Astroids

Imagine a ladder falls down so that the top of the ladder slides down the wall and the base of the ladder slides along the ground away from the wall.  If you were to record the falling ladder from the side and overlay each frame onto a single picture, the ladders from the different frames would form a curve.

Falling Ladder

At first glance it may appear that this curve is circular.  After all, the top of the curve is measured when the ladder is leaning straight up against the wall, and the bottom of the curve is measured when the ladder is lying flat on the ground, which would appear to show a radius that is the same length as the ladder.  However, if we compare our falling ladder curve with a curve from a circle, we can see that the circular curve is slightly steeper.

Falling Ladder Curve vs. Circular Curve

Obviously, some more calculations will be needed to find an equation for this curve.  We can start by calling the wall the y-axis and the ground the x-axis.  Then if we call the length of the ladder w, and the angle between the ladder and the ground θ, then the distance from the top of the ladder to the origin (along the wall) is w sin θ and the distance from the bottom of the ladder to the origin (along the ground) is w cos θ.  We can also label a point (x, y) somewhere on the ladder.

There are then two similar triangles that we can set in proportion to each other: the large triangle with sides w sin θ and w cos θ, the smaller triangle with sides y and w cos θ – x.  Therefore:

y / w cos θ – x = w sin θ / w cos θ	(similar triangles)
yw cos θ = w sin θ (w cos θ – x)	(cross multiply)
yw cos θ = w2 sin θ cos θ – wx sin θ	(distribute)
y = w sin θ – x tan θ	(divide by w cos θ)

This means that at any vertical line at x, the ladder intersects it at a height of y = w sin θ – x tan θ.  The maximum height of this intersection would be when the derivative ^dy/_dθ = w cos θ – x sec²θ equals zero.  Therefore:

dy/dθ = w cos θ – x sec2θ = 0	(derivative equal to zero)
w cos θ – x/cos2θ = 0	(sec θ = 1/cos θ)
w cos3θ – x = 0	(multiply by cos2θ)
w cos3θ = x	(add x)
cos3θ = x/w	(divide by w)
cos θ = 3√x/w	(cube root)

If cos θ = ³√^x/_w, we can use trigonometric identities to show that sin θ = √(1 – cos²θ) = √(1 – (³√^x/_w)²) and tan θ = sin θ / cos θ = √(1 – (³√^x/_w)²) / ³√^x/_w = ³√^w/_x√(1 – (³√^x/_w)²).  Combining this with our previous equation y = w sin θ – x tan θ, we get:

y = w sin θ – x tan θ	(previous equation)
y = w√(1 – (3√x/w)2) – x 3√w/x√(1 – (3√x/w)2)	(substitute sin θ and tan θ)
y = w√(1 – x2/3w-2/3) – xw1/3x-1/3√(1 – x2/3w-2/3)	(rewrite as exponents)
y = w√(1 – x2/3w-2/3) – x2/3w1/3√(1 – x2/3w-2/3)	(x1x-1/3 = x2/3)
y = w2/3w1/3√(1 – x2/3w-2/3) – x2/3w1/3√(1 – x2/3w-2/3)	(w = w2/3w1/3)
y = w2/3√(w2/3(1 – x2/3w-2/3)) – x2/3√(w2/3(1 – x2/3w-2/3))	(w1/3 = √(w2/3))
y = w2/3√(w2/3 – x2/3) – x2/3√(w2/3 – x2/3)	(distribute, and w2/3w-2/3 = 1)
y = (w2/3 – x2/3)√(w2/3 – x2/3)	(combine like terms)
y = (w2/3 – x2/3)(w2/3 – x2/3)1/2	(rewrite as exponents)
y = (w2/3 – x2/3)3/2	(add exponents)
y2/3 = w2/3 – x2/3	(law of exponents)
x2/3 + y2/3 = w2/3	(add x2/3)

Therefore, the path of the curve that a falling ladder makes is x^2/3 + y^2/3 = w^2/3, where w is the length of the ladder.  In mathematics, this curve is called an “astroid”, which is derived from the Greek word for “star”, because when you graph it in all four quadrants you get a star shape.

Astroid

When you were kid, you may have played with a Spirograph kit, where you can create different designs by rotating geared circles inside another geared circle.

Spirograph Kit

In the same way, an astroid can also be created by following the path of a point on the edge of a circle that is rotated inside another circle that is four times its side.

Creating an Astroid with Circles

To prove this, we can call the radius of the big circle w, the radius of the little circle r, the angle between the x-axis and the segment that contains the two centers of the circles θ, and the angle between segment that contains the two centers of the circles and the segment that contains the center of the little circle to the moving point α. 

Since the big circle is four times the size of the little circle, the center of the little circle is always w – r = 4r – r = 3r away from the origin, so the coordinates of the center of the little circle can be expressed as (3r cos θ, 3r sin θ).  Also since the big circle is four times the size of the little circle, α = 4θ, so the standard angle for (x, y) with respect to the little circle would be 2π – (4θ – θ) = 2π – 3θ.  Putting these two details together, x = 3r cos θ + r cos (2π – 3θ) and y = 3r sin θ + r sin (2π – 3θ).  Simplifying:

x = 3r cos θ + r cos (2π – 3θ)	(x component)
x = 3r cos θ + r cos 3θ	(cos (2π – x) = cos x)
x = 3r cos θ + r (4 cos3θ – 3 cos θ)	(cos 3x = 4 cos3x – 3 cos x)
x = 3r cos θ + 4r cos3θ – 3r cos θ	(distribute)
x = 4r cos3θ	(simplify)
x = w cos3θ	(w = 4r)
x/w = cos3θ	(divide by w)
cos θ = 3√x/w	(cube root both sides)

and:

y = 3r sin θ + r sin (2π – 3θ)	(y component)
y = 3r sin θ – r sin 3θ	(sin (2π – x) = -sin x)
y = 3r sin θ – r (3 sin θ – 4 sin3θ)	(sin 3x = 3 sin x – 4 sin3x)
y = 3r sin θ – 3r sin θ + 4r sin3θ	(distribute)
y = 4r sin3θ	(simplify)
y = w sin3θ	(w = 4r)
y/w = sin3θ	(divide by w)
sin θ = 3√y/w	(cube root both sides)

Finally, substituting cos θ = ³√^x/_w and sin θ = ³√^y/_w into the trigonometric identity sin²θ + cos²θ = 1:

sin2θ + cos2θ = 1	(trigonometric identity)
(3√x/w)2 + (3√y/w)2 = 1	(substitute cos θ and sin θ)
(x/w)2/3 + (y/w)2/3 = 1	(rewrite as exponents)
x2/3 + y2/3 = w2/3	(multiply by w2/3)

which is the equation of an astroid.

There are two remarkable points that should be made about astroids.  First of all, it is amazing that the same curve can be used to describe two completely unrelated situations: a curve made by a falling ladder and a Spirograph design made between two circles.  Secondly, this is yet another geometric formula in the form of xⁿ + yⁿ = zⁿ, where n = ²/₃ for the astroid, n = 2 for Pythagorean’s Theorem, and n = -½ for the radii of three kissing circles and a line (see here).  As mentioned in a previous article, we must admit that so many different formulas of the same form does not describe a chaotic universe of random chance, but rather an orderly universe of intelligent design.

Diagonals of a Polygon

Recently one of my co-workers was teaching a Geometry class on the topic of polygons, when the students asked if there was a formula for finding the number of diagonals according to the number of sides.  They drew out different polygons on the board and counted the number of diagonals in each:

Unfortunately, nobody was able to derive an equation on the spot, but fortunately, we were able to Google it.  The formula is  

d = ½n(n – 3)

where d is the number of diagonals and n is the number of sides.

The reason the formula is so difficult to find is because the relation is quadratic (the highest exponent of the variable is a square, when the formula is re-written as d = ¹/₂n² – ³/₂n).  It is much more intuitive to find an equation in which the relation is linear instead (like when converting units).  This and the fact that the formula also requires fractions made it difficult to find.

Instead of trying to find a pattern in the numbers, a geometrical approach should be used for this problem.  Surprisingly, it is easier to see the relationship in a polynomial with more sides than with less.  Consider all the diagonals of a hexagon from a single vertex:

From that vertex, a diagonal can be made to all the other vertices in the hexagon, with the exception of itself and the two adjacent vertices.  In other words, that vertex connects to all the other vertices in the hexagon except for 3 of them, for a total of 6 – 3 = 3 diagonals.  The same argument can be made 6 times, one for each vertex, so now we have counted 6(6 – 3) = 18 diagonals.  But since each diagonal has 2 vertices each, we have actually counted each diagonal in the hexagon twice.  To correct this, we half our answer, which means there are ½(6)(6 – 3) = 9 diagonals.

The same argument can be made for all polygons.  A polygon with n sides also has n vertices.  From each of these vertices, a diagonal can be made to all the other vertices except for 3 of them (itself and the two adjacent vertices), for a total count of n(n – 3) diagonals.  But since each diagonal has 2 vertices each we have counted each diagonal in the polygon twice, so we must half our answer, for the final and generalized formula of d = ½n(n – 3).

Euler Line

Most Geometry classes go over the different properties of triangles, including the orthocenter, centroid, and circumcenter.  The orthocenter is the point of concurrency of the three altitudes (a line segment through a vertex and perpendicular to a line containing the opposite side).

Orthocenter O of ∆ABC

The centroid is the point of concurrency of the three medians (a line segment through a vertex and through the midpoint of the opposite side).  One property of a centroid is that it cuts each median in a 2:1 ratio (see here for the proof).

Centroid P of ∆ABC

The circumcenter is the point of concurrency of the three perpendicular bisectors (a line segment through a vertex and through the perpendicular bisector of the opposite side). 

Circumcenter Q of ∆ABC

A little known fact about these three points of concurrency is that the orthocenter, centroid, and circumcenter always lie on the same line (with the exception of an equilateral triangle, in which the orthocenter, centroid, and circumcenter are actually the same point).  This was proved in 1765 by the Swiss mathematician Leonhard Euler and is consequently named after him.  Here are some examples of the Euler Line in different triangles:

Orthocenter O, Centroid P, and

Circumcenter Q of Acute ∆ABC

Orthocenter O, Centroid P, and

Circumcenter Q of Right ∆ABC

Orthocenter O, Centroid P, and

Circumcenter Q of Obtuse ∆ABC

A clever proof for the Euler Line can be found here and involves proving the similarity of two triangles containing the orthocenter O, centroid P, and circumcenter Q, or more specifically, the similarity of ∆OAP and ∆QDP in the following diagram:

First, we must recognize that in the next diagram, Q is both the circumcenter of ∆ABC and the orthocenter of ∆DEF.  Also, because D, E, and F are midpoints of the sides of ∆ABC , ∆ABC is similar to ∆DEF at a 2:1 ratio.

Therefore, since ∆ABC and ∆DEF are at 2:1 ratio, the vertex to orthocenter of ∆ABC (segment AO) and the vertex to orthocenter of ∆DEF (segment DQ) are also at a 2:1 ratio (side: AO ~ DQ).  Second, as mentioned previously, the centroid cuts the median at a 2:1 ratio, which means segments AP and DP are also at a 2:1 ratio (side: AP ~ DP).  Finally, ÐOAP is congruent to ÐQDP because they are alternate interior angles to the parallel lines formed by AO and DQ (angle: ÐOAP @ ÐQDP).  This is sufficient to prove that ∆OAP is similar to ∆QDP by SAS similarity.  Since the triangles are similar, ÐAOP is congruent to ÐDQP, which makes OP and QP segments of the same transversal OQ.  In other words, O, P, and Q (the orthocenter, centroid, and circumcenter) lie on the same line.

Centroid Proof

The centroid is the point of concurrency of the three medians (a line segment through a vertex and through the midpoint of the opposite side).  One property of a centroid is that it cuts each median in a 2:1 ratio.

                                                                       

Centroid P of ∆ABC

This property can be derived by considering the area of each triangle formed in the centroid diagram.  Since D is the midpoint of BC, the area of ∆BPD is equal to the area of ∆CPD (which we will call x), since they share the same height and have equal bases.  Similarly, since E is the midpoint of AC, the area of ∆BPE is equal to the area of ∆CPE (which we will call y); and since F is the midpoint of AB, the area of ∆BPF is equal to the area of ∆APF (which we will call z).

Now, since D is the midpoint of BC, it also means that the area of ∆BAD is equal to the area of ∆CAD, since they also share the same height and also have equal bases.  So according to our diagram, x + z + z = x + y + y, which simplifies to y = z.  We can use the same reasoning to show that x = y, and x = z, which means that x = y = z. 

So when we consider a triangle with any median as one of its sides, say ∆BEA, we see that it is composed of three equal areas (in this case z, z, and y), in which ∆BPA is composed of two of those equal areas (z and z) and ∆EPA is composed of the other area (y).  This means BP is twice as long as PE, proving that the centroid cuts the median in a 2:1 ratio.