Centroids of Inscribable Polygons

The term centroid, or center of gravity, of a triangle usually refers to the center of gravity obtained by treating the vertices as point masses with a mass of one unit. This leads to the centroid noramally found in geometry by intersecting the three medians. If we have points (xA,yA), (xB,yB), and (xC,yC) it is well known, and easily derived, that the centroid would be .

There are other types of centroids, and one of them has an interesting relation to the class of polygons mentioned in the title. Another center of gravity can be found if we use the boundary of the polygon and assume it is made up of uniform rods. For a triangle, this produces a center of gravity that is at the intersection of the angle bisectors of the medial triangle (the triangle with vertices at the midpoints of the original triangle). This point is also called the Spieker Center of the triangle. To distinguish these two types of centroids, I will call the center of gravity of the vertices G_p (for point) and the center of gravity of the boundary G_b.

Some interesting sidebars about the angle bisectors of the medial triangle before I continue about centers of gravity.... The angle bisectors of the medial triangle are always parallel to the angle bisector of the opposite vertex. Also, these lines are each a "cleaver" of the triangle; a term created by Dov Avishalom in a Mathematics Magazine article for a line that passes through the midpoint of one side and divides the perimeter of the triangle into two equal halves.

I recently came across a paper by Tom Apostle and Mamikon A. Mnatsakanian that pointed out that in any triangle the two centroids and the incenter were always collinear. They also show that the ratio of the distances of the two centroids from the incenter is always in a ratio of 3/2. The ratio G_b/ G_p= 3/2.


Then in a really nice generalization, they showed that the same property applies to any inscribable polygon, one in which each edge is tangent to a common circle, such as the quadrilateral below. They use the clever term "circumgons" in the paper for such shapes. This is really a nice term since "inscribable" and "circumscribable" are used both to describe the circle inscribed in a polygon, or the polygon circumscribed about the circle... even I get confused. . Since it is often much easier to find G_p and the incenter of inscribable polygons, this provides a clever way to find G_b.

Students may be interested to know that Archimedes used the formula Area= 1/2 P r for the area of a circle, where P is the perimeter (circumference). This is, of course, simply and alternate form of the formula they learned in grade school. They may be even more surprised to find that the Archmedian formula gives the area of any inscribable polygon if P is the perimeter of the polygon, and r is the radius of the inscribed circle. The proof is very easy, and is often provided in high school textbooks for the limited case of regular polygons which would always be inscribable. The image below provides non-subtle clues to the proof.

Dollar Bill Trigonometry

Just in time for my introduction to trig identities, Kate Nowak at f(t) reminded me of a beautiful geometric property of dollar bills.

Kate shows how to fold the dollar step by step, transforming it along the way from a rectangle to an isosceles trapezoid, then to a rhombus, and finally an equilateral triangle. And you can tease out a pretty decent regular tetrahedron with no overlap.

All this goes to show that the ratio of the height and width of a dollar bill is ...
But that is an important number.. that is the area of an equilateral triangle with unit sides. So if you made an equilateral triangle with all three sides the width of a dollar bill, then it would have the same area as the dollar bill... And the tetrahedron made from the dollar fold has the same surface area as well.


The first two folds are pretty interesting for even a more advanced class... they form a 30-60-90 triangle, and it might be kind of nice to challenge students to explain WHY??? If it is not a 30 degree right triangle, then the whole equilateral triangle thing is a mess. The key is actually in recognizing that we have done what students/teachers may have heard is impossible, we have trisected an angle.

I was first introduced to this little dollar folding trick by a fourteen year old in my algebra II class some years ago..."Hey, Mr Ballew, bet you can't fold a dollar into an equilateral triangle." After I figured it out... (it took me a while).. I challenged him to explain why the first fold made a 30 degree right triangle. I can't tell you how proud it made me when seconds later he said, "Oh, it's because all three angles are equal." Quite clearly the folded portion is the same angle as the triangle where it would be "unfolded", so those two angles are congruent...and the third angle, well the length of the hypotenuse is one dollar bill height, and the opposite side is 1/2 a dollar bill height, so it is a 30 degree angle...which means the other two congruent angles must each be 30 degrees...


SPOOKY fun with folding money

Ok, it works with a fifty or a hundred bill too, all US bills are the same dimension (not true in most countries)...but hey, I'm on a teacher's pay, so I use ones...

Truth in Blogging Details.... AS wonderful as I would like this to be..the dollar seems NOT to be in a ratio of exactly... Most places I have searched say the US dollar bill is 6.6294cm (2.61") wide, by 15.5956cm (6.14") long.. that is a ratio of 0.425081433... the square root of three divided by four turns out to be about 0.433012702... a difference between the two ratios of .008....

Complex and Other Conjugates

We had covered the Fundamental Theorem of Algebra in pre-calc and after a few days of trying to make sure they understood how to decide how many real and non-real solutions there were to a real valued polynomials and that they could be factored over linear and quadratic factors , I was trying to get them to see how easy it was to work back and forth from the quadratic polynomial and the complex zeros. Finally I told them, "It's almost like the Pythagorian Theorem. a²+ b² = c " [They will later learn that this is really just the product of two equal pythagorean results for each of the conjugates, but not yet]

I showed them that the complex solutions always came in conjugate pairs, a + bi and a - bi... and that if they knew the zeros, they could create a quadratic which had those solutions. It seems that students are no longer exposed to what I call the sum and product property of quadratic zeros... that if the zeros of a quadratic are r and s, a quadratic with those zeros is given by x² - (r+s)x + (rs)... (If you knew the zeros were at x=3 and x= -1, then a quadratic with those zeros would be x² - (3-1)x + (3*-1) or x² - (2)x -3).

They had to get it somewhere if they were ever taught to factor, but it never seems to find purchase in their mathematical memory. I pointed out that for complex conjugates, r+s was real since the complex +bi and - bi eliminated each other, and so r+s was just twice the real coefficient of either conjugate. I connected this to the graph by showing that this "a" in the a+bi was the axis of symmetry of the graph. For example a quadratic with complex zeros at 3+2i and 3-2i will have a linear term of -(3+3)x [assuming we use the easiest case with A=1]. This also means that the of the quadratic will have its vertex at a point with an x value of 3.

It is also easy to see the constant term of the quadratic. If you multiply (a+bi)(a-bi) the distribution gives a² +abi - abi - b²i²... As with the sum, the two imaginary terms will cancel out, and since i²=-1, we see that the product of the two complex conjugates will always be a² + b². So for the example 3+2i and 3-2i, the product of the conjugates is just 9+4=13. Putting this in as the constant term, we see that the quadratic x² - 6x + 13 has zeros at 3+2i and 3-2i.

This gives another way to solve a quadratic equation with complex roots. If we have a function x² - B x + C that we know has complex solutions, we know that the real coefficient, a, of the conjugate pair will be -B/2 (shades of the quadratic formula)... and we know that a² + b² = C, so we can figure out b pretty simply. If we had a quadratic such as x² - 8 x + 25 we would quickly see that the real coefficient is -(-8)/2 = 4... and know that 4² + b² = 25, so b must be 3, and the two solutions are at 4 +/- 3i. I finally admit to the students that this is just a variant of the quadratic formula, and sooner or later someone will ask if it is not possible to do the same thing with a quadratic with real zeros... I let them suggest a quadratic, and only press for the use of an even B in the linear coefficient for ease of explanation. For example with x² - 6 x + 5 (ok, I picked an easy one).. we would see that the "a" part was three again... but now we have 3² + b^2 = 5.

If we remember that b was the imaginary coefficient, it makes sense that the b^2 would be negative if the zeros were real... Kids translate that into, "just subtract instead of add, and take the square root of that for the "b" value. So 9 - 4 = 5, and the square root of 4 is two. Our real solutions are 3 +/- 2; or 5 and 1.
At this point I show them that with real rational coefficients, not only do non-real zeros come in conjugate pairs, but so do the real irrational zeros. If we had x² - 8 x + 10; the solutions of 4 +/- sqrt(16-10) would be the conjugate irrational solutions.

If we graph y=x² - 6x + 13, we notice that not only is the vertex at x=3, but its y coordinate is at 4... the square of 2, which was the b part of our complex conjugates... It seems that functions with complex solutions a +/- bi, the vertex is at (a, b²). This also works for real valued zeros if we with the a +/- b approach above. The vertex of x² -6x + 5 would be at (3, - 2²).

I'm not sure I would try to drag a really weak class throuh all this, but it often helps bright kids see how interconnected the zeros, factors, and graphs of all quadratics are, not just the complex numbers.

A Catalog of Quadratic Equations

It is the time of year when I try to enchant my Pre-calc students with the beauty of complex solutions to polynomials, and particularly quadratics.. A good time to remind them of a beautiful result created by a high school student.....

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The Mandelbrot set can be thought of as a catalog of closed Julia Sets over the complex plain. Each point on the plain representing a complex number, and the coloring representing whether the Julia set for that point had a closed or open orbit.

The equation y= ¹/₄ x² on the coordinate plain can, in similar fashion, serve as a catalog of all possible quadratic equations. The method was discovered by a high school age British student by the name of Anthony Bayes around 1955 as reported in the Mathematical Gazette in September of that year by H. Gebert, the young man's teacher.

If every point (b,c) on the coordinate plain is representative of a quadratic equation x²+bx+c=0, then the set of all points in the cup of the parabola have only complex solutions [for example, the point (2,5) would represent the equation x²+2x+5=0 which has only complex roots]. Those points beneath the curve have two real distinct solutions, and the points on the curve specify equations which have a double root[ again, (2,1) would represent the perfect square trinomial x²+2x+1=0]. For any point on the curve, the solutions are -1/2 the x-coordinate. In the case of (2,1) the double root of x²+2x+1=0 is at -1/2 of 2 = -1.

Any line tangent to the cuve passes only through points representing equations which share at least one solution. The line y=x-1 which is tanget to the curve at the point (2,1). Since the solutions for x²+2x+1=0 are both -1; all the points on this tangent represent quadratic equations which have x=-1 as one solution. As a case in point, (or a point in this case), (5,4) is on the tangent line, and the solutions of the quadratic equation x²+5x+4=0 are x=-1 and x=-4. If you drew the other tangent to the parabola passing through the point (5,4) it would contain the point (8,16) on the parabola representing the quadratic with a double root at x=-4, that is, the quadratic x²+8x+16=0.

This would allow a simple approximation method to find the solutions of any quadratic with reasonably small values using only a straightedge and a printed graph of the curve y=1/4 x². Simply pick the point (b,c) corresponding to the values of the equation, and then lay a straight edge through the point and tangent to the parabola. The solutions would be -1/2 the x values of the points of tangency, as given above.