Sliding Rectangles and the Lam-Ca Construction

In the quiet elegance of geometric construction, sometimes a simple configuration reveals a profound truth. This post explores one such configuration, part of a legacy left behind by my late Vietnamese friend, Nguyen Tan Tai, who referred to it as the Lam-Ca construction — from the Vietnamese words meaning “do it all.” And that’s exactly what these diagrams aim to do: to explain, prove, and illuminate all at once.

What Nguyen Tan Tai created is more than a diagram: it is a visual proof that brings together trigonometric identities, reciprocal relationships, and the power of constant area.

Click the image or open the interactive Geogebra diagram here.

Unit Diameter Circle Instead of Unit Radius

Most trigonometric constructions begin with a circle with radius 1, where sines and cosines are represented as vertical and horizontal projections of a point on the circle. But in this configuration, the circle has unit diameter.

In a circle of unit diameter:

• The angle α is inscribed.

• The opposite chord has a length equal to sin α.

• For an inscribed right triangle (i.e. with hypotenuse = 1) The adjacent chord has a length equal to cos α.

• The projection of the opposite and adjacent chords on the hypotenuse have respective lengths sin²α and cos²α.

We therefore use chord lengths derived directly from the geometry of the circle. They reflect the sine and cosine values as segments formed within the circle, preserving symmetry and proportions naturally. The squares (blue and red) and rectangles (green and orange) of the upper part of the figure illustrate this nicely. I have already posted this construction in a previous post.

The Insightful Twist: Sliding Rectangles of Constant Area

What Nguyen Tan Tai introduced was an original twist on the classical representation of the inscribed rectangle: instead of constructing squares on the chord segments, he formed rectangles (shown in grey in the lower part of the figure) defined by:

• Left rectangle: one side is sin α, the other is 1/sin α

• Right rectangle: one side is cos α, the other is 1/cos α

The result? Each rectangle has a constant area of exactly 1, even as the angle α changes. In this way we see a whole family of unit-area rectangles. Their lower vertices trace a horizontal line anchored at y = –1. The shape of the rectangle dynamically adapts, while its area remains invariant.

Why This Matters

This construction transforms reciprocal trigonometric identities from abstract formulas into tangible, visual objects. The construction reveals them as side lengths in a dynamic geometry.

This approach:

• Makes reciprocal relationships intuitive

• Shows how geometry can express trigonometry with visual clarity

• Bridges sine and cosine with their multiplicative inverses

• Highlights a rarely explored link between circular geometry and constant-area transformations

The Lam-Ca Perspective

What makes the Lam-Ca construction special is its simplicity: no coordinates, no algebra, just pure geometry. Using circles of unit diameter and basic projections, Nguyen Tan Tai’s approach recovers deep identities from nothing but straightedge-and-compass logic.

This is not just a visual trick. It’s a philosophy: that geometry, when done right, can do it all.

A Quiet Tribute

To our knowledge, this method of constructing unit-area rectangles with sides defined by chord-based trigonometric values and their reciprocals has not been formally documented elsewhere. It stands as a tribute to Nguyen Tan Tai’s creativity and insight—a quietly powerful expression of geometry’s poetic depth.

This visualization is a piece of his legacy, and I’m honored to share it. What if some of the most beautiful trigonometric identities could be seen rather than derived algebraically?

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In the next post, I’ll extend this same diagram to reveal tripling angle identities like cos(3α) and sin(3α) — all through sliding rectangles and simple symmetry. Stay tuned.

Geometric representations of higher degree binomials

A binomial is a mathematical expression with two terms, say x and y. The simplest binomial is the sum of those two terms x+y. Powers of this binomial (x+y)ⁿ are ruled by the so-called binomial theorem, which states that the coefficients of the expansion terms x^py^q are given by the Pascal triangle entries. So we have:
(x+y)⁰ = 1
(x+y)¹ = 1.x + 1.y

(x+y)² = 1.x²+ 2.xy + 1.y²

(x+y)³ = 1.x³+ 3.x²y + 3.xy² + 1.y³

(x+y)⁴ = 1.x⁴+ 4.x³y + 6.x²y² + 4.xy³ + 1.y⁴

and so forth ...

But where do these coefficients come from? You can just be satisfied by the result algebraic calculation, or you could look for geometric representations. For instance, a nice geometric explanation is given at the www.mathaware.org site.

Geometric proof of square binomial (original at The Geometry of the Binomial Theorem, Math Awareness Month site)

Geometric proof of cubic binomial (original at The Geometry of the Binomial Theorem, Math Awareness Month site) 

For the second degree binomial, the Pascal triangle entries emerge naturally as the number of squares or rectangles with area given by the expansion terms x^py^q. For third degree binomials, they emerge as the number of cubes or parallelepipeds with volume given by the expansion terms. And for higher degree binomials, they emerge as the number of hypercubes and hyperparallelepipeds given by the expansion terms, so an Euclidean geometric representation for those higher degree binomials seems impossible.

This suggested impossibility prompted me to have a closer look at geometric representations of binomial expansions. As a matter of fact a number to the fourth power x⁴ is just a number, but it is also a square x⁴ = (x²)². And there is no impossibility in representing numbers and squares geometrically, so the suggested impossibility is only an impossibility along the common line of thought, which sees cubic expansions as volumes. Along another line of thought, which seems to have been unnoticed, cubic expansions can be seen as areas, and then geometric representations of higher degree binomials become possible. The following figures illustrate this fact for the special case where x+y is normalized to 1 (for arbitrary x+y, one just has to rescale the figure each time, the pattern remains the same).

For the binomial (x+y)¹ = 1.x + 1.y, we can divide a unit square into two rectangles, one rectangle of area x and one rectangle of area y, see Figure 1.

For the binomial (x+y)² = 1.x²+ 2.xy + 1.y², we can divide the unit square into two squares, one of area x² and one of area y², plus two rectangles of area xy, see Figure 2.

For the binomial (x+y)³ = 1.x³+ 3.x²y + 3.xy² + 1.y³, we can divide each of the previous squares and rectangles into proportions x and y, giving eight rectangles, one of area x³, one of area y³, plus three rectangles of area x²y and three rectangles of area xy², see Figure 3.

For the binomial (x+y)⁴ = 1.x⁴+ 4.x³y + 6.x²y² + 4.xy³ + 1.y⁴, we can again divide each of the previous squares and rectangles into proportions x and y, giving one square of area x⁴, one of area y⁴, four squares and two rectangles (giving six) of area x²y², four rectangles of area x³y and four rectangles of area xy³, see Figure 4.

And we could go further indefinitely, doodling areas of incrementing powers, just in 2D, without any reference to unintuitive hyperspaces.

Exhaustion of nested squares and Wallis product

When one is asked to divide a square into N equal area parts, we commonly think of dividing a square into N equal rectangles. Figure 1 shows an example for N=11.

There is another general solution to that problem, in particular we could wrap it up in an enigma: is it possible to divide a square into N square-shaped parts of equal area, N being any integer?

The nested circles structure described in Figure 4 of the Variations of circular area division into equal parts post suggests the following solution.

Step 1. Starting from a circle of radius 1, area π, draw circles of incrementing area 2π, 3π, 4π, 5π, 6π, and so on... The resulting concentric rings have all the same area π.

Step 2. For each circle, circumscribe a square. The resulting squares have area 4, 8, 12, 16, 20, 24, and so on...

Step 3. Conclusion: the areas between each nested squares are equal. The figure illustrates the case for N=11.

So here we have a compass and ruler solution for dividing a square into N equal area parts. An interesting corollary is that this division into N equal parts can be applied to any 2D-figure, whether it be a triangle, a pentagon, any polygon, or even arbitrary drawings, provided that it be circumscribed (or inscribed) in a circle. As an answer to a comment on the preceding post, why not divide Leonardo da Vinci's Vitruvian man into N nested Vitruvian men? There are certainly other applications, one of them has been suggested to me by another comment on the preceding post.

In order to calculate an approximate value of π, Archimedes used a method of exhaustion which consists in inscribing a circle between two identically shaped polygons. Increasing gradually the number of sides of the polygon, the shape of the polygon approaches the shape of a circle. This geometrical approach is at the origin of his numerical approximation of π. Many other ways have been developed to approximate π, especially using infinite series. Geometrical representations of these series are rarely available, probably because no direct link has been found between them and the circle. The benefit of the nested circle and square structure of Figure 2 is that it suggests alternative methods of exhaustion.

For example, we could try to equate the area of a circle to the area of a square in this nested structure. This yields an approximation for squaring the circle (or circling the square). For a circle of radius 1, the area of the circle is π, the area of the circumscribed square is 4. The ratio between both areas is therefore π/4. The procedure to let that ratio go to 1 then follows the following steps.

Step 1. Draw the initial circle and square with ratio π/4.

Step 2. A better fit would be to inscribe both figures in higher orders of the nested structure. But the area of the circle is too small with respect to the square. So, we must enlarge the area of the circle a little more than the square. We therefore enlarge the area of the circle by 3 while enlarging the area of the square by 2, which is illustrated in the figure below.

Step 3. But now the area of the circle is too large with respect to the square. So for this fit, we must enlarge the area of the circle a little less than the area of the square. We therefore enlarge the area of the circle by 3, while enlarging the area of the square by 4, as illustrated below.

Step 4. But now the area of the circle is again too small with respect to the square. So for the next fit, we must enlarge the area of the circle a little more than the area of the square. We therefore enlarge the area of the circle by 5, while enlarging the area of the square by 4, as illustrated below.

Step 5. As after step 2, the area of the circle is too large with respect to the square... and I become a bit exhausted by drawing at exponentially growing scales;-) but I hope I've shown enough to suggest the continuation of this exhaustion scheme. You can of course rescale the figures, keeping track of the successive numbers and their roots on the figures.

The described method of exhaustion searches for a limit by alternately letting the circular area become larger, smaller, larger, smaller, etc... than the square area, but each time with a reduced fraction. As a limit it can be expressed by the following equation:

It seems we have gone through Wallis' product, which is usually disconnected from any geometrical approach.

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Physics quote of the day :

Profound study of nature is the most fertile source of mathematical discoveries. ~ Joseph Fourier