Random question about shapes

Do you mean equal in terms of their area?  Imagine holding your scissors at the far left edge of any shape.  If you cut it there, the right piece will obviously have more area than the left.  So you slide the scissors from left to right, looking for a better place to cut.  If you slide too far, the scissors will end up towards the far right edge, and then the left piece will obviously have more area than the right piece.

If I can slide my scissors along a continuous path from left to right, so that in my starting position there's more area in the right piece and in the ending position there's more area in the left  piece, it logically has to be true that at some moment my scissors were over the location where that transition happened - that location is a place where you can cut and have two halves of equal area!
Depends on what you mean by "equal halves", like do you mean they have equal volume or area or ending up with two identical superimposable shapes ?
Even though the answer is somewhat obviously (once one had analysis) "no", the question itself is very good and should be exactly the thing talked about in high-school math imo.
You might be able to come up with a fractal shape for which you couldn't find the point to finish the cut. Like, the cut would be infinitely long.
You need to be more precise with what you mean that two shapes are equal. One proposal for equality below is area, and in that case the answer is no, you can always find a balance line on which the weight on each side of the shape is the same.

But if your notion of equality is a different one. You may even have to be clearer what you mean by cut in specific circumstances. Say I cut an R more or less horizontally between the top hole and the bottom "legs". There is a line segment coinciding with the cut. How is a cut treating geometry that coincides with it? Is it given to one or the other, is each keeping the line? Is the line removed? This changes what cut shapes you end up with and can chance what answers you can get. In other words here we are trying to be more precise about what we mean by "cut".

In short mathematics is about being careful what you are asking and becoming more and more precise about what the question really is and what the answers hence have to be.
I think it depends on what sort of constraints there are on the cutting and on the result

That is, I think you could design a shape such that if you are only allowed one straight cut, you would not be able to have exactly two pieces of equal area - i.e. you'd end up with three or more pieces
by
This is a great question.

As others have said for a smooth shape a single straight cut transitioning from the left boundary to the right boundary will increase the area.

At some point it will equal half the area and the answer is yes.

Of further interest though are non smooth shapes.
For instance An L shape.  A vertical cut transitioning from left to right will encounter a discontinuity in area and therefore it may not be possible to divide into a half area (at least verically).

This rules out any vertical(and for same reason horizontal) cut.

one can further extend this shape by introducing an angle parallel to a given cut angle. Ruling out that cut.

You can continue this process infinitely.

The question therefore arises, can a shape exist which has a side that is parallel to every possible cut angle.

If that shape exists then than answer will be no.

If it does exist what does it look like?
That shape can be a single point, since no line splits a point into two equal halves.

It can also be any odd number of points. If you split and odd number of points into two sets, one will have an even number of points, and the other will have an odd number.
For any shape in any dimensional space (2D,3D,4D etc) with a positive measure (area, volume, hyper volume etc) you can always find two smaller shapes with equal measure. For instance you could do this by putting a ball at the origin of a certain radius and thinking about the measure of the intersection of the ball with the given shape. The area of the intersection changes continuously with the radius. Now you send the radius to infinity and apply the intermediate value theorem as others have mentioned.

There’s a numberphile video called “fix a wobbly table” that uses a very similar idea but applied in a different context.
by
Equal as in same area/volume or equal as in congruent?
If first, it's always possible to cut into two "equal" halves, if second, it's not always possible to cut into two "equal" halves.
For the second case you need only look at an scalene triangle.

0 like 0 dislike