Can it be?
Can there exist positive real numbers a , b , c such that they satisfy a 2 + b 2 = c 2 and they are not the three side lengths of a triangle?
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5 solutions
Nice solution. :D
Just a typo in last equation, a 2 + 2 a b + b 2 = c 2 + 2 a b > c 2
But 1,0 and 1 are possible without being sides of triangle.
The problem said that all the numbers are positive
The solution is correct in 2 dimensions, what would happen if we are in the 3rd or 4th dimension ?
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What is the relationship between dimension and this problem?
The easiest solution is to think that if any three positive numbers a,b and c satisfy the condition a² + b² = c², then they are the sides of a right angled triangle (Also known as the converse of Pythagoras theorem). Therefore three such numbers will always form a triangle.
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I agree with you and I used the same way to choose my answer
As long as A + B > C which must be proven. Here is a proof by contradiction. Assume A+B< C Then squaring both sides (A+B)^2<C^2, because all are positive real numbers. Expanding: A^2 + 2AB + B^2<C^2 But A^2 + B^2 = C^2 Therefore 2AB<0, thus AB<0 Thus either A or B (but not both) must be negative But A and B are both positive real numbers This is a contradiction, thus assumption is false. QED
So let a,b,c be positive real numbers that satisfy a^2+b^2=c^2 I will make proof by contradiction:
I first asume that it is posible, so I must check 3 options:
1) a+b < c , since a,b,c are positive real numbers I can square both sides of the inequality.
a^2+2ab+b^2 < c^2
a^2+b^2-c^2 < -2ab and since they satisfy a^2+b^2-c^2=0 I can rewrite this as follows
0 < -2ab
ab < 0 and this is contradiction.
2) b+c < a again I can square both sides.
b^2+2bc+c^2 < a^2 now I will add b^2 to both sides of the inequality.
2b^2+2bc < a^2+b^2-c^2
2b^2+2bc < 0 and since 2b^2>0 it must be true that bc < 0 and again this is contradiction.
3) a+c < b square both sides
a^2+2ac+c^2 < b^2 add a^2
2a^2+2ac < a^2+b^2-c^2
2a^2+2ac < 0 so again ac < 0 and this is contradiction.
And we are done.
Sorry for how it is written, but when I posted it, it messed up.
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It's ok. It's so helpful. Maybe try to use latex later.
This must have happened because you left no line between two of your sentences.
Leave a line after each sentence to make sure they don't get messed up in one line
Good reply @Dan Czinege
I mean we can simply construct length a, and perpendicular to it, length b. Now, by Pythagoras theorem, the hypotenuse is bounded to be sqrt(a^2+b^2)= c (given). Hence we have given an algorithm to construct a right angled triangle for all a, b, c belonging to positive real numbers.
For given c > 0 , all points ( a , b ) with a , b > 0 and a 2 + b 2 = c 2 lie on the open circle arc with the origin as center and radius c in the first quadrant. In each case, there exists a triangle (even a right triangle) with side lengths a , b , c , viz. with vertices ( 0 , 0 ) , ( a , 0 ) , and ( a , b ) .
a 2 = c 2 − b 2 = ( c + b ) ( c − b ) = m × n
If both M and N is larger than a then m × n > a 2
If both M and N is smaller than a then \m\times n<a^2)
Hence c − b < a < c + b , the requisites to form a triangle
Assume that a 2 + b 2 = c 2 is satisfied for positive real a , b and c . Then a < c and b < c . Clearly, then, b + c > c > a and a + c > c > b . Thus, it only remains to show that a + b > c and we will have a triangle. But ( a + b ) 2 = a 2 + 2 a b + b 2 = c 2 + 2 a b > c 2 . So a + b > c .