Problem with an equally divided square

[Luke94]

Some time ago I found a mathematical problem, which at first seemed very simple to me, but which I still could not solve. Maybe someone has an idea or an approach. Many thanks in advance.

Task: Prove that the diagonal intersection point P of a square EDFG lies on the straight line g, which divides the square into two parts of equal area.

My approach: First I thought about the symmetry properties. Therefore, I first have two equations: f(x)= abs x-n and f(x)=abs x+n

with the thought that -n<x<n. Still, I couldn't get any further. Probably my approach is also little to use. Maybe someone here has an idea.

Best regards

 

[Dr.Peterson]

Can you explain what your constant n, variable x, and function f represent?

Also, the problem could be stated better, as there is more than one "straight line g which divides the square into two parts of equal area". Do you mean that P lies on EVERY line that divides into equal parts, or on SOME such line, or what? There is no "the".

 

[Luke94]

Can you explain what your constant n, variable x, and function f represent?

Also, the problem could be stated better, as there is more than one "straight line g which divides the square into two parts of equal area". Do you mean that P lies on EVERY line that divides into equal parts, or on SOME such line, or what? There is no "the".

The square has exactly two diagonals, which cross in the middle. This one point is then P. If I want to divide the area of the square (with a line g) into two equally large parts, then this straight line leads compulsorily through this point P; in any case I want to show exactly that.

Anyway, my approach is probably useless.

 

[lev888]

Let's pick a line g that's at some angle to the base of the square (easier to explain). Draw a horizontal line through P. It divides the square into 2 rectangles, which have equal area. The 2 lines also form 2 triangles that have equal areas (why?). Now, what is the difference between the rectangles, and the shapes the line g produced? The triangles. What can we conclude about the areas of those shapes if the rectangles and triangles have the same area?

 

[JeffM]

This is in my opinion a trick question because it implicitly references an infinite number of lines (all that go through a specific point) and then asks you to find the line that has a certain property when there are four such lines. Furthermore the corners are confusingly labeled; EDFG rather than DEFG is an invitation to confusion.

Let P, Q, R, and S be the midpoints of ED, DF, FG, and GE, and let X be the intersection point of line segments EF and DG, which I shall say define lines g_1 and g_2 respectively. The line segments PR and QS also include the point X, which define line segments g_3 and g_4 respectively. Each one of g_1, g_2, g_3, and g_4 divide the square into two equal areas. The first two divide it into two triangles of equal area. The last two divide it into two rectangles of equal area.

 

[JeffM]

The square has exactly two diagonals, which cross in the middle. This one point is then P. If I want to divide the area of the square (with a line g) into two equally large parts, then this straight line leads compulsorily through this point P; in any case I want to show exactly that.

Anyway, my approach is probably useless.

Try this approach, using pure geometry or analytical geometry.

First prove that an arbitrary line through the point of intersection of the diagonals does divide the square into equal areas. Next prove that any distinct line parallel to the arbitrary line either does not divide the square at all or divides it into unequal areas.

 

[Luke94]

Try this approach, using pure geometry or analytical geometry.

First prove that an arbitrary line through the point of intersection of the diagonals does divide the square into equal areas. Next prove that any distinct line parallel to the arbitrary line either does not divide the square at all or divides it into unequal areas.

I will try it, thank you very much.