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 \(\frac{1}{2}\).

**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 x

- \(Ax^2+Bx^2=C\), where A is equal to 0.

^{2}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 x

^{2}and y^{2}.
- \(Ax^2+Bx^2=C\), 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

- 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 baseHyperbola:

- 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 x

^{2}and y^{2}.
- \(Ax^2+Bx^2=C\), 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

- 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

- 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).

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