WIP ODD DICE

.vscode/settings.json

@@ -1,4 +1,5 @@
 {
 	"latex-workshop.latex.recipe.default": "latexmk (xelatex)",
-	"tinymist.formatterPrintWidth": 80
+	"tinymist.formatterPrintWidth": 80,
+	"tinymist.typstExtraArgs": ["--package-path=./lib/typst"]
 }

lib/typst/local/handout/0.1.0/handout.typ

@@ -0,0 +1,282 @@
+/// Typst handout library, used for all documents in this repository.
+
+
+/// If false, hide instructor info.
+///
+/// Compile with the following command to hide solutions:
+/// `typst compile main.typ --input show_solutions=false`
+///
+/// Solutions are shown by default. This behavior
+/// is less surprising than hiding content by default.
+#let show_solutions = {
+  if "show_solutions" in sys.inputs {
+    // Show solutions unless they're explicitly disabled
+    not (
+      sys.inputs.show_solutions == "false" or sys.inputs.show_solutions == "no"
+    )
+  } else {
+    // Show solutions by default
+    true
+  }
+}
+
+// Colors
+#let ored = rgb("D62121")
+#let ogrape = rgb("9C36B5")
+#let ocyan = rgb("2288BF")
+#let oteal = rgb("12B886")
+#let ogreen = rgb("37B26D")
+#let oblue = rgb("1C7ED6")
+
+
+
+//
+// MARK: header
+//
+
+#let make_title(
+  group,
+  quarter,
+  title,
+  subtitle,
+) = {
+  align(
+    center,
+    block(
+      width: 60%,
+      height: auto,
+      breakable: false,
+      align(
+        center,
+        stack(
+          spacing: 7pt,
+          (
+            text(size: 10pt, group) + h(1fr) + text(size: 10pt, quarter)
+          ),
+          line(length: 100%, stroke: 0.2mm),
+          (
+            text(size: 20pt, title) + linebreak() + text(size: 10pt, subtitle)
+          ),
+          line(length: 100%, stroke: 0.2mm),
+        ),
+      ),
+    ),
+  )
+}
+
+#let warn = {
+  set text(ored)
+  align(
+    center,
+    block(
+      width: 60%,
+      height: auto,
+      breakable: false,
+      fill: rgb(255, 255, 255),
+      stroke: ored + 2pt,
+      inset: 3mm,
+      (
+        align(center, text(weight: "bold", size: 12pt, [Instructor's Handout]))
+          + parbreak()
+          + align(
+            left,
+            text(
+              size: 10pt,
+              [This handout contains solutions and notes.]
+                + linebreak()
+                + [Recompile without solutions before distributing.],
+            ),
+          )
+      ),
+    ),
+  )
+}
+
+#let preparedby(name) = (
+  text(
+    size: 10pt,
+    [Prepared by ]
+      + name
+      + [ on ]
+      + datetime
+        .today()
+        .display("[month repr:long] [day padding:none], [year]"),
+  )
+)
+
+//
+// MARK: Solutions
+//
+
+#let solution(content) = {
+  if show_solutions {
+    align(
+      center,
+      stack(
+        block(
+          width: 100%,
+          breakable: false,
+          fill: ored,
+          stroke: ored + 2pt,
+          inset: 1.5mm,
+          (
+            align(left, text(fill: white, weight: "bold", [Solution:]))
+          ),
+        ),
+        block(
+          width: 100%,
+          height: auto,
+          breakable: false,
+          fill: ored.lighten(80%).desaturate(10%),
+          stroke: ored + 2pt,
+          inset: 3mm,
+          align(left, content),
+        ),
+      ),
+    )
+  }
+}
+
+#let notsolution(content) = {
+  if not show_solutions { content }
+}
+
+
+//
+// MARK: Sections
+//
+
+#let generic(t) = block(
+  above: 8mm,
+  below: 2mm,
+  text(weight: "bold", t),
+)
+
+#let _generic_base(kind, ..args) = {
+  counter("obj").step()
+  if args.pos().len() == 0 {
+    generic([
+      #kind
+      #context counter("obj").display():
+    ])
+  } else {
+    generic(
+      [
+        #kind
+        #context counter("obj").display():
+      ]
+        + " "
+        + args.pos().at(0),
+    )
+  }
+}
+
+#let problem(..args) = _generic_base("Problem", ..args)
+#let definition(..args) = _generic_base("Definition", ..args)
+#let theorem(..args) = _generic_base("Theorem", ..args)
+
+
+//
+// MARK: Misc
+//
+
+
+#let hint(content) = {
+  text(fill: rgb(100, 100, 100), style: "oblique", "Hint: ")
+  text(fill: rgb(100, 100, 100), content)
+}
+
+#let note(content) = {
+  text(fill: rgb(100, 100, 100), content)
+}
+
+#let examplesolution(content) = {
+  let c = oblue
+
+  align(
+    center,
+    stack(
+      block(
+        width: 100%,
+        breakable: false,
+        fill: c,
+        stroke: c + 2pt,
+        inset: 1.5mm,
+        (
+          align(left, text(fill: white, weight: "bold", [Example solution:]))
+        ),
+      ),
+      block(
+        width: 100%,
+        height: auto,
+        breakable: false,
+        fill: c.lighten(80%).desaturate(10%),
+        stroke: c + 2pt,
+        inset: 3mm,
+        align(left, content),
+      ),
+    ),
+  )
+}
+
+
+//
+// MARK: wrapper
+//
+
+#let handout(
+  doc,
+  group: none,
+  quarter: none,
+  title: none,
+  by: none,
+  subtitle: none,
+) = {
+  set par(leading: 0.55em, first-line-indent: 0mm, justify: true)
+  set text(font: "New Computer Modern")
+  set par(spacing: 0.5em)
+  show list: set block(spacing: 0.5em, below: 1em)
+  set heading(numbering: (..nums) => nums.pos().at(0))
+
+  set page(
+    margin: 20mm,
+    width: 8in,
+    height: 11.5in,
+    footer: align(
+      center,
+      context counter(page).display(),
+    ),
+    footer-descent: 5mm,
+  )
+
+
+  set list(
+    tight: false,
+    indent: 5mm,
+    spacing: 3mm,
+  )
+
+  show heading.where(level: 1): it => {
+    set align(center)
+    set text(weight: "bold")
+    block[
+      Section #counter(heading).display(): #text(it.body)
+    ]
+  }
+
+  make_title(
+    group,
+    quarter,
+    title,
+    {
+      if by == none { none } else { [#preparedby(by)\ ] }
+      if subtitle == none { none } else { subtitle }
+    },
+  )
+
+  if show_solutions {
+    warn
+  }
+  doc
+}
+

lib/typst/local/handout/0.1.0/typst.toml

@@ -0,0 +1,6 @@
+[package]
+name = "handout"
+version = "0.1.0"
+entrypoint = "handout.typ"
+authors = []
+license = "GPL"

src/Advanced/Tropical Polynomials/macros.typ

@@ -0,0 +1,70 @@
+#import "@local/handout:0.1.0": *
+#import "@preview/cetz:0.3.1"
+
+
+// Shorthand, we'll be using these a lot.
+#let tp = sym.plus.circle
+#let tm = sym.times.circle
+
+#let graphgrid(inner_content) = {
+  align(
+    center,
+    box(
+      inset: 3mm,
+      cetz.canvas({
+        import cetz.draw: *
+        let x = 5.25
+
+        grid(
+          (0, 0), (x, x), step: 0.75,
+          stroke: luma(100) + 0.3mm
+        )
+
+        if (inner_content != none) {
+          inner_content
+        }
+
+        mark((0, x + 0.5), (0, x + 1), symbol: ">", fill: black, scale: 1)
+        mark((x + 0.5, 0), (x + 1, 0), symbol: ">", fill: black, scale: 1)
+
+        line(
+          (0, x + 0.25),
+          (0, 0),
+          (x + 0.25, 0),
+          stroke: 0.75mm + black,
+        )
+      }),
+    ),
+  )
+}
+
+/// Adds extra padding to an equation.
+/// Used as follows:
+///
+/// #eqmbox($
+///   f(x) = -2(x #tp 2)(x #tp 8)
+/// $)
+///
+/// Note that there are newlines between the $ and content,
+/// this gives us display math (which is what we want when using this macro)
+#let eqnbox(eqn) = {
+  align(
+    center,
+    box(
+      inset: 3mm,
+      eqn,
+    ),
+  )
+}
+
+#let dotline(a, b) = {
+  cetz.draw.line(
+    a,
+    b,
+    stroke: (
+      dash: "dashed",
+      thickness: 0.5mm,
+      paint: ored,
+    ),
+  )
+}

src/Advanced/Tropical Polynomials/main.typ

@@ -0,0 +1,22 @@
+#import "@local/handout:0.1.0": *
+
+#show: doc => handout(
+  doc,
+  group: "Advanced 2",
+  quarter: link(
+    "https://betalupi.com/handouts",
+    "betalupi.com/handouts",
+  ),
+
+  title: [Tropical Polynomials],
+  by: "Mark",
+  subtitle: "Based on a handout by Bryant Mathews",
+)
+
+#include "parts/00 arithmetic.typ"
+#pagebreak()
+
+#include "parts/01 polynomials.typ"
+#pagebreak()
+
+#include "parts/02 cubic.typ"

src/Advanced/Tropical Polynomials/meta.toml

@@ -0,0 +1,7 @@
+[metadata]
+title = "Tropical Polynomials"
+
+
+[publish]
+handout = true
+solutions = true

src/Advanced/Tropical Polynomials/parts/00 arithmetic.typ

@@ -0,0 +1,293 @@
+#import "@local/handout:0.1.0": *
+#import "../macros.typ": *
+
+= Tropical Arithmetic
+
+#definition()
+The _tropical sum_ of two numbers is their minimum:
+$
+  x #tp y = min(x, y)
+$
+
+#definition()
+The _tropical product_ of two numbers is their sum:
+$
+  x #tm y = x + y
+$
+
+
+#problem()
+- Is tropical addition commutative? \
+  #note([i.e, does $x #tp y = y #tp x$?])
+- Is tropical addition associative? \
+  #note([i.e, does $(x #tp y) #tp z = x #tp (y #tp z)$?])
+- Is there a tropical additive identity? \
+  #note([i.e, is there an $i$ so that $x #tp i = x$ for all real $x$?])
+
+#solution([
+  - Is tropical addition commutative?\
+    Yes, $min(min(x,y),z) = min(x,y,z) = min(x,min(y,z))$
+  - Is tropical addition associative? \
+    Yes, $min(x,y) = min(y,x)$
+  - Is there a tropical additive identity? \
+    No. There is no $n$ where $x <= n$ for all real $x$
+])
+
+#v(1fr)
+
+#problem()
+Let's expand $#sym.RR$ to include a tropical additive identity.
+- What would be an appropriate name for this new number?
+- Give a reasonable definition for...
+  - the tropical sum of this number and a real number $x$
+  - the tropical sum of this number and itself
+  - the tropical product of this number and a real number $x$
+  - the tropical product of this number and itself
+
+#solution([
+  #sym.infinity makes sense, with
+  $#sym.infinity #tp x = x$; #h(1em)
+  $#sym.infinity #tp #sym.infinity = #sym.infinity$; #h(1em)
+  $#sym.infinity #tm x = #sym.infinity$; #h(1em) and
+  $#sym.infinity #tm #sym.infinity = #sym.infinity$
+])
+
+#v(1fr)
+#pagebreak() // MARK: page
+
+
+
+
+
+#problem()
+Do tropical additive inverses exist? \
+#note([
+  Is there an inverse $y$ for every $x$ so that $x #tp y = #sym.infinity$?
+])
+
+#solution([
+  No. Unless $x = #sym.infinity$, there is no x where $min(x, y) = #sym.infinity$
+])
+
+#v(1fr)
+
+
+
+#problem()
+Is tropical multiplication associative? \
+#note([Does $(x #tm y) #tm z = x #tm (y #tm z)$ for all $x,y,z$?])
+
+#solution([Yes, since (normal) addition is associative])
+
+#v(1fr)
+
+
+#problem()
+Is tropical multiplication commutative? \
+#note([Does $x #tm y = y #tm x$ for all $x, y$?])
+
+#solution([Yes, since (normal) addition is commutative])
+
+#v(1fr)
+
+
+#problem()
+Is there a tropical multiplicative identity? \
+#note([Is there an $i$ so that $x #tm i = x$ for all $x$?])
+
+#solution([Yes, it is 0.])
+
+#v(1fr)
+
+
+#problem()
+Do tropical multiplicative inverses always exist? \
+#note([
+  For every $x != #sym.infinity$, does there exist an inverse $y$ so that $x #tm y = i$, \
+  where $i$ is the additive identity?
+])
+
+#solution([Yes, it is $-x$. For $x != 0$, $x #tm (-x) = 0$])
+
+#v(1fr)
+#pagebreak() // MARK: page
+
+
+#problem()
+Is tropical multiplication distributive over addition? \
+#note([Does $x #tm (y #tp z) = x #tm y #tp x #tm z$?])
+
+#solution([Yes, $x + min(y,z) = min(x+y, x+z)$])
+
+#v(1fr)
+
+#problem()
+Fill the following tropical addition and multiplication tables
+
+#let col = 10mm
+
+#notsolution(
+  table(
+    columns: (1fr, 1fr),
+    align: center,
+    stroke: none,
+    table(
+      columns: (col, col, col, col, col, col),
+      align: center,
+      table.header(
+        [$#tp$],
+        [$1$],
+        [$2$],
+        [$3$],
+        [$4$],
+        [$#sym.infinity$],
+      ),
+
+      box(inset: 3pt, $1$), [], [], [], [], [],
+      box(inset: 3pt, $2$), [], [], [], [], [],
+      box(inset: 3pt, $3$), [], [], [], [], [],
+      box(inset: 3pt, $4$), [], [], [], [], [],
+      box(inset: 3pt, $#sym.infinity$), [], [], [], [], [],
+    ),
+    table(
+      columns: (col, col, col, col, col, col),
+      align: center,
+      table.header(
+        [$#tm$],
+        [$0$],
+        [$1$],
+        [$2$],
+        [$3$],
+        [$4$],
+      ),
+
+      box(inset: 3pt, $0$), [], [], [], [], [],
+      box(inset: 3pt, $1$), [], [], [], [], [],
+      box(inset: 3pt, $2$), [], [], [], [], [],
+      box(inset: 3pt, $3$), [], [], [], [], [],
+      box(inset: 3pt, $4$), [], [], [], [], [],
+    ),
+  ),
+)
+
+#solution(
+  table(
+    columns: (1fr, 1fr),
+    align: center,
+    stroke: none,
+    table(
+      columns: (col, col, col, col, col, col),
+      align: center,
+      table.header(
+        [$#tp$],
+        [$1$],
+        [$2$],
+        [$3$],
+        [$4$],
+        [$#sym.infinity$],
+      ),
+
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $1$),
+
+      box(inset: 3pt, $2$),
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $2$),
+      box(inset: 3pt, $2$),
+      box(inset: 3pt, $2$),
+      box(inset: 3pt, $2$),
+
+      box(inset: 3pt, $3$),
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $2$),
+      box(inset: 3pt, $3$),
+      box(inset: 3pt, $3$),
+      box(inset: 3pt, $3$),
+
+      box(inset: 3pt, $4$),
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $2$),
+      box(inset: 3pt, $3$),
+      box(inset: 3pt, $4$),
+      box(inset: 3pt, $4$),
+
+      box(inset: 3pt, $#sym.infinity$),
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $2$),
+      box(inset: 3pt, $3$),
+      box(inset: 3pt, $4$),
+      box(inset: 3pt, $#sym.infinity$),
+    ),
+    table(
+      columns: (col, col, col, col, col, col),
+      align: center,
+      table.header(
+        [$#tm$],
+        [$0$],
+        [$1$],
+        [$2$],
+        [$3$],
+        [$4$],
+      ),
+
+      box(inset: 3pt, $0$),
+      box(inset: 3pt, $0$),
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $2$),
+      box(inset: 3pt, $3$),
+      box(inset: 3pt, $4$),
+
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $1$),
+      box(inset: 3pt, $2$),
+      box(inset: 3pt, $3$),
+      box(inset: 3pt, $4$),
+      box(inset: 3pt, $5$),
+
+      box(inset: 3pt, $2$),
+      box(inset: 3pt, $2$),
+      box(inset: 3pt, $3$),
+      box(inset: 3pt, $4$),
+      box(inset: 3pt, $5$),
+      box(inset: 3pt, $6$),
+
+      box(inset: 3pt, $3$),
+      box(inset: 3pt, $3$),
+      box(inset: 3pt, $4$),
+      box(inset: 3pt, $5$),
+      box(inset: 3pt, $6$),
+      box(inset: 3pt, $7$),
+
+      box(inset: 3pt, $4$),
+      box(inset: 3pt, $4$),
+      box(inset: 3pt, $5$),
+      box(inset: 3pt, $6$),
+      box(inset: 3pt, $7$),
+      box(inset: 3pt, $8$),
+    ),
+  ),
+)
+
+
+#v(2mm)
+
+#problem()
+Expand and simplify $f(x) = (x #tp 2)(x #tp 3)$, then evaluate $f(1)$ and $f(4)$ \
+Adjacent parenthesis imply tropical multiplication
+
+#solution([
+  $
+    (x #tp 2)(x #tp 3)
+    &= x^2 #tp 2x #tp 3x #tp (2 #tm 3) \
+    &= x^2 #tp (2 #tp 3)x #tp (2 #tm 3) \
+    &= x^2 #tp 2x #tp 5
+  $
+
+  Also, $f(1) = 2$ and $f(4) = 5$.
+])
+
+#v(1fr)

src/Advanced/Tropical Polynomials/parts/01 polynomials.typ

@@ -0,0 +1,387 @@
+#import "@local/handout:0.1.0": *
+#import "../macros.typ": *
+#import "@preview/cetz:0.3.1"
+
+= Tropical Polynomials
+
+#definition()
+A _polynomial_ is an expression formed by adding and multiplying numbers and a variable $x$. \
+Every polynomial can be written as
+#align(
+  center,
+  box(
+    inset: 3mm,
+    $
+      c_0 + c_1 x + c_2 x^2 + ... + c_n x^n
+    $,
+  ),
+)
+for some nonnegative integer $n$ and coefficients $c_0, c_1, ..., c_n$. \
+The _degree_ of a polynomial is the largest $n$ for which $c_n$ is nonzero.
+
+#theorem()
+The _fundamental theorem of algebra_ implies that any non-constant polynomial with real coefficients
+can be written as a product of polynomials of degree 1 or 2 with real coefficients.
+
+#v(2mm)
+
+For example, the polynomial $-160 - 64x - 2x^2 + 17x^3 + 8x^4 + x^5$ \
+can be written as $(x^2 + 2x+5)(x-2)(x+4)(x+4)$
+
+#v(2mm)
+A similar theorem exists for polynomials with complex coefficients. \
+These coefficients may be found using the roots of this polynomial. \
+As it turns out, there are formulas that determine the roots of quadratic, cubic, and quartic #note([(degree 2, 3, and 4)]) polynomials. There are no formulas for the roots of polynomials with larger degrees---in this case, we usually rely on approximate roots found by computers.
+
+#v(2mm)
+In this section, we will analyze tropical polynomials:
+- Is there a fundamental theorem of tropical algebra?
+- Is there a tropical quadratic formula? How about a cubic formula?
+- Is it difficult to find the roots of tropical polynomials with large degrees?
+
+
+#definition()
+A _tropical_ polynomial is a polynomial that uses tropical addition and multiplication. \
+In other words, it is an expression of the form
+#align(
+  center,
+  box(
+    inset: 3mm,
+    $
+      c_0 #tp (c_1 #tm x) #tp (c_2 #tm x^2) #tp ... #tp (c_n #tm x^n)
+    $,
+  ),
+)
+where all exponents represent repeated tropical multiplication.
+
+#pagebreak() // MARK: page
+
+
+
+
+#problem()
+Draw a graph of the tropical polynomial $f(x) = x^2 #tp 1x #tp 4$. \
+#hint([$1x$ is not equal to $x$.])
+
+#notsolution(graphgrid(none))
+
+#solution([
+  $f(x) = min(2x , 1+x, 4)$, which looks like:
+
+  #graphgrid({
+    import cetz.draw: *
+    let step = 0.75
+
+    dotline((0, 0), (4 * step, 8 * step))
+    dotline((0, 1 * step), (7 * step, 8 * step))
+    dotline((0, 4 * step), (8 * step, 4 * step))
+
+    line(
+      (0, 0),
+      (1 * step, 2 * step),
+      (3 * step, 4 * step),
+      (7.5 * step, 4 * step),
+      stroke: 1mm + oblue,
+    )
+  })
+])
+
+
+#problem()
+Now, factor $f(x) = x^2 #tp 1x #tp 4$ into two polynomials with degree 1. \
+In other words, find $r$ and $s$ so that
+#align(
+  center,
+  box(
+    inset: 3mm,
+    $
+      x^2 #tp 1x #tp 4 = (x #tp r)(x #tp s)
+    $,
+  ),
+)
+
+we will call $r$ and $s$ the _roots_ of $f$.
+
+#solution([
+  Because $(x #tp r)(x #tp s) = x^2 #tp (r #tp s)x #tp s r$, we must have $r #tp s = 1$ and $r #tm s = 4$. \
+  In standard notation, we need $min(r, s) = 1$ and $r + s = 4$, so we take $r = 1$ and $s = 3$:
+
+  #v(2mm)
+
+  $
+    f(x) = x^2 #tp 1x #tp 4 = (x #tp 1)(x #tp 3)
+  $
+])
+
+#v(1fr)
+
+#problem()
+How can we use the graph to determine these roots?
+
+#solution([The roots are the corners of the graph.])
+
+#v(0.5fr)
+#pagebreak() // MARK: page
+
+
+
+
+
+
+#problem()
+Graph $f(x) = -2x^2 #tp x #tp 8$. \
+#hint([Use half scale. 1 box = 2 units.])
+
+#notsolution(graphgrid(none))
+
+#solution([
+  #graphgrid({
+    import cetz.draw: *
+    let step = 0.75
+
+    dotline((0, 0), (8 * step, 8 * step))
+    dotline((0.5 * step, 0), (4 * step, 8 * step))
+    dotline((0, 4 * step), (8 * step, 4 * step))
+
+    line(
+      (0.5 * step, 0),
+      (1 * step, 1 * step),
+      (4 * step, 4 * step),
+      (7.5 * step, 4 * step),
+      stroke: 1mm + oblue,
+    )
+  })
+])
+
+#problem()
+Find a factorization of $f$ in the form $a(x #tp r)(x#tp s)$.
+
+#solution([
+  We (tropically) factor out $-2$ to get
+
+  #eqnbox($
+    f(x) = -2(x^2 #tp 2x #tp 10)
+  $)
+
+
+  by the same process as the previous problem, we get
+  #eqnbox($
+    f(x) = -2(x #tp 2)(x #tp 8)
+  $)
+])
+
+#v(1fr)
+
+#problem()
+Can you see the roots $r$ and $s$ in the graph? \
+How are the roots related to the coefficients of $f$? \
+#hint([look at consecutive coefficients: $0 - (-2) = 2$])
+
+#solution([
+  The roots are the differences between consecutive coefficients of $f$:
+  - $0-(-2) = 2$
+  - $8 - 0 = 8$
+])
+
+#v(0.5fr)
+
+
+#problem()
+Find a tropical polynomial that has roots $4$ and $5$ \
+and always produces $7$ for sufficiently large inputs.
+
+#solution([
+  We are looking for $f(x) = a x^2 #tp b x #tp c$. \
+  Since $f(#sym.infinity) = 7$, we know that $c = 7$. \
+  Using the pattern from the previous problem, we'll subtract $5$ from $c$ to get $b = 2$, \
+  and $4$ from $b$ to get $a = -2$.
+
+  And so, $f(x) = -2x^2 #tp 2x #tp 7$
+
+  #v(2mm)
+
+  Subtracting roots in the opposite order does not work.
+])
+
+#v(1fr)
+#pagebreak() // MARK: page
+
+
+#problem()
+Graph $f(x) = 1x^2 #tp 3x #tp 5$.
+
+#notsolution(graphgrid(none))
+
+#solution([
+  The graphs of all three terms intersect at the same point:
+
+  #graphgrid({
+    import cetz.draw: *
+    let step = 0.75
+
+    dotline((0, 1 * step), (3.5 * step, 8 * step))
+    dotline((0, 5 * step), (8 * step, 5 * step))
+    dotline((0, 3 * step), (5 * step, 8 * step))
+
+    line(
+      (0, 1 * step),
+      (2 * step, 5 * step),
+      (7.5 * step, 5 * step),
+      stroke: 1mm + oblue,
+    )
+  })
+])
+
+
+#problem()
+Find a factorization of $f$ in the form $a(x #tp r)(x#tp s)$.
+
+#solution(
+  eqnbox($
+    f(x) = 1x^2 #tp 3 x #tp 5 = 1(x #tp 2)^2
+  $),
+)
+
+#v(1fr)
+
+#problem()
+How is this graph different from the previous two? \
+How is this polynomial's factorization different from the previous two? \
+How are the roots of $f$ related to its coefficients?
+
+#solution([
+  The factorization contains the same term twice. \
+  Also note that the differences between consecutive coefficients of $f$ are both two.
+])
+
+#v(0.5fr)
+#pagebreak() // MARK: page
+
+
+#problem()
+Graph $f(x) = 2x^2 #tp 4x #tp 4$.
+
+#notsolution(graphgrid(none))
+
+#solution(
+  graphgrid({
+    import cetz.draw: *
+    let step = 0.75
+
+    dotline((0, 2 * step), (3 * step, 8 * step))
+    dotline((0, 4 * step), (5 * step, 8 * step))
+    dotline((0, 4 * step), (8 * step, 4 * step))
+
+    line(
+      (0, 2 * step),
+      (1 * step, 4 * step),
+      (7.5 * step, 4 * step),
+      stroke: 1mm + oblue,
+    )
+  }),
+)
+
+
+#problem()
+Find a factorization of $f$ in the form $a(x #tp r)(x#tp s)$, or show that one does not exist.
+
+#solution([
+  We can factor out a 2 to get $f(x) = 2(x^2 #tp 2x #tp 2)$,
+  but $x^2 #tp 2x #tp 2$ does not factor. \
+  There are no $a$ and $b$ with minimum 2 and sum 2.
+])
+
+#v(1fr)
+
+#problem()
+Find a polynomial that has the same graph as $f$, but can be factored.
+
+#solution([
+  $
+    2x^2 #tp 3x #tp 4 = 2(x #tp 1)^2
+  $
+])
+
+
+#v(1fr)
+
+#pagebreak() // MARK: page
+
+
+#theorem()
+The _fundamental thorem of tropical algebra_ states that for every tropical polynomial $f$, \
+there exists a _unique_ tropical polynomial $accent(f, macron)$ with the same graph that can be factored \
+into linear factors.
+
+#v(2mm)
+
+Whenever we say "the roots of $f$", we really mean "the roots of $accent(f, macron)$." \
+$f$ and $accent(f, macron)$ might be the same polynomial.
+
+#problem()
+If $f(x) = a x^2 #tp b x #tp c$, then $accent(f, macron)(x) = a x^2 #tp B x #tp c$ for some $B$. \
+Find a formula for $B$ in terms of $a$, $b$, and $c$. \
+#hint([there are two cases to consider.])
+
+#solution([
+  If we want to factor $a(x^2 #tp (b-a)x #tp (c-a))$, we need to find $r$ and $s$ so that
+  - $min(r,s) = b-a$, and
+  - $r + s = c - a$
+
+  #v(2mm)
+
+  This is possible if and only if $2(b-a) <= c-a$, \
+  or equivalently if $b <= (a+c) #sym.div 2$
+
+  #v(8mm)
+
+  *Case 1:* If $b <= (a + c #sym.div) 2$, then $accent(f, macron) = f$ and $b = B$.
+
+  #v(2mm)
+
+  *Case 2:* If $b > (a + c #sym.div) 2$, then
+  $
+    accent(f, macron)(x)
+    &= a x^2 #tp ((a+c)/2)x #tp c \
+    &= a(x #tp (c-a)/2)^2
+  $
+  has the same graph as $f$, and thus $B = (a+c) #sym.div 2$
+
+  #v(8mm)
+
+  We can combine these results as follows:
+  $
+    B = min(b, (a+c)/2)
+  $
+])
+
+
+#v(1fr)
+
+#problem()
+Find a tropical quadratic formula in terms of $a$, $b$, and $c$ \
+for the roots $x$ of a tropical polynomial $f(x) = a x^2 #tp b x #tp c$. \
+#hint([
+  again, there are two cases. \
+  Remember that "roots of $f$" means "roots of $accent(f, macron)$".
+])
+
+#solution([
+  *Case 1:* If $b <= (a+c) #sym.div 2$, then $accent(f, macron) = f$ has roots $b-a$ and $c-b$, so
+  $
+    accent(f, macron)(x) = a(x #tp (b-a))(x #tp (c-b))
+  $
+
+  #v(8mm)
+
+  *Case 2:* If $b > (a+c) #sym.div 2$, then $accent(f, macron)$ has root $(c-a) #sym.div$ with multiplicity 2, so
+  $
+    accent(f, macron)(x) = a(x #tp (c-a)/2)^2
+  $
+
+  #v(8mm)
+
+  It is interesting to note that the condition $2b < a+ c$ for there to be two distinct roots becomes $b^2 > a c$ in tropical notation. This is reminiscent of the discriminant condition for standard polynomials!
+])
+
+#v(1fr)

src/Advanced/Tropical Polynomials/parts/02 cubic.typ

@@ -0,0 +1,212 @@
+#import "@local/handout:0.1.0": *
+#import "../macros.typ": *
+#import "@preview/cetz:0.3.1"
+
+= Tropical Cubic Polynomials
+
+#problem()
+Consider the polynomial $f(x) = x^3 #tp x^2 #tp 3x #tp 6$. \
+- sketch a graph of this polynomial
+- use this graph to find the roots of $f$
+- write (and expand) a product of linear factors with the same graph as $f$.
+
+#notsolution(graphgrid(none))
+
+#solution([
+  - Roots are 1, 2, and 3.
+  - $accent(f, macron)(x) = x^3 #tp 1x^2 #tp 3x #tp 6 = (x #tp 1)(x #tp 2)(x #tp 3)$
+
+
+  #graphgrid({
+    import cetz.draw: *
+    let step = 0.75
+
+    dotline((0, 0), (2.66 * step, 8 * step))
+    dotline((0, 1 * step), (3.5 * step, 8 * step))
+    dotline((0, 3 * step), (5 * step, 8 * step))
+    dotline((0, 6 * step), (8 * step, 6 * step))
+
+    line(
+      (0, 0),
+      (1 * step, 3 * step),
+      (2 * step, 5 * step),
+      (3 * step, 6 * step),
+      (7.5 * step, 6 * step),
+      stroke: 1mm + oblue,
+    )
+  })
+])
+
+
+
+#v(1fr)
+#pagebreak() // MARK: page
+
+#problem()
+Consider the polynomial $f(x) = x^3 #tp x^2 #tp 6x #tp 6$. \
+- sketch a graph of this polynomial
+- use this graph to find the roots of $f$
+- write (and expand) a product of linear factors with the same graph as $f$.
+
+#notsolution(graphgrid(none))
+
+#solution([
+  - Roots are 1, 2.5, and 2.5.
+  - $accent(f, macron)(x) = x^3 #tp 1x^2 #tp 3.5x #tp 6 = (x #tp 1)(x #tp 2.5)^2$
+
+  #graphgrid({
+    import cetz.draw: *
+    let step = 0.75
+
+    dotline((0, 0), (2.66 * step, 8 * step))
+    dotline((0, 1 * step), (3.5 * step, 8 * step))
+    dotline((0, 6 * step), (2 * step, 8 * step))
+    dotline((0, 6 * step), (8 * step, 6 * step))
+
+    line(
+      (0, 0),
+      (1 * step, 3 * step),
+      (2.5 * step, 6 * step),
+      (7.5 * step, 6 * step),
+      stroke: 1mm + oblue,
+    )
+  })
+])
+
+
+#v(1fr)
+
+#problem()
+Consider the polynomial $f(x) = x^3 #tp 6x^2 #tp 6x #tp 6$. \
+- sketch a graph of this polynomial
+- use this graph to find the roots of $f$
+- write (and expand) a product of linear factors with the same graph as $f$.
+
+#notsolution(graphgrid(none))
+
+#solution([
+  - Roots are 2, 2, and 2.
+  - $accent(f, macron)(x) = x^3 #tp 2x^2 #tp 4x #tp 6 = (x #tp 2)^3$
+
+  #graphgrid({
+    import cetz.draw: *
+    let step = 0.75
+
+    dotline((0, 0), (2.66 * step, 8 * step))
+    dotline((0, 6 * step), (1 * step, 8 * step))
+    dotline((0, 6 * step), (2 * step, 8 * step))
+    dotline((0, 6 * step), (8 * step, 6 * step))
+
+    line(
+      (0, 0),
+      (2 * step, 6 * step),
+      (7.5 * step, 6 * step),
+      stroke: 1mm + oblue,
+    )
+  })
+])
+
+
+#v(1fr)
+#pagebreak() // MARK: page
+
+
+#problem()
+If $f(x) = a x^3 #tp b x^2 #tp c x #tp d$, then $accent(f, macron)(x) = a x^3 #tp B x^2 #tp C x #tp d$ for some $B$ and $C$. \
+Using the last three problems, find formulas for $B$ and $C$ in terms of $a$, $b$, $c$, and $d$.
+
+#solution([
+
+  $
+    B = min(b, (a+c)/2, (2a+d)/2)
+  $
+  $
+    C = min(c, (b+d)/2, (a+2d)/2)
+  $
+])
+
+#v(1fr)
+#pagebreak() // MARK: page
+
+#problem()
+What are the roots of the following polynomial?
+
+#align(
+  center,
+  box(
+    inset: 3mm,
+    $
+      3 x^6 #tp 4 x^5 #tp 2 x^4 #tp x^3 #tp x^2 #tp 4 x #tp 5
+    $,
+  ),
+)
+
+#solution([
+  We have
+  $
+    accent(f, macron)(x) = 3x^6 #tp 2x^5 #tp 1x^4 #tp x^3 #tp 1x^2 #tp 3x #tp 5
+  $
+  which has roots $-1$, $-1$, $-1$, $1$, $2$, $2$
+])
+
+
+#v(1fr)
+#pagebreak() // MARK: page
+
+#problem()
+If
+$
+  f(x) = c_0 #tp c_1 x #tp c_2 x^2 #tp ... #tp c_n x^n
+$
+then
+$
+  accent(f, macron)(x) = c_0 #tp C_1 x #tp C_2 x^2 #tp ... #tp C_(n-1) x^(n-1) #tp c_n x^n
+$
+
+#v(2mm)
+
+Find a formula for each $C_i$ in terms of $c_0, c_1, ..., c_n$.
+
+#solution([
+  $
+    A_j
+    &= min_(l<=j<k)( (a_l - a_k) / (k-l) (k-j) + a_k ) \
+    &= min_(l<=j<k)( a_l (k-j) / (k-l) + a_k (j-l) / (k-l) )
+  $
+
+  #v(2mm)
+
+  which is a weighted average of some $a_l$ and $a_k$, with $l<=j<k$
+])
+
+#v(1fr)
+
+
+#problem()
+With the same setup as the previous problem, \
+find formulas for the roots $r_1, r_2, ..., r_n$.
+
+#solution([
+  The roots are the differences between consecutive coefficients of $accent(f, macron)$:
+  $
+    r_i = A_i - A_(i-1)
+  $
+  where we set $A_n = a_n$ and $A_0 = a_0$.
+])
+
+#v(1fr)
+
+#problem()
+Can you find a geometric interpretation of these formulas \
+in terms of the points $(-i, c_i)$ for $0 <= i <= n$?
+
+#solution([
+  The inequality #note([(for $l <= k < k$)])
+  $
+    A_j <= (a_l - a_k) / (k-l) (k-j)+a_k
+  $
+  states that the point $(-j,A_j)$ must lie on or below the line segment between the points $(-k, a_k)$ and $(-l, a_l)$.
+  This makes it easy to find the $A_j$ using a graph of the points $(-i, a_i)$ for $0 <= i <=n$.
+])
+
+#v(0.5fr)

src/Warm-Ups/A Familiar Concept/main.tex

@@ -1,35 +0,0 @@
-\documentclass[
-	solutions,
-	hidewarning,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-
-
-\title{Warm-Up: A Familiar Concept}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today}
-
-
-\begin{document}
-
-	\maketitle
-
-
-	\problem{}<one>
-	Let $v = [-5, -2, 0, 1, 4, 1000]$. Find all $x$ that minimize the following metric. \par
-	$$
-		\sum_{\forall i} |v_i - x| = |v_1 - x| + |v_2 - x| + ... + |v_6 - x|
-	$$
-	\vfill
-
-	\problem{}
-	Let $v = [-5, -2, 0, 1, 4, 1000, 1001]$. Find all $x$ that minimize the metric in \ref{one}.
-	\vfill
-
-	\problem{}
-	What is this metric usually called?
-
-\end{document}

src/Warm-Ups/A Familiar Concept/main.typ

@@ -0,0 +1,39 @@
+#import "@local/handout:0.1.0": *
+
+#show: doc => handout(
+  doc,
+  quarter: link(
+    "https://betalupi.com/handouts",
+    "betalupi.com/handouts",
+  ),
+
+  title: [Warm-Up: A Familiar Concept],
+  by: "Mark",
+)
+
+#problem()
+Let $v = [-5, -2, 0, 1, 4, 1000]$. Find all $x$ that minimize the following metric:
+
+#align(
+  center,
+  box(
+    inset: 3mm,
+    $
+      sum_(#sym.forall i) |v_i - x| = |v_1 - x| + |v_2 - x| + ... + |v_6 - x|
+    $,
+  ),
+)
+
+#v(1fr)
+
+#problem()
+Let $v = [-5, -2, 0, 1, 4, 1000, 1001]$. Find all $x$ that minimize the metric in the previous problem.
+
+#v(1fr)
+
+
+#problem()
+What is this metric usually called?
+
+
+#v(0.25fr)

src/Warm-Ups/Fuse Timers/main.tex

@@ -1,31 +0,0 @@
-\documentclass[
-	solutions,
-	hidewarning,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-
-
-\title{Warm-Up: Fuse Timers}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today.}
-
-\begin{document}
-
-	\maketitle
-
-
-	\problem{}
-	Suppose we have two strings and a lighter. Each string takes an hour to fully burn. \par
-	However, we do not know how fast each part of the string burns:
-	half might burn in 1 minute, and the rest could take 59.
-
-	\vspace{2mm}
-
-	How would we measure exactly 45 minutes using these strings?
-
-	\vfill
-
-\end{document}

src/Warm-Ups/Fuse Timers/main.typ

@@ -0,0 +1,21 @@
+#import "@local/handout:0.1.0": *
+
+#show: doc => handout(
+  doc,
+  quarter: link(
+    "https://betalupi.com/handouts",
+    "betalupi.com/handouts",
+  ),
+
+  title: [Warm-Up: Fuse Timers],
+  by: "Mark",
+)
+
+#problem()
+Suppose we have two strings and a lighter. Each string takes exactly an hour to fully burn. \
+However, we do not know how fast each part of the string burns:
+half might burn in 1 minute, and the rest could take 59.
+
+#v(2mm)
+
+How can we measure exactly 45 minutes using these two strings?

src/Warm-Ups/Mario Kart/main.tex

@@ -1,54 +0,0 @@
-\documentclass[
-	solutions,
-	hidewarning,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-
-\title{Warm-Up: Mario Kart}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today}
-
-
-\begin{document}
-
-	\maketitle
-
-	\problem{}
-	A standard Mario Kart cup consists of 12 players and four races. \par
-	Each race is scored as follows:
-	\begin{itemize}
-		\item 15 points are awarded for first place;
-		\item 12 for second;
-		\item and $(13 - \text{place})$ otherwise.
-	\end{itemize}
-	In any one race, no players may tie.
-	A player's score at the end of a cup is the sum of their scores for each of the four races.
-
-	\vspace{2mm}
-
-	An $n$-way tie occurs when the top $n$ players have the same score at the end of a round. \par
-	What is the largest possible $n$, and how is it achieved?
-
-	\begin{solution}
-		A 12-way tie is impossible, since the total number of point is not divisible by 12.
-
-		\vspace{2mm}
-
-		A 11-way tie is possible, with a top score of 28:
-		\begin{itemize}
-			\item Four players finish $1^\text{st}$, $3^\text{ed}$, $11^\text{th}$, and $12^\text{th}$;
-
-			% spell:off
-			\item Four players finish $2^\text{nd}$, $4^\text{th}$, $9^\text{th}$, and $10^\text{th}$;
-			% spell:on
-
-			\item Two players finish fifth twice and seventh twice,
-			\item One player finishes sixth in each race.
-		\end{itemize}
-		The final player always finishes eighth, with a non-tie score of 20.
-	\end{solution}
-
-\end{document}

src/Warm-Ups/Mario Kart/main.typ

@@ -0,0 +1,41 @@
+#import "@local/handout:0.1.0": *
+
+#show: doc => handout(
+  doc,
+  quarter: link(
+    "https://betalupi.com/handouts",
+    "betalupi.com/handouts",
+  ),
+
+  title: [Warm-Up: Mario Kart],
+  by: "Mark",
+)
+
+#problem()
+A standard Mario Kart cup consists of 12 players and four races. \
+Each race is scored as follows:
+- 15 points are awarded for first place;
+- 12 for second;
+- and $(13 - #text("place"))$ otherwise.
+
+In any one race, no players may tie. \
+A player's score at the end of a cup is the sum of their scores for each of the four races.
+
+#v(2mm)
+
+An $n$-way tie occurs when the top $n$ players have the same score at the end of a round. \
+What is the largest possible $n$, and how is it achieved?
+
+#solution([
+  A 12-way tie is impossible, since the total number of point is not divisible by 12.
+
+  #v(2mm)
+
+  A 11-way tie is possible, with a top score of 28:
+  - Four players finish $1^#text("st")$, $3^#text("ed")$, $11^#text("th")$, and $12^#text("th")$;
+  - Four players finish $2^#text("nd")$, $4^#text("th")$, $9^#text("th")$, and $10^#text("th")$; // spell:disable-line
+  - Two players finish fifth twice and seventh twice,
+  - One player finishes sixth in each race.
+  The final player always finishes eighth, with a non-tie score of 20.
+
+])

src/Warm-Ups/Odd Dice/main.tex

@@ -1,132 +0,0 @@
-\documentclass[
-	nosolutions,
-	hidewarning,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-
-
-\usepackage{tikz}
-\usetikzlibrary{arrows.meta}
-\usetikzlibrary{shapes.geometric}
-
-% We put nodes in a separate layer, so we can
-% slightly overlap with paths for a perfect fit
-\pgfdeclarelayer{nodes}
-\pgfdeclarelayer{path}
-\pgfsetlayers{main,nodes}
-
-% Layer settings
-\tikzset{
-	% Layer hack, lets us write
-	% later = * in scopes.
-	layer/.style = {
-		execute at begin scope={\pgfonlayer{#1}},
-		execute at end scope={\endpgfonlayer}
-	},
-	%
-	% Arrowhead tweak
-	>={Latex[ width=2mm, length=2mm ]},
-	%
-	% Nodes
-	main/.style = {
-		draw,
-		circle,
-		fill = white,
-		line width = 0.35mm
-	}
-}
-
-\title{Warm Up: Odd Dice}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today}
-
-
-\begin{document}
-
-	\maketitle
-
-	\problem{}
-
-	We say a set of dice $\{A, B, C\}$ is \textit{nontransitive}
-	if, on average, $A$ beats $B$, $B$ beats $C$, and $C$ beats $A$.
-	In other words, we get a counterintuitive \say{rock - paper - scissors} effect.
-
-	\vspace{2mm}
-
-	Create a set of nontransitive six-sided dice. \par
-	\hint{All sides should be numbered with positive integers less than 10.}
-
-	\begin{solution}
-		One possible set can be numbered as follows:
-		\begin{itemize}
-			\item Die $A$: $2, 2, 4, 4, 9, 9$
-			\item Die $B$: $1, 1, 6, 6, 8, 8$
-			\item Die $C$: $3, 3, 5, 5, 7, 7$
-		\end{itemize}
-
-		\vspace{4mm}
-
-		Another solution is below:
-		\begin{itemize}
-			\item Die $A$: $3, 3, 3, 3, 3, 6$
-			\item Die $B$: $2, 2, 2, 5, 5, 5$
-			\item Die $C$: $1, 4, 4, 4, 4, 4$
-		\end{itemize}
-	\end{solution}
-
-	\vfill
-
-	\problem{}
-	Now, consider the set of six-sided dice below:
-	\begin{itemize}
-		\item Die $A$: $4, 4, 4, 4, 4, 9$
-		\item Die $B$: $3, 3, 3, 3, 8, 8$
-		\item Die $C$: $2, 2, 2, 7, 7, 7$
-		\item Die $D$: $1, 1, 6, 6, 6, 6$
-		\item Die $E$: $0, 5, 5, 5, 5, 5$
-	\end{itemize}
-	On average, which die beats each of the others? Draw a graph. \par
-
-	\begin{solution}
-		\begin{center}
-		\begin{tikzpicture}[scale = 0.5]
-			\begin{scope}[layer = nodes]
-				\node[main] (a) at (-2, 0.2) {$a$};
-				\node[main] (b) at (0, 2) {$b$};
-				\node[main] (c) at (2, 0.2) {$c$};
-				\node[main] (d) at (1, -2) {$d$};
-				\node[main] (e) at (-1, -2) {$e$};
-			\end{scope}
-
-			\draw[->]
-				(a) edge (b)
-				(b) edge (c)
-				(c) edge (d)
-				(d) edge (e)
-				(e) edge (a)
-
-				(a) edge (c)
-				(b) edge (d)
-				(c) edge (e)
-				(d) edge (a)
-				(e) edge (b)
-			;
-		\end{tikzpicture}
-		\end{center}
-	\end{solution}
-
-	\vfill
-
-	Now, say we roll each die twice. What happens to the graph above?
-
-	\begin{solution}
-		The direction of each edge is reversed!
-	\end{solution}
-
-	\vfill
-	\pagebreak
-
-\end{document}

src/Warm-Ups/Odd Dice/main.typ

@@ -0,0 +1,87 @@
+#import "@local/handout:0.1.0": *
+
+#show: doc => handout(
+  doc,
+  quarter: link(
+    "https://betalupi.com/handouts",
+    "betalupi.com/handouts",
+  ),
+
+  title: [Warm-Up: Odd Dice],
+  by: "Mark",
+)
+
+#problem()
+We say a set of dice ${A, B, C}$ is _nontransitive_
+if, on average, $A$ beats $B$, $B$ beats $C$, and $C$ beats $A$.
+In other words, we get a counterintuitive "rock - paper - scissors" effect.
+
+#v(2mm)
+
+Create a set of nontransitive six-sided dice. \
+#hint([All sides should be numbered with positive integers less than 10.])
+
+#solution([
+  One possible set can be numbered as follows:
+  - Die $A$: $2, 2, 4, 4, 9, 9$
+  - Die $B$: $1, 1, 6, 6, 8, 8$
+  - Die $C$: $3, 3, 5, 5, 7, 7$
+
+  #v(4mm)
+
+  Another solution is below:
+  - Die $A$: $3, 3, 3, 3, 3, 6$
+  - Die $B$: $2, 2, 2, 5, 5, 5$
+  - Die $C$: $1, 4, 4, 4, 4, 4$
+
+])
+
+#v(1fr)
+
+#problem()
+Now, consider the set of six-sided dice below:
+- Die $A$: $4, 4, 4, 4, 4, 9$
+- Die $B$: $3, 3, 3, 3, 8, 8$
+- Die $C$: $2, 2, 2, 7, 7, 7$
+- Die $D$: $1, 1, 6, 6, 6, 6$
+- Die $E$: $0, 5, 5, 5, 5, 5$
+On average, which die beats each of the others? Draw a diagram.
+
+#solution([
+  /*
+  \begin{tikzpicture}[scale = 0.5]
+    \begin{scope}[layer = nodes]
+      \node[main] (a) at (-2, 0.2) {$a$};
+      \node[main] (b) at (0, 2) {$b$};
+      \node[main] (c) at (2, 0.2) {$c$};
+      \node[main] (d) at (1, -2) {$d$};
+      \node[main] (e) at (-1, -2) {$e$};
+    \end{scope}
+
+    \draw[->]
+      (a) edge (b)
+      (b) edge (c)
+      (c) edge (d)
+      (d) edge (e)
+      (e) edge (a)
+
+      (a) edge (c)
+      (b) edge (d)
+      (c) edge (e)
+      (d) edge (a)
+      (e) edge (b)
+    ;
+  \end{tikzpicture}
+  */
+])
+
+#v(1fr)
+
+#problem()
+Now, say we roll each die twice. What happens to the graph fromE the previous problem?
+
+#solution([
+  The direction of each edge is reversed!
+])
+
+#v(1fr)

src/Warm-Ups/Partition Products/main.tex

@@ -1,57 +0,0 @@
-\documentclass[
-	solutions,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-
-\title{Warm-Up: Partition Products}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today.}
-
-\begin{document}
-
-	\maketitle
-
-	\problem{}
-	Take any positive integer $n$. \par
-	Now, write it as sum of smaller positive integers: $n = a_1 + a_2 + ... + a_k$ \par
-	Maximize the product $a_1 \times a_2 \times ... \times a_k$
-
-
-
-	\begin{solution}
-
-		\textbf{Interesting Solution:}
-
-		Of course, all $a_i$ should be greater than $1$. \par
-		Also, all $a_i$ should be smaller than four, since $x \leq x(x-2)$ if $x \geq 4$. \par
-		Thus, we're left with sequences that only contain 2 and 3. \par
-		\note{Note that two twos are the same as one four, but we exclude fours for simplicity.}
-
-		\vspace{2mm}
-
-		Finally, we see that $3^2 > 2^3$, so any three twos are better repackaged as two threes. \par
-		The best sequence $a_i$ thus consists of a maximal number of threes followed by 0, 1, or 2 twos.
-
-		\linehack{}
-
-
-
-		\textbf{Calculus Solution:}
-
-		First, solve this problem for equal, non-integer $a_i$:
-
-		\vspace{2mm}
-
-		We know $n = \prod{a_i}$, thus $\ln(n) = \sum{\ln(a_i)}$. \par
-		If all $a_i$ are equal, we get $\ln(n) = k \times \ln(n / k)$. \par
-		Derive wrt $k$ and set to zero to get $\ln(n / k) = 1$ \par
-		So $k = n / e$ and $n / k = e \approx 2.7$
-
-		\vspace{2mm}
-
-		If we try to approximate this with integers, we get the same solution as above.
-	\end{solution}
-\end{document}

src/Warm-Ups/Partition Products/main.typ

@@ -0,0 +1,47 @@
+#import "@local/handout:0.1.0": *
+
+#show: doc => handout(
+  doc,
+  quarter: link(
+    "https://betalupi.com/handouts",
+    "betalupi.com/handouts",
+  ),
+
+  title: [Warm-Up: Partition Products],
+  by: "Mark",
+)
+
+#problem()
+Take any positive integer $n$. \
+Now, write it as sum of smaller positive integers: $n = a_1 + a_2 + ...  a_k$ \
+Maximize the product $a_1 #sym.times a_2 #sym.times ... #sym.times a_k$
+
+
+#solution([
+  *Interesting Solution:*
+
+  Of course, all $a_i$ should be greater than $1$. \
+  Also, all $a_i$ should be smaller than four, since $x <= x(x-2)$ if $x >= 4$. \
+  Thus, we're left with sequences that only contain 2 and 3. \
+  #note([Note that two twos are the same as one four, but we exclude fours for simplicity.])
+
+  #v(2mm)
+
+  Finally, we see that $3^2 > 2^3$, so any three twos are better repackaged as two threes. \
+  The best sequence $a_i$ thus consists of a maximal number of threes followed by 0, 1, or 2 twos.
+
+  #v(8mm)
+
+  *Calculus Solution:*
+
+  First, solve this problem for equal, real $a_i$:
+  #v(2mm)
+  We know $n = product(a_i)$, thus $ln(n) = sum(ln(a_i))$. \
+  If all $a_i$ are equal, we get $ln(n) = k #sym.times ln(n / k)$. \
+  Derive wrt $k$ and set to zero to get $ln(n / k) = 1$ \
+  So $k = n / e$ and $n / k = e #sym.approx 2.7$
+
+  #v(2mm)
+
+  If we try to approximate this with integers, we get the same solution as above.
+])

src/Warm-Ups/Prime Factors/main.tex

@@ -1,34 +0,0 @@
-\documentclass[
-	solutions,
-	singlenumbering,
-	nopagenumber,
-	hidewarning
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-\title{Warm-Up: Prime Factors}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today.}
-
-\begin{document}
-
-	\maketitle
-
-	\problem{}
-	What proportion of integers have $2$ as their smallest prime factor?
-	% 1^2
-	\vfill
-
-
-	\problem{}
-	What proportion of integers have $3$ as their second-smallest prime factor?
-	% 1/6
-	\vfill
-
-
-	\problem{}
-	What is the median second-smallest prime factor?
-	% 37
-	\vfill
-
-\end{document}

src/Warm-Ups/Prime Factors/main.typ

@@ -0,0 +1,29 @@
+#import "@local/handout:0.1.0": *
+
+#show: doc => handout(
+  doc,
+  quarter: link(
+    "https://betalupi.com/handouts",
+    "betalupi.com/handouts",
+  ),
+
+  title: [Warm-Up: Prime Factors],
+  by: "Mark",
+)
+
+#problem()
+What proportion of integers have $2$ as their smallest prime factor?
+#solution([$1 div 2$])
+#v(1fr)
+
+
+#problem()
+What proportion of integers have $3$ as their second-smallest prime factor?
+#solution([$1 div 6$])
+#v(1fr)
+
+
+#problem()
+What is the median second-smallest prime factor?
+#solution([37])
+#v(1fr)

src/Warm-Ups/Regex/main.tex

@@ -1,153 +0,0 @@
-\documentclass[
-	solutions,
-	hidewarning,
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-
-\usepackage{xcolor}
-\usepackage{soul}
-\usepackage{hyperref}
-
-\definecolor{Light}{gray}{.90}
-\sethlcolor{Light}
-\newcommand{\htexttt}[1]{\texttt{\hl{#1}}}
-
-
-\title{The Regex Warm-Up}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today}
-
-\begin{document}
-
-	\maketitle
-
-
-	Last time, we discussed Deterministic Finite Automata. One interesting application of these mathematical objects is found in computer science: Regular Expressions. \par
-	This is often abbreviated \say{regex}, which is pronounced like \say{gif.}
-
-	\vspace{2mm}
-
-	Regex is a language used to specify patterns in a string. You can think of it as a concise way to define a DFA, using text instead of a huge graph. \par
-
-	Often enough, a clever regex pattern can do the work of a few hundred lines of code.
-
-	\vspace{2mm}
-
-	Like the DFAs we've studied, a regex pattern \textit{accepts} or \textit{rejects} a string. However, we don't usually use this terminology with regex, and instead say that a string \textit{matches} or \textit{doesn't match} a pattern.
-
-	\vspace{5mm}
-
-	Regex strings consist of characters, quantifiers, sets, and groups.
-
-	\vspace{5mm}
-
-	\textbf{Quantifiers} \par
-	Quantifiers specify how many of a character to match. \par
-	There are four of these: \htexttt{+}, \htexttt{*}, \htexttt{?}, and \htexttt{\{ \}}
-
-	\vspace{2mm}
-
-	\htexttt{+} means \say{match one or more of the preceding token} \par
-	\htexttt{*} means \say{match zero or more of the preceding token}
-
-	For example, the pattern \htexttt{ca+t} will match the following strings:
-	\begin{itemize}
-		\item \texttt{cat}
-		\item \texttt{caat}
-		\item \texttt{caaaaaaaat}
-	\end{itemize}
-	\htexttt{ca+t} will \textbf{not} match the string \texttt{ct}. \par
-	The pattern \htexttt{ca*t} will match all the strings above, including \texttt{ct}.
-	\vspace{2mm}
-
-
-	\htexttt{?} means \say{match one or none of the preceding token} \par
-	The pattern \htexttt{linea?r} will match only \texttt{linear} and \texttt{liner}.
-	\vspace{2mm}
-
-	Brackets \htexttt{\{min, max\}} are the most flexible quantifier. \par
-	They specify exactly how many tokens to match: \par
-	\htexttt{ab\{2\}a} will match only \texttt{abba}. \par
-	\htexttt{ab\{1,3\}a} will match only \texttt{aba}, \texttt{abba}, and \texttt{abbba}. \par
-	% spell:off
-	\htexttt{ab\{2,\}a} will match any \texttt{ab...ba} with at least two \texttt{b}s.
-	% spell:on
-
-	\vspace{5mm}
-
-	\problem{}
-	Write the patterns \htexttt{a*} and \htexttt{a+} using only \htexttt{\{ \}}.
-	\vfill
-
-	\problem{}
-	Draw a DFA equivalent to the regex pattern \htexttt{01*0}.
-	\vfill
-
-	\pagebreak
-
-
-
-
-
-
-	\textbf{Characters, Sets, and Groups} \par
-	In the previous section, we saw how we can specify characters literally: \par
-	\texttt{a+} means \say{one or more \texttt{a} character}
-
-	\vspace{2mm}
-
-	There are, of course, other ways we can specify characters.
-
-	\vspace{2mm}
-
-	The first such way is the \textit{set}, denoted \htexttt{[ ]}. A set can pretend to be any character inside it. \par
-	For example, \htexttt{m[aoy]th} will match \texttt{math}, \texttt{moth}, or \texttt{myth}. \par
-	\htexttt{a[01]+b} will match \texttt{a0b}, \texttt{a111b}, \texttt{a1100110b}, and any other similar string. \par
-	You may negate a set with a \htexttt{\textasciicircum}. \par
-	\htexttt{[\textasciicircum abc]} will match any character except \texttt{a}, \texttt{b}, or \texttt{c}, including symbols and spaces.
-
-	\vspace{2mm}
-
-	If we want to keep characters together, we can use the \textit{group}, denoted \htexttt{( )}. \par
-
-	Groups work exactly as you'd expect, representing an atomic\footnotemark{} group of characters. \par
-	\htexttt{a(01)+b} will match \texttt{a01b} and \texttt{a010101b}, but will \textbf{not} match \texttt{a0b}, \texttt{a1b}, or \texttt{a1100110b}.
-
-	\footnotetext{In other words, \say{unbreakable}}
-
-
-	\problem{}<regex>
-	You are now familiar with most of the tools regex has to offer. \par
-	Write patterns that match the following strings:
-	\begin{enumerate}[itemsep=1mm]
-		\item An ISO-8601 date, like \texttt{2022-10-29}. \par
-		\hint{Invalid dates like \texttt{2022-13-29} should also be matched.}
-
-		\item An email address. \par
-		\hint{Don't forget about subdomains, like \texttt{math.ucla.edu}.}
-
-		\item A UCLA room number, like \texttt{MS 5118} or \texttt{Kinsey 1220B}.
-
-		\item Any ISBN-10 of the form \texttt{0-316-00395-7}. \par
-		\hint{Remember that the check digit may be an \texttt{X}. Dashes are optional.}
-
-		\item A word of even length. \par
-		\hint{The set \texttt{[A-z]} contains every english letter, capitalized and lowercase. \\
-		\texttt{[a-z]} will only match lowercase letters.}
-
-		\item A word with exactly 3 vowels. \par
-		\hint{The special token \texttt{\textbackslash w} will match any word character. It is equivalent to \texttt{[A-z0-9\_]} \\ \texttt{\_} stands for a literal underscore.}
-
-		\item A word that has even length and exactly 3 vowels.
-
-		\item A sentence that does not start with a capital letter.
-	\end{enumerate}
-
-	\vfill
-
-	\problem{}
-	If you'd like to know more, check out \url{https://regexr.com}. It offers an interactive regex prompt, as well as a cheatsheet that explains every other regex token there is. \par
-	You will find a nice set of challenges at \url{https://alf.nu/RegexGolf}.
-	I especially encourage you to look into this if you are interested in computer science.
-\end{document}

src/Warm-Ups/Regex/main.typ

@@ -0,0 +1,141 @@
+#import "@local/handout:0.1.0": *
+
+#show: doc => handout(
+  doc,
+  quarter: link(
+    "https://betalupi.com/handouts",
+    "betalupi.com/handouts",
+  ),
+
+  title: [The Regex Warm-Up],
+  by: "Mark",
+)
+
+
+Last time, we discussed Deterministic Finite Automata. One interesting application of these mathematical objects is found in computer science: Regular Expressions. \
+This is often abbreviated "regex," which is pronounced like "gif."
+
+#v(2mm)
+
+Regex is a language used to specify patterns in a string. You can think of it as a concise way to define a DFA, using text instead of a huge graph. \
+
+Often enough, a clever regex pattern can do the work of a few hundred lines of code.
+
+#v(2mm)
+
+Like the DFAs we've studied, a regex pattern _accepts_ or _rejects_ a string. However, we don't usually use this terminology with regex, and instead say that a string _matches_ or _doesn't match_ a pattern.
+
+#v(5mm)
+
+Regex strings consist of characters, quantifiers, sets, and groups.
+
+#v(5mm)
+
+
+
+*Quantifiers* \
+Quantifiers specify how many of a character to match. \
+There are four of these: `+`, `*`, `?`, and `{ }`.
+
+#v(4mm)
+
+`+` means "match one or more of the preceding token" \
+`*` means "match zero or more of the preceding token"
+
+For example, the pattern `ca+t` will match the following strings:
+- `cat`
+- `caat`
+- `caaaaaaaat`
+`ca+t` will *not* match the string `ct`. \
+The pattern `ca*t` will match all the strings above, including `ct`.
+
+
+#v(4mm)
+
+
+`?` means "match one or none of the preceding token" \
+The pattern `linea?r` will match only `linear` and `liner`.
+
+#v(4mm)
+
+Brackets `{min, max}` are the most flexible quantifier. \
+They specify exactly how many tokens to match: \
+`ab{2}a` will match only `abba`. \
+`ab{1,3}a` will match only `aba`, `abba`, and `abbba`. \
+`ab{2,}a` will match any `ab...ba` with at least two `b`s. // spell:disable-line
+
+#problem()
+Write the patterns `a*` and `a+` using only `{ }`.
+#v(1fr)
+
+#problem()
+Draw a DFA equivalent to the regex pattern `01*0`.
+#v(1fr)
+
+#pagebreak()
+
+
+
+
+
+
+*Characters, Sets, and Groups* \
+In the previous section, we saw how we can specify characters literally: \
+`a+` means "one or more `a` characters" \
+There are, of course, other ways we can specify characters.
+
+#v(4mm)
+
+The first such way is the _set_, denoted `[ ]`. A set can pretend to be any character inside it. \
+For example, `m[aoy]th` will match `math`, `moth`, or `myth`. \
+`a[01]+b` will match `a0b`, `a111b`, `a1100110b`, and any other similar string. \
+
+#v(4mm)
+
+We can negate a set with a `^`. \
+`[^abc]` will match any single character except `a`, `b`, or `c`, including symbols and spaces.
+
+#v(4mm)
+
+If we want to keep characters together, we can use the _group_, denoted `( )`. \
+
+Groups work exactly as you'd expect, representing an atomic#footnote([In other words, "unbreakable"]) group of characters. \
+`a(01)+b` will match `a01b` and `a010101b`, but will *not* match `a0b`, `a1b`, or `a1100110b`.
+
+#problem()
+You are now familiar with most of the tools regex has to offer. \
+Write patterns that match the following strings:
+
+- An ISO-8601 date, like `2022-10-29`. \
+  #hint([Invalid dates like `2022-13-29` should also be matched.])
+
+- An email address. \
+  #hint([Don't forget about subdomains, like `math.ucla.edu`.])
+
+- A UCLA room number, like `MS 5118` or `Kinsey 1220B`.
+
+- Any ISBN-10 of the form `0-316-00395-7`. \
+  #hint([Remember that the check digit may be an `X`. Dashes are optional.])
+
+- A word of even length. \
+  #hint([
+    The set `[A-z]` contains every english letter, capitalized and lowercase. \
+    `[a-z]` will only match lowercase letters.
+  ])
+
+- A word with exactly 3 vowels. \
+  #hint([
+    The special token `\w` will match any word character. \
+    It is equivalent to `[A-z0-9_]`. `_` represents a literal underscore.
+  ])
+
+- A word that has even length and exactly 3 vowels.
+
+- A sentence that does not start with a capital letter.
+#v(1fr)
+
+#problem()
+If you'd like to know more, check out `https://regexr.com`.
+It offers an interactive regex prompt,
+as well as a cheatsheet that explains every other regex token there is. \
+You can find a nice set of challenges at `https://alf.nu/RegexGolf`.

src/Warm-Ups/Travellers/main.tex

@@ -1,30 +0,0 @@
-\documentclass[
-	solutions,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-\title{Warm-Up: Travellers}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today}
-
-\begin{document}
-
-	\maketitle
-
-	\problem{}
-	Four travellers are on a plane, each moving along a straight line at an arbitrary constant speed. \par
-	No two of their paths are parallel, and no three intersect at the same point. \par
-	We know that traveller A has met travelers B, C, and D, \par
-	and that traveller B has met C and D (and A). Show that C and D must also have met. \par
-
-	\begin{solution}
-		When a body travels at a constant speed, its graph with respect to time is a straight line. \par
-		So, we add time axis in the third dimension, perpendicular to our plane. \par
-		Naturally, the projection of each of these onto the plane corresponds to a road.
-
-		Now, note that two intersecting lines define a plane and use the conditions in the problem to show that no two lines are parallel.
-	\end{solution}
-
-\end{document}

src/Warm-Ups/Travellers/main.typ

@@ -0,0 +1,26 @@
+#import "@local/handout:0.1.0": *
+
+#show: doc => handout(
+  doc,
+  quarter: link(
+    "https://betalupi.com/handouts",
+    "betalupi.com/handouts",
+  ),
+
+  title: [Warm-Up: Travellers],
+  by: "Mark",
+)
+
+#problem()
+Four travellers are on a plane, each moving along a straight line at an arbitrary constant speed. \
+No two of their paths are parallel, and no three intersect at the same point. \
+We know that traveller A has met travelers B, C, and D, \
+and that traveller B has met C and D (and A). Show that C and D must also have met.
+
+#solution([
+  When a body travels at a constant speed, its graph with respect to time is a straight line. \
+  So, we add time axis in the third dimension, perpendicular to our plane. \
+  Naturally, the projection of each of these onto the plane corresponds to a road.
+
+  Now, note that two intersecting lines define a plane and use the conditions in the problem to show that no two lines are parallel.
+])

src/Warm-Ups/fmod/main.tex

@@ -1,22 +0,0 @@
-\documentclass[
-	solutions,
-	hidewarning,
-	singlenumbering,
-	nopagenumber
-]{../../../lib/tex/ormc_handout}
-\usepackage{../../../lib/tex/macros}
-
-
-\title{Warm-Up: \texttt{fmod}}
-\uptitler{\smallurl{}}
-\subtitle{Prepared by Mark on \today.}
-
-\begin{document}
-
-	\maketitle
-
-	\problem{}
-	I'm sure you're all familiar with how \texttt{mod(a, b)} and \texttt{remainder(a, b)} work with integers. \par
-	Devise an equivalent for floats (i.e, real numbers).
-
-\end{document}

src/Warm-Ups/fmod/main.typ

@@ -0,0 +1,16 @@
+#import "@local/handout:0.1.0": *
+
+#show: doc => handout(
+  doc,
+  quarter: link(
+    "https://betalupi.com/handouts",
+    "betalupi.com/handouts",
+  ),
+
+  title: [Warm-Up: `fmod`],
+  by: "Mark",
+)
+
+#problem()
+I'm sure you're all familiar with how `mod(a, b)` and `remainder(a, b)` \ work when `a` and `b` are integers.
+Devise an equivalent for floats (i.e, real numbers).

tools/scripts/build.py

@@ -143,10 +143,12 @@ def build_typst(source_dir: Path, out_subdir: Path) -> IndexEntry | None:
 		[
 			TYPST_PATH,
 			"compile",
+			"--package-path",
+			f"{ROOT}/lib/typst",
 			"--ignore-system-fonts",
-			"main.typ",
 			"--input",
 			"show_solutions=false",
+			"main.typ",
 			f"{out}/{handout_file}",
 		],
 		cwd=source_dir,
@@ -164,6 +166,8 @@ def build_typst(source_dir: Path, out_subdir: Path) -> IndexEntry | None:
 			[
 				TYPST_PATH,
 				"compile",
+				"--package-path",
+				f"{ROOT}/lib/typst",
 				"--ignore-system-fonts",
 				"main.typ",
 				f"{out}/{solutions_file}",