Can Four Shapes Share Sides Without Gaps?

[Flexington]

1. Found this question in a past paper i am looking at in preperation for a midterm.
Draw a diagram that shows four shapes of equal dimensions with each shape having at least one side in common with the other three shapes. The picture should be 2d containing no gaps or shapes.

I do not know of any mathmatical theories so far that can help solve this problem. I assume its one of logic reasoning.

3. I have no idea where to start but to guess at random shapes and see if they fufil the criteria. However this has been unsuccesfull. Please help.

 

[willem2]

What does "containing no gaps nor shapes" mean? No gaps probably means that all the shapes together should form a convex shape, but I have no idea what "containing no shapes" is.

If you take 4 identical triangles or rectangles there are many ways you can combine them in a convex shape, most of them very simple.

 

[Flexington]

Apologies. It should read no gaps or holes.

 

[Flexington]

I am assuming the shapes must be asymmetrical or be combined to form an asymmetrical picture. This is because the intersect of the axis of symmetry due to the four shapes will form a point between the four shapes, preventing each individual shape from having at least one common border with the other three.

 

[willem2]

do you mean each shape must have one side in common with all the other shapes?

 

[Flexington]

yeah.

 

[D H]

Flexington, this appears is homework, so at a minimum you need to show some work. You have not done that.

Moreover, you claim this is from an exam question. Schools typically have rules against getting outside help on exams. Students who violate these rules can end up in very big trouble. You need to assure yourself that asking for help on this particular problem is within the bounds of your school's rules on outside help.

 

[Flexington]

i am new to the forum and am not farmiliar with the syntax for drawring shapes as of yet. i have been assuming that this problem can be solved, however it maybe that it can't and the answer is a proof that the criteria can't be met. Like i said i have no idea how to approach this except for attempting random shapes, and so far have been unsuccesful. Any advice to push me in the right direction would be most appreciated.

 

[fzero]

One way to approach this question is with graph theory. The corners of the shapes are vertices, V, the lines connecting the vertices are edges, E, and the number of shapes that make up the tiled object are the faces, F. Since edges can't cross, the object is a simple, connected graph and must satisfy Euler's condition

V-E+F=2.

Now the shapes (polygons) that are being used as tiles are also simple and connected, so their vertices, v, and edges, e, must satisfy (they have one face, so f=1)

v-e =1.

Now you need to use the rules about sharing faces and having no gaps to compute V and E in terms of v and e. If we find a contradiction with Euler's condition, we can conclude that we can't build an object subject to the rules.