Problem:

A vertex of one square is pegged to the centre of an identical square, and the overlapping area is blue. One of the squares is then rotated about the vertex and the resulting overlap is red.

Which area is greater?Let the area of each large square be exactly 1 unit squared. Then, the area of the blue square is exactly 1/4 units squared. The same would apply to the red area if you were to rotate the square k⋅45 degrees for a natural number k.

Thus, I am assuming that no area is greater, and that it is a trick question − although the red area might appear to be greater than the blue area, they are still the same: 1/4.

But how can it be

proven?I know the area of a triangle with a base b and a height h⊥b is bh÷2. Since the area of each square is exactly 1 unit squared, then each side would also have a length of 1.

Therefore, the height of the red triangle area is 1/2, and so Red Area=b(12)2=b4.

According to the diagram, the square has not rotated a complete 45 degrees, so b<1. It follows, then, that Red Area<14⇔Red Area<Blue Area.

Assertion:To conclude, the blue area is greater than the red area.

Is this true? If so, is there another way of proving the assertion?

Thanks to users who commented below, I did not take account of the fact that the red area is

nota triangle − it does not have three sides! This now leads back to my original question on whether my hypothesis was correct.This question is very similar to this post.

Source:The Golden Ratio (why it is so irrational) − Numberphile from 14:02.

**Answer**

The four numbered areas are congruent.

[Added later] The figure below is from a suggested edit by @TomZych, and it shows the congruent parts more clearly. Given all the upvotes to the (probably tongue-in-cheek) comment “This answer also deserves the tick for artistic reasons,” I’m leaving my original “artistic” figure but also adding Tom’s improved version to my answer.

**Attribution***Source : Link , Question Author : Mr Pie , Answer Author : Steve Kass*