Progression of a Trig Solution

On my last birthday, one of my more mathematical-minded friends posted the following picture on my Facebook page:

Since I love solving trigonometry problems, I immediately tried to solve it.

First Attempt: Law of Sines

At first glance, I noticed that if I assigned one of the missing angles with a variable, I could use some angle sum properties to label every other angle in terms of that variable.  Letting ∠AKC = t, I found that ∠DKG = 60° – t, ∠ACK = 150° – t, ∠KCD = t + 30°, ∠KDC = 90° – t, and ∠KDG = t + 90°.

I also noticed that since ∠A = ∠G = 30°, △AKG is an isosceles triangle, so I labeled AK = KG = y.

Since I had a lot of angles and sides opposite to each other, I used the law of sines on △AKC and △DKG to obtain two equations with two variables:

which can be solved numerically to t ≈ 7.589 and y ≈ 60.044.

Then since AK cos 30° = ½ AG, by substitution 60.044 cos 30° = ½ (13 + x + 48), and this can be solved to x = 43.

While the answer amusingly worked out to my age, the process involving complicated decimals through numerical methods was less than satisfactory.  I wondered if there was a better way to arrive at the solution.

Second Attempt: Law of Cosines

The next logical attempt after trying the law of sines was to try the law of cosines.  With the knowledge that t works out to a random decimal, I decided to try and avoid that and stick to the integer angles of 30° and 60° instead.

By the law of cosines on △ACK, CK² = AK² + AC² – 2 AK AC cos 30° = y² + 169 – 13√3y.

By the law of cosines on △DGK, DK² = GK² + DG² – 2 GK DG cos 30° = y² + 2304 – 48√3y.

By the law of cosines on △CDK, CD² = CK² + DK² – 2 CK DK cos 60° = CK² + DK² – CK DK.

By substitution on the last equation, x² = y² + 169 – 13√3y + y² + 2304 – 48√3y – √(y² + 169 – 13√3y) √(y² + 2304 – 48√3y).

Also, since AK cos 30° = ½ AG then y cos 30° = ½ (13 + x + 48) or y = (x + 61) / √3.

After substitution, the equation can be (eventually) manipulated to (x – 43)(x + 43)(61x + 1849) = 0, which solves to x = 43 for x > 0.

Although this method was long, it did eliminate the need for numerical methods with complicated decimals, so that was a little better.  I noticed that there were several iterations of the form 13 and 48 throughout this method, so I wondered if a general solution where a = 13 and b = 48 could shed some light onto why the solution worked out to such a nice integer.

Third Attempt: Law of Cosines with General Variables

Using the same steps from before, but with a = 13 and b = 48, then:

By the law of cosines on △ACK, CK² = AK² + AC² – 2 AK AC cos 30° = y² + a² – a√3y.

By the law of cosines on △DGK, DK² = GK² + DG² – 2 GK DG cos 30° = y² + b² – b√3y.

By the law of cosines on △CDK, CD² = CK² + DK² – 2 CK DK cos 60° = CK² + DK² – CK DK.

By substitution on the last equation, x² = y² + a² – a√3y + y² + b² – b√3y – √(y² + a² – a√3y) √(y² + b² – b√3y).

Also, since AK cos 30° = ½ AG then y cos 30° = ½ (a + x + b) or y = (x + a + b) / √3.

After substitution, the equation can be (eventually) manipulated to (x – √(a² + b² – ab))(x + √(a² + b² – ab))((a + b)x + (a² + b² – ab)) = 0, which solves to x = √(a² + b² – ab) for x > 0.

Now I recognized that the general solution x = √(a² + b² – ab) can be re-written as x = √(a² + b² – 2 ab cos 60°), the law of cosines for a 60° triangle!  But why are the segments a, b, and c linear in the diagram, and not part of a 60° triangle?  Can the diagram be manipulated to make solving it easier?

Fourth Attempt: Diagram Manipulation

I noticed that since AK = KG, △ACK can be rotated so that AK matches up to KG:

Then ∠KGC’ = ∠DGK + ∠KGC’ = 30° + 30° = 60°, and ∠DKC’ = ∠DKG + ∠GKC’ = (60° – t) + t = 60°.

Now here’s something interesting: △DKC ≅ △DKC’ by SAS (DK = DK, ∠DKC = ∠DKC’ = 60°, and KC = KC’), so that DC’ = CD = x.

Then by the law of cosines on △DGC’, x = √(DG² + GC’² – 2 DG GC’ cos 60°) = √(48² + 13² – 48 ⋅ 13) = 43.

This method was much easier and much more elegant than the other methods!

Extending the Problem

I’m always looking to find new trig problems to either give to my math students or to post on math forums, especially one with a clever solution like this one.  There are plenty of other 60° triangle integer triples that could have been used, like (3, 8, 7), (5, 8, 7), (7, 15, 13), etc. (see here for more), but I liked the following problem because of its unexpected solution:

This can be manipulated as:

so that by the law of cosines on △DGC’, √2 = √(x² + 1² – 2 ⋅ x ⋅ 1 cos 60°), which for x > 0 solves to x = ½(1 + √5) = φ, the golden ratio!

Similar problems with different angles can also be created.  For example, this problem features an equilateral triangle, where the base angles are 60° and the top middle angle is 30°:

This can be manipulated as follows:

so that by the law of cosines on △DGC’, x = √(3² + 5² – 2 ⋅ 3 ⋅ 5 cos 120°) = 7.

Finally, this problem features an isosceles right triangle, where the base angles and the top middle angle are all 45°:

Its diagram can be manipulated as follows:

so that by the Pythagorean Theorem on △DGC’, x = √(5² + 12²) = 13.  (Interestingly, since this solution only uses triangle congruency and the Pythagorean Theorem, this problem could have been solved by the ancient Greeks!)

Conclusion

There are often different ways to solve a trig problem, with some more difficult than others.  It can be extremely satisfying if you are able to find a clever and easy solution, especially after several attempts at a problem.

Epicycloids and Hypocycloids

In the book Trigonometric Delights by Eli Maor, there is a chapter on epicycloids and hypocycloids and their many fascinating properties.

These properties are illustrated with both descriptions and diagrams, which is the best a book can do, but require an active imagination to visualize each moving part.  Fortunately in the last two decades since Trigonometric Delights was written, programs for making animated diagrams have become more and more accessible, and I was able to use the free websites www.desmos.com and www.gifsmos.com to make some animated epicycloids and animated hypocycloids to better illustrate these theorems.

Hypocycloids

A hypocycloid can be formed by tracing the path of a point on a small circle that is rolling on the inside of a larger circle, similar to a design that can be made with spirograph. 

Although different hypocycloids can be made by varying the radii of each circle, all of them can be described by parametric equations.  Consider the following diagram, where R is the radius of the large circle and r is the radius of the small circle:

The coordinates of P relative to C are P(r cos Φ, -r sin Φ) and the coordinates of C relative to O are C((R – r) cos θ, (R – r) sin θ) so that the coordinates of P relative to O are:

x = (R – r) cos θ + r cos Φ

y = (R – r) sin θ – r sin Φ

However, from the rolling motion, arc PQ’ = arc QQ’, and since PQ’ = Rθ and QQ’ = r(θ + Φ), that means Rθ = r(θ + Φ), which rearranges to Φ = (R – r)/r · θ.  Substituting Φ into the equations above gives the parametric equations for a hypocycloid:

x = (R – r) cos θ + r cos (R – r)/r · θ

y = (R – r) sin θ – r sin (R – r)/r · θ

If the larger circle is twice as big as the smaller circle, then R = 2r, and the equations become:

x = 2r cos θ

y = 0

which traces a horizontal line segment along the diameter:

If the larger circle is four times as big as the smaller circle, then R = 4r, and the equations become:

x = 3r cos θ + r cos 3θ

y = 3r sin θ – r sin 3θ

which traces an astroid:

Double Generation Theorem of Hypocycloids

In 1725, Daniel Bernoulli discovered the double generation theorem of hypocycloids: a circle of radius r₁ rolling on the inside of a fixed circle of radius R generates the same hypocycloid as does a circle of radius r₂ = R – r₁ rolling inside the same fixed circle.

For example, here is the astroid again, the hypocycloid made by circles with radii R = 1 and r₁ = ¼:

and the following is the same astroid formed by circles with radii R = 1 and r₂ = R – r₁ = 1 – ¼ = ¾, but with the smaller circle moving in the opposite direction: 

and here are both animations at the same time:

Here is the double generation theorem illustrated by circles with radii R = 1, r₁ = 2/5, and r₂ = 3/5: 

and here it is illustrated by circles with radii R = 1, r₁ = 2/7, and r₂ = 3/7:

The double generation theorem of hypocycloids can be proved by algebraic manipulation.  The hypocycloid formed by the circles with radii R and r₂ has the parametric equations:

x = (R – r₂) cos θ₂ + r₂ cos (R – r₂)/r₂ · θ₂

y = (R – r₂) sin θ₂ – r₂ sin (R – r₂)/r₂ · θ₂

Replacing r₂ = R – r₁ and θ₂ = 2πp – (R – r₁)/r₁ · θ₁ (where p is an integer) and then simplifying gives: 

x = (R – r₁) cos θ₁ + r₁ cos (R – r₁)/r₁ · θ₁

y = (R – r₁) sin θ₁ – r₁ sin (R – r₁)/r₁ · θ₁

which are the parametric equations for the hypocycloid formed by the circles with radii R and r₁.

Epicycloid

An epicycloid can be formed by tracing the path of a point on a small circle that is rolling on the outside of a larger circle: 

Using a similar argument as the hypocycloid, but adding r to R instead of subtracting r from R, the parametric equations for the epicycloid can be found to be: 

x = (R + r) cos θ – r cos (R + r)/r · θ

y = (R + r) sin θ – r sin (R + r)/r · θ

An epicycloid can also be formed by tracing the path of a point on a large circle that is rolling around a fixed smaller circle in its interior:

If r is the radius of the large circle and R is the radius of the fixed smaller circle, the parametric equations for this type of epicycloid can be found to be: 

x = r cos (r – R)/r · θ – (r – R) cos θ

y = r sin (r – R)/r · θ – (r – R) sin θ

Double Generation Theorem of Epicycloids

As alluded to in the epicycloid animations above, there is a double generation theorem for epicycloids: a circle of radius r₁ rolling on the outside of a fixed circle of radius R generates the same epicycloid as does a circle of radius r₂ = R + r₁ rolling around the same fixed circle in its interior. 

 When the circles have radii R = 1, r₁ = 1, and r₂ = 2, the epicycloid traced is a cardioid:

And here is an epicycloid traced by circles with radii R = 1, r₁ = 2/5, and r₂ = 7/5:

Just like hypocycloids, the double generation theorem for epicycloids can be proved by algebraic manipulation.  The epicycloid formed by a large circle with a radius of r₂ rotated around a fixed circle with a radius of R has the parametric equations:

x = r₂ cos (r₂ – R)/r₂ · θ₂ – (r₂ – R) cos θ₂

y = r₂ sin (r₂ – R)/r₂ · θ₂ – (r₂ – R) sin θ₂

Replacing r₂ = R + r₁ and θ₂ = (R + r₁)/r₁ · θ₁ and then simplifying gives:

x = (R + r₁) cos θ₁ – r₁ cos (R + r₁)/r₁ · θ₁

y = (R + r₁) sin θ₁ – r₁ sin (R + r₁)/ r₁ · θ₁

which are the parametric equations for the epicycloid formed by a circle with radius r₁ rotating around a fixed circle with a radius of R.

Conclusion

Epicycloids and hypocycloids have many fascinating properties, as described by Eli Maor in his book Trigonometric Delights.  My hope is that these animated diagrams will help people understand these properties better, and give a better appreciation for mathematics in general.

Proving Common Trig Identities with Circle Theorems

I just re-read Trigonometric Delights by Eli Maor, and in the middle of his historical account of mathematics, he gives creative proofs for some common trigonometric identities based solely on Greek circle theorems.

Using the inscribed angle theorem, Ptolemy’s Theorem, and Thales’s Theorem, Maor is able to prove the law of sines, the angle addition and subtraction formulas for sine, the Pythagorean Theorem, and the Pythagorean identity of sine and cosine.  His proofs are outlined below.

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Consider the following triangle that is inscribed in a circle:

By the inscribed angle theorem, ∠AOB = 2∠ACB = 2γ.

Since OM bisects ∠AOB = 2γ, ∠MOB = γ.

Then from ΔOMB,

which rearranges to:

By a similar argument,

and

Therefore,

which proves the law of sines.

If 2r = 1, then a = sin α, b = sin β, and c = sin γ.  In other words, the side of an inscribed triangle in a circle with a unit diameter is equal to the sine of its opposite angle.

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Now consider the following diagram, where AC = 1:

Since 2r = 1, then as shown above BD is equal to the sine of its opposite angle, so BD = sin(α + β).

From ΔABC, cos α = AB/AC = AB/1 = AB and sin α = BC/AC = BC/1 = BC. 

Similarly from ΔADC, AD = cos β and CD = sin β.

But by Ptolemy’s Theorem, 

AC · BD = AB · CD + BC · DA

and after substituting the sine and cosine values, 

1 · sin(α + β) = cos α · sin β + sin α · cos β

which proves the angle addition formula for sine. 

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Similarly, consider the following diagram, where AD = 1:

Now BC is equal to the sine of its opposite angle, so BC = sin(α – β). 

From ΔABD, cos α = AB/AD = AB/1 = AB and sin α = BD/AD = BD/1 = BD. 

Similarly from ΔACD, AC = cos β and CD = sin β.

But by Ptolemy’s Theorem, 

AC · BD = AB · CD + BC · DA

and after substituting the sine and cosine values,

cos β · sin α = cos α · sin β + sin (α – β) · 1

which rearranges to:

sin(α – β) = sin α · cos β – cos α · sin β

which proves the angle subtraction formula for sine.

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Lastly, consider the following diagram of rectangle ABCD inscribed in a circle:

By the properties of a rectangle, AB = CD and AD = BC.

By Thales’s Theorem, AC and BD are diameters, so AC = BD.

By Ptolemy’s Theorem, 

AC · BD = AB · CD + BC · DA

and after substituting AB = CD, AD = BC, and AC = BD,

(AC)² = (AB)² + (BC)²

which proves the Pythagorean Theorem.

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Furthermore, if AC = 1, then cos α = AB/AC = AB/1 = AB and sin α = BC/AC = BC/1 = BC.

Substituting AC = 1, AB = cos α, and BC = sin α into the Pythagorean Theorem gives:

1 = cos²α + sin²α

which proves the Pythagorean identity of sine and cosine.

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Today’s math textbooks also have proofs for these trigonometric identities, but they are typically long and cumbersome.  Maor’s proofs are much more elegant, and also logically flow from a historical perspective.

Circles of Apollonius of a Triangle

Let’s investigate all triangles in which one side length is twice another side length and the third side length is a constant length of 9. Some triangles that fit these requirements would include ones with integer side lengths (4, 8, 9), (5, 10, 9), (6, 12, 9), (7, 14, 9), and (8, 16, 9):

Placing each of these triangles one on top of the other, it would appear that the third vertex follows a circular path:

and can be animated as follows:

Is this path always a circle, or is this just a coincidence with the values that we picked? 

This is no coincidence.  The Greek mathematician Apollonius showed that any circle can be defined as the set of points with a constant ratio of distances to two set points.  Conversely, we can prove that a set of points which maintains a constant non-one ratio of distances to two fixed points will always form a circle. 

If the constant ratio is k = ^b/_a and the distance between the two fixed points is c so that a < b < c, then we can make △ABC with side lengths a, b, and c and with opposite angles ∠A, ∠B, and ∠C, so that ∠B is at the origin and side c is along the x-axis so that the coordinate of A is (c, 0).

Then the locus of points C that are ak away from B and bk away from A can be expressed by the distance equation as x² + y² = a²k² and (x – c)² + y² = b²k².  Rearranging and combining gives a²b²k² = b²x² + b²y² = a²(x – c)² + a²y², and further rearranging gives (x + ^{a^2 c}/_{b^2 – a^2})² + y² = (^abc/_{b^2 – a^2})², a circle equation with a radius of r = ^abc/_{b^2 – a^2}. Therefore, vertex C will always lie on the path of a circle.

In fact, three circles of Apollonius can be drawn for any scalene triangle, one for each vertex:

Using the same argument as above, the radii of the three circles of Apollonius will be r₁ = ^abc/_{c^2 – a^2}, r₂ = ^abc/_{c^2 – b^2}, and r₃ = ^abc/_{b^2 – a^2}.  

Since ¹/_r1 = ¹/_abc(c² – a²), ¹/_r2 = ¹/_abc(c² – b²), and ¹/_r3 = ¹/_abc(b² – a²), and since ¹/_abc(c² – a²) = ¹/_abc(b² – a²) + ¹/_abc(c² – b²), we can derive the elegantly compact relation:

where r₁ < r₂ and r₁ < r₃.

Another way to write this equation is r₂^-1 + r₃^-1 = r₁^-1, which places it in an increasing list of geometric equations that are in the form of xⁿ + yⁿ = zⁿ, the most famous being a² + b² = c², the Pythagorean Theorem.  Other equations in that form include the relationship between the major axis, minor axis, and focal length of ellipses and hyperbolas (b² + c² = a² and a² + b² = c²), the radii of three mutually tangent circles and a line  (r₁^-½ + r₂^-½ = r₃^-½), and, of course, the addition of segments (a¹ + b¹ = c¹).

It is amazing that over a thousand years ago Apollonius showed that any circle can be defined as the set of points with a constant ratio of distances to two set points.  It is also amazing that the simple relationship of ¹/_r1 = ¹/_r2 + ¹/_r3 is true for the radii of the three circles of Apollonius for any scalene triangle, despite the complicated steps in between.

Perspective, Silhouettes, and Taxi-Cab Geometry

In a drawing with one-point perspective, all lines perpendicular to the picture plane converge to a single point on the horizon called the vanishing point.  In this picture of some railroad tracks, for example, it would appear that the railroad tracks meet at some point far off in the distance, even though they are actually parallel.

one-point perspective of railroad tracks

Now imagine that you are looking down at some cubes on a level surface, and that you draw them using one-point perspective.  The top of each cube will not only appear larger than the bottom of each cube, but also be drawn further away from the vanishing point.

one-point perspective of some level cubes

  

Next, you create a silhouette of each cube by coloring inside each of the drawn outlines.  The resulting shapes are either irregular hexagons, pentagons, or quadrilaterals.

the silhouettes of some level cubes drawn with one-point perspective

Amazingly, the area of each silhouette can be determined by just four variables – the side length a of the top square of the drawn cube, the side length b of the bottom square of the drawn cube, the horizontal shift x of the cube from a cube directly over the vanishing point, and the vertical shift y of the cube from a cube directly over the vanishing point.

The center of the top square of a cube in the first quadrant would be drawn at (ax, ay) and its vertices at (a(x ± ½), a(y ± ½)).

Likewise, the center of the bottom square of a cube in the first quadrant would be drawn at (bx, by) and its vertices at (b(x ± ½), b(y ± ½)). 

Each silhouette is a hexagon (or a degenerate hexagon) that is the difference between the area of a rectangle with sides c = a(x + ½) – b(x – ½) and d = a(y + ½) – b(y – ½) and the areas of the two opposite triangles, one with sides c – a and d – b, and the other with sides c – b and d – a. 

In other words, the area of the silhouette is A = cd – ½(c – a)(d – b) – ½(c – b)(d – a), and letting p = ½(a² – b²) and q = ½(a² + b²), this simplifies to A = p(x + y) + q.  A similar argument can be made for the other quadrants to give the general formula of A = pd + q, where d = |x| + |y|. 

Surprisingly, d = |x| + |y| is also the equation for distances in taxicab geometry, a seemingly unrelated mathematical topic.  In taxicab geometry, segments are limited to being either vertical or horizontal, just like roads in a city block.  Since a taxicab cannot drive through city blocks along a hypotenuse, the distance it must travel to get to an intersection that is 4 blocks east and 3 blocks north is at least |3| + |4| = 7 blocks (whereas a flying bird not limited to the roads can do it in √(3² + 4²) = 5). 

taxicab geometry paths from (0, 0) to (4, 3)

In general, the distance a taxicab must travel to get to an intersection that is x blocks east or west and y blocks north or south is at least d = |x| + |y|.

Therefore, there is a surprising connection between one-point perspective drawings and taxicab geometry.  The area of the silhouette of a cube drawn in one-point perspective is A = pd + q, where d = |x| + |y|, the same distance a taxicab must at least travel to get to an intersection that is x blocks east or west and y blocks north or south.  This means that the area of the silhouette is a linear function of its own taxicab distance.