Also he hath demonstrated the proportion of the Diameter to the Circumference and shews what was the reason why Archimedes did use these two numbers ...

John Aubrey
Brief Lives, Penguin Books, 2000, p. 378

The Nature of π and Its Determination


The customary game plan for determination of π that tabulates measurements of several circular objects appears to simultaneously serve two ends: establishing existence of π and approximating its value. Both goals are better served by other means.

In the framework of similarity, existence of π is self-evident. Since all circles are similar, their relative (or shape) attributes are the same for every circle.

Approximation of π, too, does not require several objects. π could as well be approximated by taking measurements of a single circular object with different tools and different technics.

The Fundamentals

Objects around us have shape and size. Two objects of the same shape are similar(#). If, in addition, they have the same size, they are congruent (or simply equal). It might be difficult to give a definition of the two concepts that would withstand close scrutiny. However, everyone has an intuitive idea of both. "A child will draw or build a model of a house without having heard of the word similar. And models, which are continually constructed to design buildings, ships, planes, and automobiles, are truly similar to the physical objects they portray." [Kline, p. 74]

In geometry, whose subject is relationships between formal objects, the task of defining the size and shape may be expected to be more tractable, but may not be so unless restrictions are imposed on the kind of objects the definitions are meant to apply to. (The famous example by Banach and Tarski of the sphere decomposition into 5 sets, of which the first two put together form an equal sphere, and the same is true for the remaining 3 sets, shatters any hope that the task may be easy.)

The definitions we discuss below lend themselves to empirical verification and are thus fostered by our intuition.

Some properties of geometric objects are strictly individual and only hold for congruent figures. The distance between two opposite vertices of a square (the length of its diagonal) is one such example. Other properties hold for all figures of a given shape. These are called shape properties. For example, the magnitude of the vertex angle of every square is 90°. The relative distance is another important example. In every square, the length of the diagonal relative to its side is always . Relative distance is a shape, rather than an individual, feature.

What does a discussion on shape and size has to do with the famous universal constant π? Simple. Circles are geometric figures of a certain (circular) shape. Circles, like squares, but unlike triangles or parallelograms, only differ in size. As long as we consider only shape properties, the numerical values we obtain are bound to be the same for all circles. In particular, the outer length of a circle (its circumference) relative to its diameter is the same for all circles. It's this number that is nowadays invariably denoted by π.

Existence of π as a term in the identity P = πD, where P is the circumference of a circle of diameter D, follows directly from the fact that all circles are similar [Dunham, p. 23], i.e. of the same shape. The value of π could be estimated with a single measurement or repeated measurements of the same circular object. Such an experiment is no less valid than running and tabulating measurements of a dozen of different circular objects. It's more meaningful because, beyond the immediate need to estimate the value of π, it also conveys the fact that the result of measurements depends on the procedure of measurement as much as on the object being measured.

NCTM Principles and Standards

In NCTM Principles and Standards we read,

P. 172: Students in grades 3-5 should encounter the notion that measurements in the real world are approximate, in part because of the instruments used and because of human error in reading the scales of these instruments... Each pair of students will find slightly different measurements, even though they are measuring the same object using the same kind of measurement tools.

The customary way to approximate π that requires a collection of circular objects masquarades the fact that existence of π is self-evident. It may also mislead students into thinking of measurements as associated solely with the objects being measured, while missing their dependence on the tools and the manner of measurement. Taking measurements of the same object repeatedly using different tools and by different teams of students underscores the essence of measurements as a way of approximating an existent quantity.

P. 241: Students should also become proficient at measuring angles and using ratio and proportion to solve problems involving scaling, similarity, and derived measures.

There are two competing, but equivalent, definitions of similarity. One, transformational, based on the notion of scaling, the other is based on the idea of shape preservation.

According to the former, two figures are similar if one could be obtained from the other by a dilation transformation (possibly accompanied by a rotation or reflection.) Under dilations, distances between corresponding points all change by the same factor. This definition is associated with the notion of scaling. Dilations are scaling transformations.

According to the second definition, two figures are similar if angles and relative distances defined by corresponding points are equal.

P. 245: It is important that middle-grades students understand similarity, which is closely related to their more general understanding of proportionality and to the idea of correspondence. Students can use measurement to explore the meaning of similarity and later to apply the concept to solve problems. The important observation that the measurements of the corresponding angles of similar shapes are equal is often a starting point for work with similarity. Measurement is also useful for determining the relationships between the side lengths and the perimeters and areas of similar shapes and the surface areas and volumes of similar objects. Students need to understand that the perimeters of pairs of similar shapes are proportional to their corresponding side lengths but that their areas are proportional to the squares of the corresponding side lengths.

As an example, the latter means that, for similar shapes,


P1/P2 = a1/a2,

where a1 and a2 are the (charactersitic) side lengths, whereas P1 and P2 are the perimeters of the two figures. (1) underscores the idea of scaling. All corresponding linear attributes of similar figures vary by the same factor. The relationship (1) could be rewritten as


P1/a1 = P2/a2,

which simly means that the ratio of the perimeter to a side is a shape property and has the same value for all similar figures of the same shape regardless of their size. In other words, P/a is an invariant of the shape! (2) embodies the idea of shape preservation. Relative (or shape) attributes of similar figures are always equal.

Equivalence of (1) and (2) implies that two definitions of similarity - as a result of scaling and shape preservation - are equivalent.

How To

First, investigate the similarity. A most convenient tool for an investigation of similar shapes is the overhead projector. Take any piece of paper and make three small holes - A, B, C - in it. Place the paper on a projector and adjust the focus. Take care to get an orthogonal projection. Otherwise, the shapes won't be similar and distortions will be possible. Let the three points project onto A', B' and C'. Measure the distances on the original and its projection. Verify that, say, AB/AC = A'B'/A'C'. Verify that angles ABC and A'B'C' are equal. Verify, for example, that (AB + BC + AC)/AB = (A'B' + B'C' + A'C')/A'B'. Be creative, consider more than three points.

No special objects may be needed for further investigation. In any classroom, there are great many rectangles: walls, windows, doors, drawers, posters, desks, shelves, books, notebooks. Are they all similar?

For rectangles, similarity is established by verifying that the ratio of the long sides is the same as the ratio of the short sides. If no two physical rectangles are similar, several may be cut out of a piece of paper. Given a point on one of the similar rectangles, what is the corresponding point on the other? To find that point, scale the distances from the given point to the sides of the first rectangle and scale them in proportion equal to the ratio of the corresponding sides of the two rectangles. Similar experiments could be run with an overhead projector.

Next, argue that any two circles are similar. If some two have radii r and R, respectively, then distances between any pair of the corresponding points is in the ratio r/R.

Finally, choose any circular object and estimate the ratio of its circumference to the radius. This could be accomplished in several ways:

  1. By wrapping a thread around the object and then measuring its length. (To improve(*) the accuracy of measurements, the thread may be made to wrap around the object several times.)

  2. By rolling the object with one point marked on a flat surface and measuring the distance between the marks left on the surface. (To improve(*) the accuracy of measurement, roll the object until several marks are left on the surface.)

  3. By using a measuring tape. It's very useful to be able to make measurements in both inches and centimeters. The ratio P/D does not depend on the units of measurement. (To improve(*) ..., etc.)

Finally, after an estimate to π has been found, which probably will be in the range of 3.14, it must be mentioned that the real ratio P/D may be only represented by an infinite decimal. It's good to have its expansions to the 10th or 20th digit handy.


Following are four entertaining problems that involve π, circumference, the formula P = πD and measurements. The first is rather well known. The second ([Mason], p. 65) is as beautiful as its solution unexpected. The other two are nice curiosities.

  1. Imagine the Earth a perfect sphere with a long long rope tightly wrapped all along the equator. The length of the rope has been increased by 10 feet and the rope was magically lifted off the Earth to form a perfect circle over the equator. Will an average cat be able to stroll under the rope without touching it? (Solution).

    Interestingly, Brian Eno in [This Explains Everything (ed. J. Brockman), 210-211] observes that, excepting mathematicians and dressmakers, most of the people find the answer to the question counterintuitive.

  2. If a bicycle rider goes over a puddle of water, both of its wheels will leave several water marks on the road. Obviously, each of the wheels will leave a periodic pattern. How the two patterns are related? Do they overlap? Does their relative position depend on the length of the puddle? The bicycle? The circumference/radius of the wheels? (Solution).

  3. Draw a semicircle and on its diameter a few (two or more or infinitely many) smaller semicircles that abut each other in sequence. By construction, the sum of the diameters of the small semicircles equals the diameter of the large one. What about their perimeters, or, say, their total arc lengths? (Solution).

  4. Cut a circular shape into four parts of equal area with lines of equal lengths. (Solution).


As was already mentioned, all squares are similar and the same is true of all equilateral (but not more general) triangles. By the same token, all regular n-gons are similar. Among smooth curves, all circles are similar, while the logarithmic spiral is self-similar. (Which means that all logarithmic spirals come in the same size!) Further, in a book whose place is on a bookshelf of every math teacher, Howard Eves writes ([Eves], p. 5):

Now the shape of a conic is determined by its eccentricity. That is, all conics of the same eccentricity are similar, differing from one another merely in size. Since all parabolas have eccentricity 1, all parabolas are similar to one another, differing from one another only in size.


  1. J. Aubrey, Brief Lives, Penguin Books, 2000
  2. W. Dunham, The Mathematical Universe, John Wiley & Sons, Inc., 1994
  3. H. Eves, Mathematical Reminiscences, MAA, 2001
  4. M. Kline, Mathematics and the Physical World, Dover, 1959
  5. J. Mason, Thinking Mathematically, Addison-Wesley, 1985
  6. I. M. Yaglom, Geometric Transformations II, MAA, 1968

π: Applications and Calculations

Related material

  • Angle Bisectors In Rectangle
  • Segment Trisection Induced by Parallels to Medians
  • Parallelogram and Similar Triangles
  • Two Triples of Similar Triangles
  • Three Similar Triangles
  • Similar Triangles on Sides and Diagonals of a Quadrilateral
  • Point on Bisector in Right Angle
  • (*) Why will wrapping a thread around a circular object several times improve the accuracy of measurements? All other factors (like the degree of attention) being equal, the accuracy of measurements depends on the calibration of the measuring device. For example, if the smallest subdivision on a tape is 1/16", then the best measurement one may expect is within 1/32" of the real quantity. If the exact quantity is S and the reading from the tape is S', then S = S' + Δ, where Δ is the error of measurements. Δ is about the same for every measurement reading. Now, if we wrap a thread about a circular object 10 times, the exact quantity that is expected to be approximated is 10P, 10 times the circumference of the circle. After a measurement has been obtained, the approximation P' is derived at by dividing the measurement by 10, or 10P = 10P' + Δ, so that P = P' + Δ/10.

    (#) Thanks are due to Cynthia Lanius for pointing out an inaccuracy in the original version.

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    Solution to problem 1. Assume P is the orginal length of the rope. Then P = πD, where D is the diameter of the Earth. For the uplifted circle we have P + 10 = πD', from where 10 = π(D' - D). Which means that the diameter of the circle will grow by 10/π, whereas the radius will change by the quantity twice as small, which amounts to about 1.59 feet - more than enough for an average cat to pass under.

    Solution to problem 2. The marks left by one of the wheels will come out so that the distance between the corresponding points on two successive marks is exactly the circumference of the wheel. The puddle itself could be thought of as the first mark. It follows that the pattern left by one wheel depends only on the circumference of that wheel. Since this is true of any two wheels, even the ones attached to the same bicycle, the marks they leave will coincide, not just overlap (provided of course that the wheels have the same radius.)

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    Copyright © 1996-2017 Alexander Bogomolny


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