# Squares in squares

Squares are drawn inside the previously drawn squares as in the figure where the vertex of each square is at the mid point of the side of the previous square. It can be obtained that the areas of subsequent squares (greater to smaller) are in a GP. My question is, will this be true for any symmetric figure or any other figures? Will there be any relation between the number of sides of the figure and the common ratio of the GP?

Note by Maharnab Mitra
5 years, 10 months ago

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All regular polygons will have the same continuation. but with different G.P. ratios.. Suppose you have an n-sided regular polygon with vertices $$A_1,A_2,A_3,...,A_n$$. Say $$B_1$$ is the midpoint of line segment $$A_1A_2$$ and $$B_2$$ of $$A_2A_3$$, then if we join $$A_1$$ to $$A_3$$, then midpoint theorem gives us $$B_1B_2$$=$$1/2$$*$$A_1A_3$$. Also, $$A_1A_3$$=$$2n$$ $$\sin \theta/2$$, where n is the side of the polygon.

So $$B_1B_2$$= $$n$$ $$\sin \theta/2$$

also, in a polygon of n-sides(regular), $$\theta$$= $$\frac {180(n-2)}{n}$$, so $$B_1B_2$$=$$n$$ $$\sin \frac {90(n-2)}{n}$$

and the ratio of the sides of the new polygon to the old one, is thus $$\sin \frac {90(n-2)}{n}$$ and ther ratio of the areas is thus the required G.P. ratio, which is nothing but $$(\sin \frac {90(n-2)}{n})^2$$. It can be easily verified in all cases. For n=4, i.e., for a square, this becomes 1/2, for a triangle it becomes 1/4, etc.

- 5 years, 10 months ago

It is easy to see that this property is preserved under any non-degenerate affine transformation, because such transformations preserve the ratios of areas of figures. So if a given starting polygon is the image of some regular polygon under some affine transformation, then its "successive midpoint inscribed polygons" have areas in geometric progression.

However, while the above is a sufficient condition, it is not a necessary one: for $$n = 4$$, the family of kite quadrilaterals (even nonconvex ones) have this property; yet kites are not affine transformations of squares. A proof is quite straightforward, and I will leave it to the reader as an elementary exercise in geometry.

One should also note that not all quadrilaterals possess this property. I have not given much thought to finding a necessary and sufficient condition for the general case of $$n$$ sides.

- 5 years, 10 months ago

for regular shapes may be existed ratio according to the number of sides

- 5 years, 10 months ago

i think it is valid for square and have a doubt for rhombus

- 5 years, 10 months ago

yes, it cannot be valid for rhombus.

- 5 years, 9 months ago

yes

- 5 years, 10 months ago

- 5 years, 10 months ago

valid for circles also.....

- 5 years, 10 months ago

4

- 5 years, 10 months ago

4

- 5 years, 10 months ago

4

- 5 years, 9 months ago

no. of sides will not be change. Regarding to length of the sides, it is infinite GP whose common ratio is 1 / sqrt of 2.

- 5 years, 9 months ago

It is obvious that this holds true for any regular polygon. To observe this, let the circumcenter of a regular polygon be O, let three consecutive points be A,B,C. Let midpoint of AB be X and of BC be Y. Observe that OX and AB, OY and BC are perpendicular. It is easy to see that angle BCO and angle CBO are both equal to angle ABC/2.

Thus the resulting 'child' polygon has the same circumcenter and is similar to its 'parent' - its just a shrunk and rotated version of its parent, and a 'grandchild' is parallel to its 'grandparent'. The ratio of area between a child and parent polygon is just the ratio of sides, squared = [sin(180/n)]^2 where n is the number of sides

- 5 years, 9 months ago

1/4

- 5 years, 9 months ago

the lenght of the vertix of the last square is : (aaa*a)/98

- 5 years, 10 months ago

THIS WILL BE SYMMETRIC FOR FIGURES OF OPPOSITE SIDES EQUAL AND OPPOSITE ANGLES EQUAL. THE COMMON RATIO IS 1/2!!!

- 5 years, 9 months ago