is the answer intuitive or do you need to know geometry theorems, because i havent had geometry in like 9 years
just tell us if you're gonna keep bumping, but rest assured noone really cares. i hate to break it to ya
make a new point X 6 units from P on PC, and you'll have two isosceles triangles BXP and EXP. When you combine both you get the trangle BXE, with the angle X = 90 degrees. Since BP = 7 = PE, BXE is also isosceles. Hence you get the angles on BPC, CPE, EPF, FPB, all 90 degrees. Once you know BE is perpendicular to FC, everything sleep is trivial.
Lithium381
Lifer
- May 12, 2001
- 12,452
- 2
- 0
yeah wow, that would take a while, i remember i USED to know how to do those, but i don't have time to look for those now
No good.
You can not assume angle X (in triangle BXE) = 90 degrees.
EDIT: Nevermind. My advanced math from over a decade ago is catching up with me now. :disgust:
EDIT2: But how do you know that BXE is also isosceles?
43st
Diamond Member
- Nov 7, 2001
- 3,197
- 0
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Bump...
You don't need any angles. Basically there's only one way for a triangle shade to surround a 6 pointed object with all points touching the outer triangle. Assume that the 20 length intersects the other lines and you can only have one answer (see above). It's a simple (or not so simple) matter of drawing the object so all points meet the walls of the triangle.
Hint: adjusting the angle between the two 12 lengths is what allows the 20 length to fit properly. The trick is that you can't adjust the 20 length until after you've changed your angle because it will effect the outer triangle.
You don't need any angles. Basically there's only one way for a triangle shade to surround a 6 pointed object with all points touching the outer triangle. Assume that the 20 length intersects the other lines and you can only have one answer (see above). It's a simple (or not so simple) matter of drawing the object so all points meet the walls of the triangle.
Hint: adjusting the angle between the two 12 lengths is what allows the 20 length to fit properly. The trick is that you can't adjust the 20 length until after you've changed your angle because it will effect the outer triangle.
BP=6=PE (just a typo, I presume). But I don't see how you get bx=xe to make the triangle isosceles. As far as I can tell, we've got one side and one angle on BXE. We don't even have PX as the true height, because we don't know that PX bisects the right angle, right?
solve for triangle BEC and FBC. Solve for angle B, then C. The rest is easy.
Edit: First solve CEp Then BFp. These give you FB and EC. With these, BEC and FBC are solved. A little more geometry yields BC being 13 and angles B & C being approximately 67.38 degrees. Angle A is therefore 45.24 degrees.
Questions, comments, words of wisdom?
Edit: First solve CEp Then BFp. These give you FB and EC. With these, BEC and FBC are solved. A little more geometry yields BC being 13 and angles B & C being approximately 67.38 degrees. Angle A is therefore 45.24 degrees.
Questions, comments, words of wisdom?
I don't think that you can assume triangle BXE is isosceles.
I kind of follow what you're saying. I understand that there can only be one solution given the information, but I don't see how to derive that solution.
Could you explain how you do this? I'm at a loss.
Any hints on how you approached the problem for the rest of us?
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