Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ ¹ ² ³ °
 

You are not logged in. #1 20051214 14:00:35
5Con Triangles?Im looking for the angles/sides of a 5con triangles #2 20051214 15:25:53
Re: 5Con 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). Character is who you are when no one is looking. #3 20051214 15:42:34
Re: 5Con Triangles?a 3 4 5 right triangle is not a 30 60 90 triangle, so it would not work for that... #4 20051214 18:20:55
Re: 5Con Triangles?Thats right! I realized the mistake after I posted and logged out. Character is who you are when no one is looking. #5 20051215 03:46:17
Re: 5Con 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. Why did the vector cross the road? It wanted to be normal. #6 20051215 10:56:20
Re: 5Con 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. #8 20051216 03:41:13
Re: 5Con 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. Why did the vector cross the road? It wanted to be normal. #9 20051221 15:03:21
Re: 5Con Triangles?in mathsyperson's generalization, variable b must be closer to 1 than the golden ratio or its reciprocal, 1.618 and .618 igloo myrtilles fourmis #10 20051221 21:24:35
Re: 5Con 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. Why did the vector cross the road? It wanted to be normal. #12 20131217 06:38:07
Re: 5Con Triangles?The approximate values are: ∠A = Arccos ≈ 20.74°;∠B = Arccos ≈ 127.17°; ∠C = Arccos ≈32.09° 