Hard Geometry Challenge

1. Sep 4, 2012

henrique_p

Well, I found this challenge in another forum (not about math) on the internet, and, originally, there were no 'w', 'z' or 'y' drawn on the pic, it was just "find the x", but I put them on because I know you guys probably would create other variables to solve the problem.
Another problem I had to deal with is if the Quadrangle is, by fact, a square, I mean, if 'w' plus 'z' really is 1. How the wreck do I solve this?

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2. Sep 4, 2012

mathman

(y/1) = (1+y)/(1+x) (both are secants of the same angle)

1 + (1+x)² = (1+y)² (Pthag. theorem)

I'll let you do the algebra.

3. Sep 4, 2012

acabus

Could you do (y/1) = (1/x) instead?

4. Sep 4, 2012

henrique_p

Yes, in this case x=y$^{-1}$.

5. Sep 4, 2012

henrique_p

Just another doubt, how can I know if the biggest triangle (1, 1+x, 1+y) has a 90º base angle? Because it wasn't specified in the problem if there is a square.

6. Sep 5, 2012

mathman

It is still OK even if is not a square. If it is a rhombus, the triangles involved are similar, so the side ratios {y:1 = (1+y):(1+x)} are equal, even if not the secant of the angle.

If it is not a rhombus, then I suspect there is no solution.

7. Sep 7, 2012

mathman

Added note: In case of a rhombus, you need to know the angle, so that you can use the law of cosines to get the secon equation.

8. Sep 7, 2012

henrique_p

Ok, by doing the algebra I found the equation on the pic. I don't know if this was right, so I went to Wolfram to solve this.

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9. Sep 7, 2012

willem2

No real solutions, so it wasn't right.

combining

a) w+z = 1
b) x^2 + z ^2 = 1
c) w^2 + 1 = y^2
d) y/1 = 1/x

use c amd d to get

e) w^2 + 1 = 1/x^2

and then a and e to get

f) z^2 - 2z +2 = 1/x^2

and finally b and f to get

g) (z^2 - 2z+2)(1-z^2) = 1

feeding this to wolfram alpha:

http://www.wolframalpha.com/input/?i=%28z^2+-+2z+%2B2%29%281+-+z^2%29+%3D+1%2C+sqrt%281-+z^2%29

gets you x ≈ 0.883204

10. Sep 8, 2012

mathman

I got: y4 + 2y3 - y2 -2y -1 = 0.

A quick inspection shows there is a root between 1 and √2, which is expected.

Last edited: Sep 8, 2012
11. Sep 8, 2012

henrique_p

Am I wrong or I can divide this equation by (y-1)?

12. Sep 9, 2012

mathman

1 is not a root, so you can't divide by y-1.

13. Sep 9, 2012

henrique_p

Yeah, sure, I miscounted the coefficients.