does this formula work if the lines between cartesian points are not a straight line? For instance, if you have a kidney shaped pool in a backyard and the backyard is 10 x 10, how much precision is needed to plot points to determine the area of the kidney shaped pool?
If you want to figure out the area from an arbitrary shape you found somewhere you need to work out the grid first to use that method, right? It is possible the grid get very tiny to the point you calculate the shape like you would do it “normally”. But anyway it is still very cool thing to know.
if you use the formula for a shape with n holes, it will also work, but for a video I think is simpler to explain using examples with an exact number of holes
So you're telling me that if it has an infinite amount of holes, the area would be infinite... I don't get the fact that the more holes in the figure, the bigger it will be.
You only proved that the Pick's Theorem is valido for Lattice-Aligned Right Triangles without boundary points in the hypotenuse, it is not clear how to generalise the argument for general triangles.
Tl;dr: It suffices to consider lattice aligned right triangles, since any lattice triangle can be rotated and then subdivided into two lattice-aligned right triangles by drawing a height from one of the vertices. This means every lattice polygon is the nonoverlapping union of lattice-aligned right triangles, with any two distinct triangles sharing at most one side. The proof follows the merging argument in the video. Suppose the original triangle has B boundary points and I interior points.There are exactly 2 boundary points which lies on the height. Suppose also there are C interior points which lies on the height. Those C points become boundary points when we subdivide the triangle. The areas of the two right triangles are given by the formula, which counts a total of B+2C+2 boundary points and I-C interior points. The sum of their areas is (B/2+C+1)+(I-C)-2= B/2+I-1, but this sum is exactly the area of the original triangle, so the formula does work for any lattice triangle.
@@divisix024 I see 2 problems with this argument. 1. there is no reason for the triangle to keep being latice aligned after being rotated(if for example none of their sides have integers length). 2. even if they do, you would still need to prove that after the rotation the triangle will have te same amount of points inside and on the border. I think the better argument is to take te smallest rectangle that encloses the triangle and observe that it can be separated in to 3 latice aligned triangles and the original triangle
You overcomplicated your explanation Once you’ve explained the chain of logic you don’t need to reexplain it every time you can just hop to the end If it requires going through the process again then go through only the pertinent parts of the process
![The most beautiful equation in math, explained visually [Euler’s Formula]](http://i.ytimg.com/vi/f8CXG7dS-D0/mqdefault.jpg)
Why doesn't this channel have more views? This is very educational!
1: early start
2: .. algorithm hates him
Yess@@ErdemtugsC
What is really good doesn't get views.
it is unnoticeable
Awesome video! The explanation was very clear and helpful. You deserve a lot more views!
The quality of these videos are insane compared to its number of views, keep up the good work
How is this channel this small!? You absolutely deserve my sub. 👍
This will come in handy! Thank you!
Thank you, so much
really calming audio :)
what program do you use to make these videos? i want to try making some myself
This video is going to get Millions of views in the future
I forwarding it to increase the views
I feel big brain now
YoFeArIr
does this formula work if the lines between cartesian points are not a straight line?
For instance, if you have a kidney shaped pool in a backyard and the backyard is 10 x 10, how much precision is needed to plot points to determine the area of the kidney shaped pool?
Continue making videos!
If you want to figure out the area from an arbitrary shape you found somewhere you need to work out the grid first to use that method, right? It is possible the grid get very tiny to the point you calculate the shape like you would do it “normally”.
But anyway it is still very cool thing to know.
It's such an ingenious formula that one Russian mathematician even built a career on it
решил по формуле Пика за.... хотя, подождите.
Underated
0:16 what happened
I love your videos!!!
How do you edit your videos?
Great video! But I want a example where the lettuce polygon is very very big and you a very very big hole there too
I didn't understand how we used specific cases (like 1 or 3 holes) to demonstrate the formula for n holes.
if you use the formula for a shape with n holes, it will also work, but for a video I think is simpler to explain using examples with an exact number of holes
Okay, but what about disconnected shapes?
just find the area of both and add together
So you're telling me that if it has an infinite amount of holes, the area would be infinite...
I don't get the fact that the more holes in the figure, the bigger it will be.
i feel cursed. the universe looked upon me.
You only proved that the Pick's Theorem is valido for Lattice-Aligned Right Triangles without boundary points in the hypotenuse, it is not clear how to generalise the argument for general triangles.
Tl;dr: It suffices to consider lattice aligned right triangles, since any lattice triangle can be rotated and then subdivided into two lattice-aligned right triangles by drawing a height from one of the vertices. This means every lattice polygon is the nonoverlapping union of lattice-aligned right triangles, with any two distinct triangles sharing at most one side.
The proof follows the merging argument in the video. Suppose the original triangle has B boundary points and I interior points.There are exactly 2 boundary points which lies on the height. Suppose also there are C interior points which lies on the height. Those C points become boundary points when we subdivide the triangle.
The areas of the two right triangles are given by the formula, which counts a total of B+2C+2 boundary points and I-C interior points. The sum of their areas is (B/2+C+1)+(I-C)-2= B/2+I-1, but this sum is exactly the area of the original triangle, so the formula does work for any lattice triangle.
@@divisix024 I see 2 problems with this argument.
1. there is no reason for the triangle to keep being latice aligned after being rotated(if for example none of their sides have integers length).
2. even if they do, you would still need to prove that after the rotation the triangle will have te same amount of points inside and on the border.
I think the better argument is to take te smallest rectangle that encloses the triangle and observe that it can be separated in to 3 latice aligned triangles and the original triangle
It’s very annoying that you’re Not providing The solutions for all The examples ( ? )
You overcomplicated your explanation
Once you’ve explained the chain of logic you don’t need to reexplain it every time you can just hop to the end
If it requires going through the process again then go through only the pertinent parts of the process
aka shoelace
If you could just speak clearly it would be perfect.
It sounds like you are an adult on charlie brown, LMAO!