Squares Inscribed In Circles

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Emigdio Binet

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Aug 4, 2024, 7:55:20 PM8/4/24

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Acircle is inscribed in a square of side , then a square is inscribed in that circle, then a circle is inscribed in the latter square, and so on. If is the sum of the areas of the first circles so inscribed, then, as grows beyond all bounds, approaches:

Now we note that the picture is self-similar; if we erase the outer square, erase the outer circle, rotate the picture, and dilate the picture from the center in both the x- and y-directions by an equal scaling factor, we will get the original picture. Therefore, the side lengths of successive squares form a geometric sequence with common ratio , as do the radii of the circles. On the other hand, the areas of the squares (and areas of the circles) form a geometric sequence with common ratio .

Since the area of the first circle is , and the common ratio of areas is , the sum of all the areas of the circles is the sum of the infinite geometric sequence, which is . This is for side length , and as noted before, there should be an factor in addition to this number to generalize it from the unit square. This gives answer .

In geometry, an inscribed planar shape or solid is one that is enclosed by and "fits snugly" inside another geometric shape or solid.[1] To say that "figure F is inscribed in figure G" means precisely the same thing as "figure G is circumscribed about figure F". A circle or ellipse inscribed in a convex polygon (or a sphere or ellipsoid inscribed in a convex polyhedron) is tangent to every side or face of the outer figure (but see Inscribed sphere for semantic variants). A polygon inscribed in a circle, ellipse, or polygon (or a polyhedron inscribed in a sphere, ellipsoid, or polyhedron) has each vertex on the outer figure; if the outer figure is a polygon or polyhedron, there must be a vertex of the inscribed polygon or polyhedron on each side of the outer figure. An inscribed figure is not necessarily unique in orientation; this can easily be seen, for example, when the given outer figure is a circle, in which case a rotation of an inscribed figure gives another inscribed figure that is congruent to the original one.

Familiar examples of inscribed figures include circles inscribed in triangles or regular polygons, and triangles or regular polygons inscribed in circles. A circle inscribed in any polygon is called its incircle, in which case the polygon is said to be a tangential polygon. A polygon inscribed in a circle is said to be a cyclic polygon, and the circle is said to be its circumscribed circle or circumcircle.

For an alternative usage of the term "inscribed", see the inscribed square problem, in which a square is considered to be inscribed in another figure (even a non-convex one) if all four of its vertices are on that figure.

The purpose of this task is to apply knowledge about triangles, circles, and squares in order to calculate and compare two different areas. This can be done in several ways: the solutions below use definition of congruence, the Pythagorean Theorem, and symmetry considerations. The symmetry argument requires auxiliary lines (the two diagonals of the inner square) while the Pythagorean Theorem can be applied to the picture as is after labeling some key points.

This task was adapted from problem #25 on the 2011 American Mathematics Competition (AMC) 8 Test. The question on the exam was different, asking which fraction, from a multiple choice list, is closest to the ratio of the shaded area to the difference of the areas of the two squares: in particular, the exam question could be answered correctly by studying the picture carefully without calculating any of the areas. For the 2011 AMC 8, which was taken by 153,485 students, the multiple choice answers for the problem had the following distribution:

The diagonals divide the larger square into 4 smaller, congruent squares because they join the midpoints of the sides of the larger square and reflecting across the vertical one (or horizontal, respectively) exchanges the left side of the big square and the right side (or the top and the bottom, respectively). The sides of the small squares are radii of the circle, so the side lengths are 1 and the area of the small squares is $1^2=1$.

Since all squares are similar to all squares and all circles are similar to all circles, we can make the same argument with any radius $r$ because the areas of the figure with radius 1 will all scale by a factor of $r^2$. In other words, the area of the circle will be $\pi r^2$, the area of the big square will be $4r^2$, the area of each of the 8 triangles will be $\frac12 r^2$, and the area of the smaller square will be $2r^2$. The area of the blue region will be $\pi r^2 - 2r^2=r^2(\pi-2)$ and if we divide the area of the blue region by the area of the grey region, we get$$ \fracr^2(\pi-2)2r^2=\frac\pi-22 \approx \frac1.142 \gt \frac12$$So the result is the same regardless of the radius of the circle.

We know $\triangle ABC$ is a right triangle because we are given that the outer quadrilateral is a square. Let $r$ denote the radius of the circle. We have $AB = CB = r$ since the side length of the large square is twice the radius of the circle and the small square touches the large square at the midpoints of its sides.By the Pythagorean theorem$$AB^2 + CB^2 = AC^2$$and so $AC = \sqrt2r$. This means that the small square has area $\sqrt2r \times \sqrt2r = 2r^2$. The large square has area $2r \times 2r = 4r^2$. So the difference in the areas of the two squares is $2r^2$.

Instead of using the Pythagorean theorem to find $AC$, we could also find the difference of areas of the two squares directly: the inner square together with 4 triangles congruent to $\triangle ABC$ make the outer square and the area of $\triangle ABC$ is $\fracr^22$. So the difference of areas of the squares is $2r^2$.

The area of the circle of radius $r$ is $\pi r^2$. So the blue shaded area is$$\pi r^2 - 2r^2 = (\pi -2)r^2.$$ We know that $\pi$ is a little bigger than $3$ so $ \pi - 2 \gt 1$ and$$(\pi - 2)r^2 \gt \frac12 \times 2r^2.$$This means that the blue shaded area is a little bigger than half of the difference of the areas of the two squares.

The area of the circle of radius $r$ is $\pi r^2$. So the blue shaded area is$$\pi r^2 - 2r^2 = (\pi -2)r^2.$$The area of the big square is $(2r)^2 = 4r^2$ since its sides have the same length as the diameter of the inscribed circle. Removing the inner square from the outer square leaves $4r^2 - 2r^2 = 2r^2$. We know that $\pi$ is a little bigger than $3$ so $ \pi - 2 \gt 1$ and$$(\pi - 2)r^2 \gt \frac12 \times 2r^2.$$The blue shaded area is a little bigger than half of the difference of the areas of the two squares.

Constant convergence arguments must have been typical in earlyGreek mathematics, for which the evidence comes mostly from non-mathematicalsources (such as Zeno the Eleatic). The indirect evidenceis that it is hard to see how else one could come to the theoremsof Euclid, Elements xii. In fact, this difficultyleads to a common complaint about Greek mathematics in the 17thcentury, that they hid their methods of discovery. Thisexample is based on Elements xii 2.

(diagram 3)If we bisect the squares, the resulting triangles are all similarand are as the squares of the diameters. Hence, the sumof the previous figure (the square) and the 8 triangles, i.e.,the octagons, are similar and are as the squares of the diameters.

(diagram 4)As we repeat this process, we get nearer and nearer to the circle. Hence, as the number of sides of the polygons goes to infinity,it is easy to see that the circles will be in the same ratio,as the squares of the diameters.

(diagram 4)As we repeat this process, we get nearer and nearer to the circle. We keep repeating this (ad infinitum).

1. Hence, as square and all the trianglesin one circle are to the square and all the triangles in the other,so is one circle to the other.

2. As square and all the trianglesin one circle are to the square and all the triangles in the other,is as square of one diameter to the other.

3. Hence, the circles are as thesquares of their diameters.

Version 1 conceives of the circle as the limit of the approximatingpolygons, as if the circle were an infinite-gon. Nonetheless,the conception of each polygon as built out of the previous andmaintaining the same ratio as the previous polygon is visuallyvery intuitive, for which see Antiphon.

Version 2 merely conceives of the circle as filled up by thesquare and the infinity of triangles. This version is tidier thanthe first, and fits more closely than Version 1 to the argumentin Euclid,Elements xii 2.

HeuristicVersion of the ArgumentGo to Elements xii 1: If similar polygons are inscribedin circles, their ratios are as the squares of the diameters of the circles.In the notes, T(AB) is stands for the square on line AB. Translation and analysis of Heiberg's Text, based mostly on Vat. gr. 190 (P)Prop. 2: Circles are to one another as the squares from the diameters.(Diagram based on Vat. gr. 190 (P) f. 215v)(general diagram)(diagram 1) Let there be circles, ABGD, EZHQ, and let there be their diameters, BD, ZH. I say that it is: as circle ABGD to circle EZHQ so the square from BD to the square from ZQ. (diagram 2) For if it is not as circle ABGD to EZHQ so the square from BD to that from ZQ, it will be: as that from BD to that from ZQ, so circle ABGD either to some region smaller than circle EZHQ or to a larger. Let it first be to a smaller, S.

Where S

S > circle EZHQ Suppose S (diagram 3) And let there be inscribed into circle EZHQ a square, EZHQ. (diagram 4) In fact the inscribed square is larger than half of circle EZHQ, since, in fact, if through points E, Z, H, Q we draw straight-lines touching the circle, square EZHQ is half the square circumscribed about the circle, but the circle is smaller than the circumscribed square. Thus, the inscribed square EZHQ is larger than half circle EZHQ. We inscribe a square and show that it is more than half the circle and so takes away more than half the circle.