3 ms·
For reference here's a GPT-3 output for the sample question: Solution: Suppose the given set of n points is A = (a 1 , a 2 , …, a n ). Between any two points
by iandanforth 6y ago
For reference here's a GPT-3 output for the sample question:
Solution:
Suppose the given set of n points is A = (a 1 , a 2 , …, a n ). Between any two points we will draw a line segment connecting them, and let the distance from a 1 to the line segment be d(a 1 ,a 2 ) and d(a 1 ,a 3 ) be the distance from a 1 to the line segment connecting a 2 and a 3.
Since any two of the points determine a line segment connecting them, there are n–2 segments connecting the n–1 points.
Let L be the common perpendicular bisector of two of these segments. So, d(a 1 ,a 2 ) = d(a 1 , L) and d(a 1 ,a 3 ) = d(a 1 , L).
Now, we consider the triangle formed by the points (a 1 , a 2 ) and (a 1 , a 3 ). If we are given the points a 1 , a 2 and a 3 , we can easily find the sides of the triangle.
Since d(a 1 ,a 2 ) + d(a 1 ,a 3 ) = d(a 1 , L) + d(a 1 , L) = 2d(a 1 , L), the triangle formed is right angled and therefore non-degenerate.
We can easily calculate the rational area of such a triangle and the side lengths.
Let us consider the simplest case when n = 3.
We have three points (a 1 , a 2 , a 3 ) such that each of a 1 , a 2 and a 3 are different.
Therefore, we can form a triangle by connecting a 1 , a 2
- pmiller2 6y agoThis is great. It looks like the confused ramblings of a student who has no clue.
- singhrac 6y agoAlways fun to grade these kinds of proofs with a bottle of wine at 2am. I never faulted anyone for doing this, but I sometimes wondered why they bothered. Throwing stuff to a wall to see what sticks doesn't give you points.
- Jtsummers 6y agoOne of my math professors had a "BS" stamp for things like this. If it were worth X points, you'd lose X-5 when it was so obviously a rambling mess. He'd rather you just say "I don't know" than waste both your time and his. Perhaps that's just an advantage of tenure, though.