Geometry on the Putnam

A lattice triangle with no lattice points other than its vertices has area exactly $\tfrac12$; a lattice polygon with $5$ interior and $8$ boundary points has area $8$. The gcd fact counts visible points and proves that a lattice line through the origin with irrational slope hits no other lattice point. Minkowski's theorem (a convex symmetric region of area $> 4$ contains a nonzero lattice point) is the Putnam's favorite "existence of a lattice point" tool.

$|1 + i|^{10} = 32$ without expanding. Rotating a point $90^\circ$ about $c$: $c + i(z - c)$. Collinearity of $a, b, c$: $\tfrac{c - a}{b - a}$ is real. Triangle $abc$ is equilateral iff $a + \omega b + \omega^2c = 0$ for a primitive cube root of unity $\omega$. Roots-of-unity sums vanishing is why regular polygons have centroid at the center.

Distance from the origin to $x + y = 4$: $\tfrac{4}{\sqrt2} = 2\sqrt2$. A convex function of a point on a convex polygon is maximized at a vertex. Helly's theorem (if every three of a family of convex sets in the plane meet, all do) and Carathéodory (a point in the convex hull of a set in $\mathbb R^d$ is a combination of $d + 1$ points) are the two named convexity theorems worth knowing.

Sweep a line of the given direction across the region; the area on the left varies continuously from $0$ to the full area, so by the IVT it equals half somewhere. (Rotating the direction and applying IVT again gives the "ham sandwich" for two regions.)

Such a triangle $T$ with edge vectors $\mathbf u, \mathbf v$ has area $\tfrac12|\det[\mathbf u\,\mathbf v]|$; the parallelogram spanned by $\mathbf u, \mathbf v$ tiles the plane by lattice translates, and it contains exactly $|\det|$ lattice points per tile counted appropriately (the index of the sublattice). If the parallelogram — two copies of $T$ — contains no lattice points beyond its vertices, that index is $1$, so $|\det| = 1$ and the area is $\tfrac12$. General Pick follows by triangulating into primitive triangles and checking additivity of $I + \tfrac B2 - 1$. $\blacksquare$ Significance: the determinant-as-lattice-index idea is the seed of the geometry of numbers.

The normal $(a, b)$ is perpendicular to the line. Take any point $Q$ on the line; the distance is the length of the projection of $P - Q$ onto the unit normal: $\tfrac{|(a,b)\cdot(P - Q)|}{\sqrt{a^2+b^2}} = \tfrac{|ax_0 + by_0 - (aq_1 + bq_2)|}{\sqrt{a^2+b^2}}$, and $aq_1 + bq_2 = c$. $\blacksquare$ Significance: the identical projection argument gives point-to-plane distances in any dimension and the margin of a linear classifier.

Step 1 — for each $x$ from $-5$ to $5$, count $y$ with $y^2 \le 25 - x^2$: $x = 0$: $|y| \le 5$, $11$ points; $x = \pm1$: $y^2 \le 24$, $|y| \le 4$, $9$ each; $x = \pm2$: $y^2 \le 21$, $9$ each; $x = \pm3$: $y^2 \le 16$, $9$ each; $x = \pm4$: $y^2 \le 9$, $7$ each; $x = \pm5$: $1$ each.

Step 2 — total $11 + 2(9 + 9 + 9 + 7 + 1) = 11 + 70 = 81$.

Step 3 — Gauss's circle problem says the count is $\pi r^2 + O(r)$: $\pi\cdot25 \approx 78.5$, and $81$ is within the error band ✓.

Step 1 — vertices $a, b, c, d$ as complex numbers; midpoints $m_1 = \tfrac{a+b}{2}$, $m_2 = \tfrac{b+c}{2}$, $m_3 = \tfrac{c+d}{2}$, $m_4 = \tfrac{d+a}{2}$.

Step 2 — a quadrilateral $m_1m_2m_3m_4$ is a parallelogram iff $m_2 - m_1 = m_3 - m_4$ (opposite sides equal as vectors).

Step 3 — $m_2 - m_1 = \tfrac{c - a}{2}$ and $m_3 - m_4 = \tfrac{c - a}{2}$ ✓.

Step 4 — the proof works for skew quadrilaterals in 3D too (replace complex numbers by vectors): Varignon's theorem is really the statement that both midpoint-sides are half the diagonal $c - a$.