Geometry Problem: Uniquely Solving for a and b | Elementary Geometry Figure

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The discussion revolves around an elementary geometry problem involving angles a and b in a figure, where the only known constraint is that a + b = 90 degrees. The contributor initially explores whether there are unique values for a and b, concluding that multiple solutions exist since any pair satisfying the constraint is valid. A proposed solution from a GRE problem suggests exaggerating angle differences to conclude that a > b, but some participants question the validity of this approach. They emphasize that without confirming the figure is a rectangle, the relationship between a and b cannot be definitively established. Ultimately, the consensus is that if the figure is indeed a rectangle, then a > b holds true.
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Hi,

This is not really a problem for the template (it is not even homework). I have come across the elementary geometry problem shown in the figure below.

http://img151.imageshack.us/img151/95/geometryproblemvc9.th.png

I am wondering whether there is some way to solve for a and b uniquely. Although I was able to come up with an expression for every single angle in the diagram in terms of a or b, the only constraint I have found so far is the obvious one: a + b = 90 degrees. That alone leads to infinitely many solutions. Wondering whether there was another constraint I had missed, I tried an arbitrary solution a = 40, b = 50, and it works. It seems more likely that any thing in the solution set to a + b = 90 is fine, rather than me having stumbled upon the only unique solution there is.

The problem as stated does not ask for the values of a and b, it asks only whether:

a > b
a < b
a = b
OR
there is not enough info to make a determination


(yes this is one of those stupid GRE "quantitative comparision" questions)

The solution proposed by the GRE people is to exaggerate the difference between the 91 degrees and the 89 degrees by drawing the former angle larger, requiring that the nearly-square rectangle be redrawn as a rectangle much wider than it is tall. Once this has been done, it becomes "clear" that a > b.

I am wondering whether there is a less crude way of arriving at this conclusion (one that makes use of geometric principles)

Thanks
 
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Am I confused? If the figure is supposed to be a rectangle, then all of the triangles are isosceles. So 2a+89=180 and 2b+91=180. a>b. Am I missing something that is making this hard?
 
We don't know that the angles at the corners are right angles.
 
If its an arbitrary quadrilateral then a>b is not necessarily true. Start drawing non-rectangular cases.
 
Oh...yeah.

Dick said:
Am I confused? If the figure is supposed to be a rectangle, then all of the triangles are isosceles. So 2a+89=180 and 2b+91=180. a>b. Am I missing something that is making this hard?

No, I don't think you are missing anything. It was supposed to be a rectangle, so that's the answer. Thanks. :redface:
 
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