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**Dionysus****Guest**

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)

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,631

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).

Character is who you are when no one is looking.

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**Dionysus****Guest**

a 3 4 5 right triangle is not a 30 60 90 triangle, so it would not work for that...

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,631

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!

Character is who you are when no one is looking.

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**mathsyperson****Moderator**- Registered: 2005-06-22
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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.

Why did the vector cross the road?

It wanted to be normal.

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**Dionysus****Guest**

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...

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 16,631

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?

Character is who you are when no one is looking.

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**mathsyperson****Moderator**- Registered: 2005-06-22
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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.

Why did the vector cross the road?

It wanted to be normal.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

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

**igloo** **myrtilles** **fourmis**

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**mathsyperson****Moderator**- Registered: 2005-06-22
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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.

Why did the vector cross the road?

It wanted to be normal.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Or it would be a straight line.

A logarithm is just a misspelled algorithm.

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**Jaf****Guest**

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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