle Y = CED
Angle Y = 50 degrees

the geometrical axiom I used was establishing a isosceles trapezoid by creating two 20-80-80 triangles, and bisecting the upper angles within the trapezoid.
 
I don't really know what you mean....I can see a trapezoid in more than one way. I know this is really asking a lot, but if you could post your solution that would be nice! If it is right...you'll be a legend in my mind!
 
I don't really know what you mean....I can see a trapezoid in more than one way. I know this is really asking a lot, but if you could post your solution that would be nice! If it is right...you'll be a legend in my mind!

I'm no legend, but here's what I sorted through in those few minutes I looked at it.

x = 20 still.
y = 110 (probably looks nothing like the drawing provided, and part of the reason why i think I'm off - not that anything is to scale)

I created one isosceles trapezoid within the 20-80-80 triangle by creating two (note quite) superimposing, smaller 20-80-80 triangles within. this gives you a top and side length that are equal in lengths, and when a diagonal is drawn (which for me was line DE), this bisects the upper left angle of the trapezoid. the angles of the isosceles trapezoid are 100 at the top, so 100/2 - angle BDC (whish is 30). that gave me 20.
 
I'm no legend, but here's what I sorted through in those few minutes I looked at it.

x = 20 still.
y = 110 (probably looks nothing like the drawing provided, and part of the reason why i think I'm off - not that anything is to scale)

I created one isosceles trapezoid within the 20-80-80 triangle by creating two (note quite) superimposing, smaller 20-80-80 triangles within. this gives you a top and side length that are equal in lengths, and when a diagonal is drawn (which for me was line DE), this bisects the upper left angle of the trapezoid. the angles of the isosceles trapezoid are 100 at the top, so 100/2 - angle BDC (whish is 30). that gave me 20.

Hmm...ok, I have a vague idea of what your talking about. I've drawn so many patterns for this problem (some pretty crazy ones as well) and probably did what you did at one point. Here is why I think you may be off....how do you know DE bisects the trapezoid such that you get equal halves? Also, I am having trouble seeing how you know that the top and sides are equal. If you look carefully at the diagram you will see that line DE is the "ghost line" or "bastard line." We are given no information regarding that line....that line seems to be independent of all the information given in the problem, and any other angles we derive also are independent of line DE. Extend line DE such that it intersects with BC. Now examine the angle that forms from the intersection....do you see what I mean? Lots of patterns I draw get stuffed because of that "bastard line."
 
Okay this is crazy long and complicated but I got it. I bet anything there's a simpler way, but this still works.

The answer is 20.

Try to follow along. Draw what I describe. Afterwards I'll create a few drawings to put into this at certain points and edit them in (if I have any time haha).

1) Create point F on AB such that BF = BC. Then angle BCE = angle BFC = 50.
2) Create point G on AC such that angle GBC = 20.
3) Since GBC = 20 and BCA = 80, we have CGB = 80. Thus triangle BCG is isosceles (BG = BC).
4) Since BF = BC and BC = BG, we have BF = BG.
5) Since CBG = 20 and ABC = 80, we have ABG = 60.
6) Draw line segment FG. Since BFG is isosceles and FBG = 60, we have BGF = GFB = 60. So triangle BFG is actually equilateral (BF = FG = BG).
7) Since FBG = 60 and ABE = 20, we have EBG = 40.
8) Since BGC = 80, we have BGA = 100.
9) Since BGA = 100 and EBG = 40, we have BEG = 40.
10) Since EBG = BEG = 40, triangle BEG is isosceles with BG = GE.
11) Since BG = FG (step 6) and BG = GE, we have FG = GE (triangle FGE is isosceles).
12) Since triangle FGE is isosceles and FGE is 40, we have GFE = FEG = 70.
13) Since BFG = 60 (step 6) and BFC = 50 (step 1), we have CFG = 10.
14) Since CFG = 10 and GFE = 70, we have CFE = 80.
15) Since BFC = 50 and CFE = 80, we have BFE = 130 (and also EFD = 50).
16) Since CFE = 80 and EFD = 50, we have CFD = 130.
17) Since BCD = 70 and ABC = 80, we have BDC = 30.
18) Since FEG = 70 (step 12) and BEG = 40 (step 9), we have FEB = 30.
19) Consider triangles BFE and CFD. In the former, angle B = 20, angle F = 130, and angle E = 30. In the latter, angle C = 20, angle F = 130, and angle D = 30. Since these are the same measurements, the two triangles are similar.
20) Since BFE and CFD are similar triangles, we note that DF/EF = CF/BF.
21) This can be arranged to read DF/CF = EF/BF.
22) Note triangles EFD and BFC. Note that in the former, angle F = 50 and in the latter, angle F = 50. Note that from step 21, DF/CF = EF/BF. So triangles EFD and BFC are also similar.
23) Since EFD and BFC are similar, their corresponding angles are equal in measurement. We already have angle F in both = 50. So in triangle EFD, we note that angle D has the same measurement as angle C does in triangle BFC. Since angle C in triangle BFC = 50, we have angle D in triangle EFD = 50.
24) Since angle EDF = 50 and angle CDF = 30, we have angle CDE = 20.

This is our desired angle.
 
What the hell...this isn't Fermat's last theorem. haha...its too much to just mentally follow. I will scrutinize it much later when I have a few hours of free time. If it is correct....you are a super genius. That problem is also known as "the problem that stumped the nation." Top mathematicians, NASA, the math people from the ETS couldn't do it.

Damn Streetwolf...you're a persistent guy. Here is something else I've been working on for awhile...you got a background in pure math or something of that nature? Give this a crack sometime.
 
nevermind it won't load....but the problem asks you to explain the missing piece in a triangle. its like some kind of conservation paradox...it will make you feel like your losing your mind.
 
was that the actual answer?
PS: everyone on this forum can attest to streetwolf's voracity for math and science.. where do you think the pseudoname came from..the guy is half man half wolf..lol!!
 
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