Morning Math - Part I

I just spent three weeks at Park City Math Institute. Morning math, the two hours that we spent working problem sets, was one of my favorite parts of the day. The problem sets were written and facilitated by Bowen Kerrins and Darryl Yong centering on “Some Applications of Geometric Thinking”. I strongly recommend checking them out here.

At the end of each week, we took some time to record our insights/connections we had made during the week as well as some questions we still had. Here’s what I came up with. If you’re thinking about doing some or all of the problem sets, I would strongly recommend not reading any further.

Rectangles, Boxes, and their Relationships to Quadratics and Unit Fractions

- The length and width of a rectangle are the solutions to the quadratic equation \(x^2-\frac {p}{2}\ +a=0\) where p represents the perimeter and a represents the area of the rectangle. A way to think about these solutions is to think about what two numbers have a sum of \(\frac{p}{2}\) and a product of a. You can find these values using the following expression: \(\frac{p}{4}\pm\sqrt{(\frac{p}{4})^2-a}\). When we set p=a, the l and w solutions satisfy the equation \(\frac{1}{l}+\frac{1}{w}=\frac{1}{2}\). A rectangle exists only if \(a\leq(\frac{p}{4})^2\), which can be restated as \(p\geq4\sqrt{a}\). A square maximizes the area and minimizes the perimeter of a rectangle.

- Similarly, we can think about the relationship between surface area and volume for rectangular prisms. \(\frac{1}{l}+\frac{1}{w}+\frac{1}{h}=\frac{1}{2}\)when SA=V. This same relationship exists for polygons that “surround” a vertex. When 3 regular polygons surround a vertex, \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}\), where a, b, and c are the number of sides of each polygon. There are ten unique whole number solutions to this equation.

- Somewhat surprisingly, if we think about four regular polygons surrounding a vertex, rather than three, we need \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1\), not \(\frac{1}{2}\).

- Consider finding all possible unique whole number solutions to the equation \(\frac{1}{a}+\frac{1}{b}=\frac{1}{n}\)for any given n.  b is related to a as follows: \(b=\frac{an}{a-n}\). You only need to check the set of solutions for whole number solutions when \(n+1\leq a\leq2n\).

Conic Sections

Parabola:

- All points are equidistant from a point (a, b) and a line \(x=n\).  Then the equation for this parabola can be written as follows: \(\sqrt{(x-a)^2+(y-b)^2}=x-n\). This is related to the equation below because when simplifying the coefficient for x² will be 0.
- \(Ax^2+Bx^2=C\), where A is equal to 0.

- Pour salt on circle that has a small circle cut out of it

- Slice a cone parallel to the side of the cone

Ellipse:

- All the points have the same sum of the distances to two points

- All points for \(0<e<1\), where e represents the ratio of the distance to the point (a, b) to the distance to the line x=n. The equation for this ellipse can be written as follows: \(\sqrt{(x-a)^2+(y-b)^2}=e(x-n)\). This is related to the equation below because when simplifying this equation, you would get coefficients of the same sign for both x² and y².

- \(Ax^2+Bx^2=C\), where A and B have the same sign.

- This ratio, e, is known as the eccentricity. It shows up in two places. As already mentioned, distance from focus to any point on the ellipse: distance from directrix to any point on the ellipse. Also, distance from the center to the focus: distance from the center to the vertex.

- Pour salt on a large piece of cardboard with a circle cut out of it
- Slice a cone from side to side

Circle:

- All the points for e = 0, where e represents to the ratio of the distance to the point (a, b) to the distance to the line x=n. The equation for this circle can be written as follows: \(\sqrt{(x-a)^2+(y-b)^2}=e(x-n)\).

- \(Ax^2+Bx^2=C\), where A and B are equal.

- Slice a cone parallel to the base

Hyperbola:

- All the points that have the same difference between the distances to two points

- All the points for some value e>1, where e represents the ratio of the distance to the point (a, b) to the distance to the line x=n. The equation for this hyperbola can be written as follows: \(\sqrt{(x-a)^2+(y-b)^2}=e(x-n)\). This is related to the equation below because when simplifying this equation, you would get a coefficients with opposite signs for x² and y².

- \(Ax^2+Bx^2=C\), where A and B have opposite signs.
- Pour salt on a large piece of cardboard with two circles cut out of it (I am still skeptical about this one)
- Slice two cones from base to base

Other Geometric Insights

- The shortest distance between any two points is a straight line. How can you use congruence to turn something into a straight line?

- Some transformations that are represented algebraically involve rotation and scaling, others just rotation. These transformations can be represented in the real plane or the complex plane

- When you square a complex number, its magnitude will also be squared (this seems trivial now that I write it, but was essential for figuring out the idea below)

- A way to create Pythagorean triples (… but not all Pythagorean triples can be created this way): Take any two number m and n. One leg is equal to \(m^2-n^2\), another leg is equal to \(2mn\), the hypotenuse is equal to \(m^2+n^2\).  If m and n share a factor, the triple will not be a primitive Pythagorean triple.

- A triangle with an incircle of radius 2 has an area that is equal to its perimeter. When you pour salt on a triangle, the center point of the incircle is the highest point. This is also the point of intersection for the three angle bisectors.

Some questions that I still have:

- Given a focus, directrix, and eccentricity, how do I accurately draw the conic section it describes?

- How can I predict what the geometric transformation will look like when given an algebraic transformation (with either real or complex numbers)?

- Why is the center point of the incircle at the intersection points of the angle bisectors in a triangle?

- How do you know there are only 10 possible boxes where the surface area and volume are the same? (I answered this on the last day, but I included it because it was the question that was most present in my mind from Day 5 on).