<![CDATA[Math Is Fun Forum / 5-Con Triangles?]]> 2013-12-16T19:38:07Z FluxBB http://www.mathisfunforum.com/viewtopic.php?id=2203 <![CDATA[Re: 5-Con Triangles?]]> 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.

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2013-12-16T19:38:07Z http://www.mathisfunforum.com/viewtopic.php?pid=293955#p293955
<![CDATA[Re: 5-Con Triangles?]]> Or it would be a straight line.

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http://www.mathisfunforum.com/profile.php?id=1041 2005-12-21T10:25:37Z http://www.mathisfunforum.com/viewtopic.php?pid=21273#p21273
<![CDATA[Re: 5-Con Triangles?]]> 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.

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http://www.mathisfunforum.com/profile.php?id=641 2005-12-21T10:24:35Z http://www.mathisfunforum.com/viewtopic.php?pid=21272#p21272
<![CDATA[Re: 5-Con Triangles?]]> 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

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http://www.mathisfunforum.com/profile.php?id=1108 2005-12-21T04:03:21Z http://www.mathisfunforum.com/viewtopic.php?pid=21245#p21245
<![CDATA[Re: 5-Con Triangles?]]> 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.

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http://www.mathisfunforum.com/profile.php?id=641 2005-12-15T16:41:13Z http://www.mathisfunforum.com/viewtopic.php?pid=20715#p20715
<![CDATA[Re: 5-Con Triangles?]]> 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? ]]>
http://www.mathisfunforum.com/profile.php?id=682 2005-12-15T04:10:19Z http://www.mathisfunforum.com/viewtopic.php?pid=20669#p20669
<![CDATA[Re: 5-Con Triangles?]]> 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...

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2005-12-14T23:56:20Z http://www.mathisfunforum.com/viewtopic.php?pid=20641#p20641
<![CDATA[Re: 5-Con Triangles?]]> 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.

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http://www.mathisfunforum.com/profile.php?id=641 2005-12-14T16:46:17Z http://www.mathisfunforum.com/viewtopic.php?pid=20602#p20602
<![CDATA[Re: 5-Con Triangles?]]> 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!

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http://www.mathisfunforum.com/profile.php?id=682 2005-12-14T07:20:55Z http://www.mathisfunforum.com/viewtopic.php?pid=20572#p20572
<![CDATA[Re: 5-Con Triangles?]]> a 3 4 5 right triangle is not a 30 60 90 triangle, so it would not work for that...

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2005-12-14T04:42:34Z http://www.mathisfunforum.com/viewtopic.php?pid=20545#p20545
<![CDATA[Re: 5-Con Triangles?]]> 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). ]]>
http://www.mathisfunforum.com/profile.php?id=682 2005-12-14T04:25:53Z http://www.mathisfunforum.com/viewtopic.php?pid=20543#p20543
<![CDATA[5-Con Triangles?]]> 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)

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2005-12-14T03:00:35Z http://www.mathisfunforum.com/viewtopic.php?pid=20515#p20515