Topic review (newest first)
The approximate values are: ∠A = Arccos ≈ 20.74°;∠B = Arccos ≈ 127.17°; ∠C = Arccos ≈32.09°
The sides are in geometric sequence: 7.2, 10.8 are the two sides. Obviously the sides are not corresponding to the angles.
I leave you to figure out the rest.
- 2005-12-21 21:25:37
Or it would be a straight line.
- 2005-12-21 21:24:35
True. I forgot about that bit. Yes, two sides of a triangle must always be greater than the other side, because otherwise they wouldn't be able to reach its two ends.
- John E. Franklin
- 2005-12-21 15:03:21
in mathsyperson's generalization, variable b must be closer to 1 than the golden ratio or its reciprocal, 1.618 and .618
So it appears. if b > 1, then a + ab > ab².
If b < 1, then a < ab + ab²
- 2005-12-16 03:41:13
It looks good to me. You've just scaled up by √2, so all the angles would be the same, and two sides match as well.
The general solution would be the first triangle having sides of a, ab and ab² and the second having sides of ab, ab² and ab³ (a>0, b ≠0 or 1) .
In ganesh's example, a = 1 and b = √2.
- 2005-12-15 15:10:19
In ΔABC, AB=1, BC=√2, AC=2.
In ΔDEF, DE=√2, EF=2, DF=2√2.
Since ΔABC and ΔDEF are similar, the three angles are equal.
And they have two sides equal.
Isn't this a solution?
- 2005-12-15 10:56:20
There is a 5 con triangle... there are things written about it, i just cant find the exact dimensions. The only thing i know about it is that there is only one, and it has some exception to the rule.
A 4 con triangle is very very possible... even in 30/60/90's... just try it.
The thing about the triangle is that the equal sides/angles dont have to be corressponding, which is why the SAS AAS etc will not prove it to be congruent...
- 2005-12-15 03:46:17
The easiest way to prove that none exist is to refer to the AAS rule. That is, if two triangles have 2 similar angles and one similar side, they are congruent.
A 5-con triangle would be a non-congruent triangle with 5 similar properties. For this to happen, it would need at least 2 similar angles and 2 similar sides, and this satisfies the AAS rule, so it would have to be congruent and so be a 6-con triangle.
In fact, I don't think you can even have 4-con triangles.
- 2005-12-14 18:20:55
Thats right! I realized the mistake after I posted and logged out.
In fact, a 30-60-90 degree may well be ruled out, as the sides would have to be in the ratio 1:√3:2.
I shall think about it and post, as of now, I ain't sure a solution exists!
- 2005-12-14 15:42:34
a 3 4 5 right triangle is not a 30 60 90 triangle, so it would not work for that...
- 2005-12-14 15:25:53
I guess a right angled triangle with angles 30, 60 and 90 degrees would be ideal to start with. Let the sides be 3, 4 and 5 units for the first traingle. The sides of the second triangle would then be 4, 5 and √(41).
- 2005-12-14 14:00:35
Im looking for the angles/sides of a 5-con triangles
What 5-Con triangles are, are any two triangles who have 5 (not necessarily corresponding) sides and angles equal.
Triangles obviously have 3 angles and 3 sides, so 5 of the 6 must be equal.
I know the 3 angles must be the same and 2 sides equal, because if all 3 sides are equal lengths, all 3 angles must be also.
So pretty much i need to find 2 triangles, who have all three angles equal, but only 2 sides equal. Any ideas?
(Ive been working with 30/60/90 triangles but cant seem to come up with it)