[proofplan] This is the two-dimensional case of [Minkowski's Theorem](/theorems/1608), proved here from scratch for convenience. The argument has two ingredients: (i) **Blichfeldt's lemma** — if a set $T$ has area strictly exceeding the area of a fundamental domain, then two points of $T$ differ by a non-zero lattice vector; (ii) a convexity-plus-symmetry trick to turn a difference of points in $T$ into a single point of $S$. We apply Blichfeldt to $T = \tfrac{1}{2} S$, which has area $\tfrac{1}{4}\operatorname{area}(S) \geq A(\Lambda)$; a limiting argument upgrades the strict inequality required by Blichfeldt to the non-strict $\geq$ in the hypothesis. The non-zero lattice point found in $S$ satisfies $|\alpha|^2 \leq r^2$ where $r^2 = A(S)/\pi$, which rearranges to $|\alpha|^2 \leq 4A(\Lambda)/\pi$. [/proofplan]

[step:State Blichfeldt's lemma for planar lattices and reduce the problem to a set of half size]Let $P = \{\lambda_1 v_1 + \lambda_2 v_2 : \lambda_1, \lambda_2 \in [0,1)\}$ be the (half-open) fundamental parallelogram for $\Lambda$. By definition, $\operatorname{area}(P) = A(\Lambda)$ and the translates $\{P + \gamma : \gamma \in \Lambda\}$ partition $\mathbb{R}^2$. **Blichfeldt's lemma (planar version).** *If $T \subseteq \mathbb{R}^2$ is a measurable set with $\operatorname{area}(T) > A(\Lambda)$, then there exist distinct $y, z \in T$ with $y - z \in \Lambda$.* Granting this, we apply it to $T = \tfrac{1}{2} S$. The map $x \mapsto \tfrac{1}{2}x$ is a linear automorphism of $\mathbb{R}^2$ with Jacobian determinant $\tfrac{1}{4}$, so \begin{align*} \operatorname{area}\!\left(\tfrac{1}{2} S\right) = \tfrac{1}{4} \operatorname{area}(S) \geq \tfrac{1}{4} \cdot 4 A(\Lambda) = A(\Lambda). \end{align*} This gives the non-strict inequality $\operatorname{area}(T) \geq A(\Lambda)$; Blichfeldt requires strict inequality. We handle the equality case by a compactness argument in Step 4 below.[/step]

[guided]The overall strategy is the standard "Minkowski trick": we want to find a non-zero $\alpha \in S \cap \Lambda$. Working with $S$ directly is hard because we have no hook to the lattice structure. The trick is to halve $S$ and use Blichfeldt's lemma — a purely measure-theoretic pigeonhole statement — to find two points of $\tfrac{1}{2} S$ differing by a lattice vector. Then convexity and symmetry of $S$ turn that difference into a point of $S$. **Setting up the fundamental parallelogram.** Since $v_1, v_2$ are $\mathbb{R}$-linearly independent, they span $\mathbb{R}^2$. The half-open parallelogram \begin{align*} P = \{\lambda_1 v_1 + \lambda_2 v_2 : \lambda_1, \lambda_2 \in [0, 1)\} \end{align*} has area $|\det(v_1, v_2)| = A(\Lambda)$ (the definition). The translates $P + \gamma$, $\gamma \in \Lambda$, partition $\mathbb{R}^2$: every point $x$ has a unique expression $x = \mu_1 v_1 + \mu_2 v_2$ with $\mu_i \in \mathbb{R}$; decomposing $\mu_i = \lfloor \mu_i \rfloor + \{\mu_i\}$ writes $x = \gamma + p$ with $\gamma = \lfloor \mu_1 \rfloor v_1 + \lfloor \mu_2 \rfloor v_2 \in \Lambda$ and $p \in P$, uniquely. **Blichfeldt's lemma.** States that any measurable $T$ with $\operatorname{area}(T) > A(\Lambda)$ contains two points $y \neq z$ with $y - z \in \Lambda$. We prove this in Step 2. Intuitively, $T$ is so large that when we wrap it around the torus $\mathbb{R}^2 / \Lambda$, it must overlap itself. **Why apply it to $\tfrac{1}{2} S$?** Directly applying Blichfeldt to $S$ would produce a lattice vector $y - z$ with $y, z \in S$, but we want a lattice vector *in* $S$ itself. The rescaling fixes this: if $y, z \in \tfrac{1}{2} S$, then $2y, -2z \in S$ (using symmetry $S = -S$), so by convexity their midpoint $\tfrac{1}{2}(2y + (-2z)) = y - z$ also lies in $S$. **Area calculation.** Dilation by $\tfrac{1}{2}$ scales areas by $(\tfrac{1}{2})^2 = \tfrac{1}{4}$: \begin{align*} \operatorname{area}(\tfrac{1}{2} S) = \tfrac{1}{4} \operatorname{area}(S) \geq \tfrac{1}{4} \cdot 4 A(\Lambda) = A(\Lambda). \end{align*} The hypothesis gives $\operatorname{area}(S) \geq 4 A(\Lambda)$ (non-strict), so we obtain $\operatorname{area}(\tfrac{1}{2} S) \geq A(\Lambda)$ (also non-strict). Blichfeldt wants strict inequality. We postpone this subtlety to Step 4, where a compactness argument on slightly larger dilates of $S$ handles the equality case.[/guided]

[step:Prove Blichfeldt's lemma via the non-injectivity of the folding map]Consider the folding map onto the fundamental parallelogram: \begin{align*} \pi: \mathbb{R}^2 &\to P \\ x &\mapsto x - \gamma(x), \quad \gamma(x) = \text{the unique } \gamma \in \Lambda \text{ with } x \in P + \gamma. \end{align*} The map $\pi$ is area-preserving: for any measurable $A \subseteq \mathbb{R}^2$ and any $\gamma \in \Lambda$, $\operatorname{area}(A \cap (P + \gamma)) = \operatorname{area}((A - \gamma) \cap P)$ by translation invariance of Lebesgue measure. Let $T \subseteq \mathbb{R}^2$ be measurable with $\operatorname{area}(T) > A(\Lambda)$. Decompose $T$ along the partition $\mathbb{R}^2 = \bigsqcup_{\gamma \in \Lambda} (P + \gamma)$: \begin{align*} \operatorname{area}(T) = \sum_{\gamma \in \Lambda} \operatorname{area}(T \cap (P + \gamma)) = \sum_{\gamma \in \Lambda} \operatorname{area}(\pi(T \cap (P + \gamma))), \end{align*} where the second equality uses that $\pi$ restricted to $P + \gamma$ is translation by $-\gamma$, which preserves area. Now suppose for contradiction that $\pi$ is injective on $T$, i.e., the sets $\pi(T \cap (P + \gamma))$ are pairwise disjoint for distinct $\gamma \in \Lambda$. Then \begin{align*} \sum_{\gamma \in \Lambda} \operatorname{area}(\pi(T \cap (P + \gamma))) = \operatorname{area}\!\left(\bigsqcup_{\gamma \in \Lambda} \pi(T \cap (P + \gamma))\right) \leq \operatorname{area}(P) = A(\Lambda), \end{align*} contradicting $\operatorname{area}(T) > A(\Lambda)$. Hence $\pi$ is not injective on $T$: there exist distinct $y, z \in T$ with $\pi(y) = \pi(z)$, equivalently \begin{align*} y - z = \gamma(y) - \gamma(z) \in \Lambda \setminus \{0\}. \end{align*}[/step]

[guided]Blichfeldt's lemma is the only place where the hypothesis $\operatorname{area}(T) > A(\Lambda)$ is used, so it sits at the heart of the Minkowski-type argument. The proof is a measure-theoretic pigeonhole: pack a large set $T$ into a small container (the torus $\mathbb{R}^2 / \Lambda$) and note that something must overlap. **The folding map $\pi$.** Since $\mathbb{R}^2 = \bigsqcup_{\gamma \in \Lambda} (P + \gamma)$ is a disjoint (partition) decomposition, every $x \in \mathbb{R}^2$ has a unique $\gamma(x) \in \Lambda$ with $x \in P + \gamma(x)$. Define $\pi(x) = x - \gamma(x) \in P$. In words, $\pi$ translates each $x$ back to $P$ by subtracting the unique lattice vector that places it there. **The folding map factors through the torus.** For $x, x' \in \mathbb{R}^2$, we have $\pi(x) = \pi(x')$ iff $x - x' = \gamma(x) - \gamma(x') \in \Lambda$. Thus $\pi$ identifies points differing by lattice vectors — it is precisely the quotient map $\mathbb{R}^2 \to \mathbb{R}^2 / \Lambda$ followed by the identification $\mathbb{R}^2 / \Lambda \cong P$. **Counting area via the fold.** Partition $T$ along the lattice translates: \begin{align*} T = \bigsqcup_{\gamma \in \Lambda} (T \cap (P + \gamma)), \qquad \operatorname{area}(T) = \sum_{\gamma \in \Lambda} \operatorname{area}(T \cap (P + \gamma)). \end{align*} On each piece $T \cap (P + \gamma)$, the folding map acts as translation by $-\gamma$: \begin{align*} \pi(T \cap (P + \gamma)) = (T \cap (P + \gamma)) - \gamma = (T - \gamma) \cap P. \end{align*} Translation preserves Lebesgue measure, so each piece has the same area as its image: \begin{align*} \operatorname{area}(\pi(T \cap (P + \gamma))) = \operatorname{area}(T \cap (P + \gamma)). \end{align*} **Pigeonhole.** Suppose $\pi|_T$ is injective. Then the images $\pi(T \cap (P + \gamma))$ are pairwise disjoint subsets of $P$, so their areas sum to at most $\operatorname{area}(P) = A(\Lambda)$: \begin{align*} \operatorname{area}(T) = \sum_{\gamma \in \Lambda} \operatorname{area}(\pi(T \cap (P + \gamma))) = \operatorname{area}\!\left(\bigsqcup_\gamma \pi(T \cap (P + \gamma))\right) \leq A(\Lambda). \end{align*} Contradiction, since we assumed $\operatorname{area}(T) > A(\Lambda)$. **Conclusion.** The map $\pi|_T$ is not injective: there exist distinct $y, z \in T$ with $\pi(y) = \pi(z)$. Unpacking, $y - z = \gamma(y) - \gamma(z) \in \Lambda$, and this lattice vector is non-zero because $y \neq z$.[/guided]

[step:Combine the two points of $\tfrac{1}{2} S$ into a single lattice point of $S$ via convexity and symmetry]For now assume the strict inequality $\operatorname{area}(S) > 4 A(\Lambda)$; we upgrade to the non-strict case $\geq$ in Step 4. Then $\operatorname{area}(\tfrac{1}{2} S) > A(\Lambda)$, and Blichfeldt's lemma (Step 2) applied to $T = \tfrac{1}{2} S$ produces distinct $y, z \in \tfrac{1}{2} S$ with \begin{align*} \alpha := y - z \in \Lambda \setminus \{0\}. \end{align*} We show $\alpha \in S$. Since $y \in \tfrac{1}{2} S$, we have $2y \in S$. Since $z \in \tfrac{1}{2} S$, we have $2z \in S$; by symmetry of $S$ about $0$ (i.e., $S = -S$, which holds because $S$ is a disc centred at $0$), also $-2z \in S$. By convexity of $S$ (discs are convex, being intersections of half-planes), the midpoint of $2y$ and $-2z$ lies in $S$: \begin{align*} \alpha = y - z = \tfrac{1}{2}\bigl(2y + (-2z)\bigr) \in S. \end{align*} Hence $\alpha \in S \cap \Lambda$ and $\alpha \neq 0$.[/step]

[guided]We now bring together the pieces: Blichfeldt gives two points of $\tfrac{1}{2} S$ differing by a lattice vector; we want the lattice vector itself to lie in $S$. This is where the hypotheses **convexity of $S$** and **symmetry of $S$ about $0$** enter, and they enter exactly here — nowhere else. **Setting up the strict-inequality case.** Temporarily strengthen the hypothesis to $\operatorname{area}(S) > 4 A(\Lambda)$ (the non-strict version of the hypothesis is handled in Step 4). Then \begin{align*} \operatorname{area}(\tfrac{1}{2} S) = \tfrac{1}{4} \operatorname{area}(S) > A(\Lambda), \end{align*} so Blichfeldt's lemma (Step 2) applies to $T = \tfrac{1}{2} S$ and yields distinct $y, z \in \tfrac{1}{2} S$ with $y - z \in \Lambda \setminus \{0\}$. Define $\alpha := y - z$. **Expressing $\alpha$ as a midpoint.** Note the algebraic identity \begin{align*} \alpha = y - z = \tfrac{1}{2}(2y) + \tfrac{1}{2}(-2z), \end{align*} which writes $\alpha$ as a convex combination of $2y$ and $-2z$ with weights $\tfrac{1}{2}, \tfrac{1}{2}$. **Showing the two endpoints lie in $S$.** - From $y \in \tfrac{1}{2} S$, by definition of the dilate there exists $s_1 \in S$ with $y = \tfrac{1}{2} s_1$, so $2y = s_1 \in S$. - From $z \in \tfrac{1}{2} S$, similarly $2z \in S$. Now apply symmetry of $S$ (the hypothesis that $S$ is a closed disc centred at $0$, and any such disc is symmetric: $s \in S \iff -s \in S$). This gives $-2z \in S$. **Applying convexity.** A set $S \subseteq \mathbb{R}^2$ is **convex** if for every $p, q \in S$ and every $t \in [0, 1]$, $tp + (1-t)q \in S$. Closed discs are convex: they are the intersection of all half-planes containing them, and each half-plane is convex. Applying convexity with $p = 2y \in S$, $q = -2z \in S$, $t = \tfrac{1}{2}$: \begin{align*} \alpha = \tfrac{1}{2}(2y) + \tfrac{1}{2}(-2z) \in S. \end{align*} **Conclusion.** $\alpha \in S \cap \Lambda$ and $\alpha \neq 0$. So $S$ contains a non-zero lattice point, under the temporarily strengthened hypothesis $\operatorname{area}(S) > 4 A(\Lambda)$.[/guided]

[step:Extend from strict to non-strict inequality via a compactness argument on dilates]Now consider the original hypothesis: $S$ is a closed disc centred at $0$ with $\operatorname{area}(S) \geq 4 A(\Lambda)$. For each integer $m \geq 1$, set \begin{align*} S_m := \left(1 + \tfrac{1}{m}\right) S, \end{align*} a closed disc centred at $0$ with $\operatorname{area}(S_m) = (1 + 1/m)^2 \operatorname{area}(S) > \operatorname{area}(S) \geq 4 A(\Lambda)$. By Step 3 applied to $S_m$, there exists a non-zero $\alpha_m \in S_m \cap \Lambda$. All $\alpha_m$ lie in $2S$ (since $1 + 1/m \leq 2$ for $m \geq 1$), which is a closed disc, hence bounded. The intersection $2S \cap \Lambda$ is **finite**: $\Lambda$ is discrete (every lattice in $\mathbb{R}^2$ is a discrete subgroup), and the intersection of a discrete set with a bounded set is finite. Hence some non-zero $\alpha \in \Lambda$ equals $\alpha_m$ for infinitely many $m$. For each such $m$, $\alpha \in S_m$, so \begin{align*} \alpha \in \bigcap_{m \geq 1} S_m. \end{align*} We claim $\bigcap_{m \geq 1} S_m = S$. One inclusion is immediate: $S \subseteq S_m$ for every $m$ (since $1 + 1/m \geq 1$), so $S \subseteq \bigcap_m S_m$. For the reverse inclusion, let $S = \overline{B}(0, r)$. Then $S_m = \overline{B}(0, r(1 + 1/m))$, and $x \in \bigcap_m S_m$ means $|x| \leq r(1 + 1/m)$ for every $m \geq 1$; letting $m \to \infty$ gives $|x| \leq r$, i.e., $x \in S$. Therefore $\alpha \in S$, and $\alpha \in \Lambda \setminus \{0\}$ as desired.[/step]

[guided]Step 3 produced a non-zero $\alpha \in S \cap \Lambda$ **assuming the strict inequality** $\operatorname{area}(S) > 4 A(\Lambda)$. The theorem hypothesis gives only $\geq$, so we need to close this gap. The standard trick — available whenever $S$ is closed — is a compactness argument on slightly larger dilates. **The dilate sequence $S_m$.** For $m \in \mathbb{Z}_{\geq 1}$, let $S_m = (1 + 1/m) S$. This is again a closed disc centred at $0$ (dilation by a positive scalar commutes with being a closed disc centred at the origin). Its area is \begin{align*} \operatorname{area}(S_m) = (1 + 1/m)^2 \operatorname{area}(S) \geq (1 + 1/m)^2 \cdot 4 A(\Lambda) > 4 A(\Lambda), \end{align*} where the last inequality is strict because $(1 + 1/m)^2 > 1$. So $S_m$ satisfies the strict-inequality hypothesis of Step 3, which yields a non-zero $\alpha_m \in S_m \cap \Lambda$. **Confining the $\alpha_m$ to a bounded set.** For $m \geq 1$, $1 + 1/m \leq 2$, so $S_m \subseteq 2S$. Hence $\alpha_m \in 2S$ for all $m$. **Finiteness of $2S \cap \Lambda$.** The set $2S$ is a closed disc (bounded). The lattice $\Lambda$ is a discrete subgroup of $\mathbb{R}^2$: by assumption $\Lambda = \mathbb{Z}v_1 + \mathbb{Z}v_2$ with $v_1, v_2$ $\mathbb{R}$-linearly independent, and any two elements $\gamma, \gamma' \in \Lambda$ differ by a non-zero lattice vector $\gamma - \gamma'$ whose length is bounded below by $\min(|v_1|, |v_2|) / C > 0$ for some geometric constant $C$ (a consequence of $v_1, v_2$ being linearly independent, hence no direction gives arbitrarily short combinations other than $0$). More directly, $\Lambda \cong \mathbb{Z}^2$ as a free abelian group of rank $2$, and this isomorphism is a bi-Lipschitz homeomorphism onto its image, so $\Lambda$ is discrete in $\mathbb{R}^2$. The intersection of a discrete subset with a compact set (closed + bounded) is finite. Hence $2S \cap \Lambda$ is a finite set. **Pigeonhole on $m$.** The infinite sequence $(\alpha_m)_{m \geq 1}$ takes values in the finite set $2S \cap \Lambda$. By pigeonhole, some value $\alpha \in 2S \cap \Lambda$ occurs infinitely often: there is an infinite subset $M \subseteq \mathbb{Z}_{\geq 1}$ with $\alpha_m = \alpha$ for all $m \in M$. For each $m \in M$, $\alpha \in S_m$. **Intersecting the dilates.** We claim \begin{align*} S = \bigcap_{m \geq 1} S_m. \end{align*} The $\subseteq$ direction: for any $m$, $S \subseteq (1 + 1/m) S = S_m$ since the origin is in $S$ (the disc is centred at $0$) and dilation by a factor $\geq 1$ maps $S$ into itself (for sets containing $0$ and star-shaped about $0$, which a disc is). The $\supseteq$ direction: writing $S = \overline{B}(0, r)$, a point $x \in \bigcap_m S_m$ satisfies $|x| \leq r(1 + 1/m)$ for every $m$; taking $m \to \infty$ gives $|x| \leq r$ by the squeeze principle on real numbers, hence $x \in S$. Since $\alpha \in S_m$ for infinitely many $m$ — hence, because the $S_m$ are nested ($S_m \supseteq S_{m+1}$), for **all** $m \geq 1$ — we have $\alpha \in \bigcap_m S_m = S$. Together with $\alpha \in \Lambda \setminus \{0\}$, this proves the first claim of the theorem. (Technical remark on nestedness: for $m < m'$, $1 + 1/m > 1 + 1/m'$, so $S_m = (1 + 1/m)S \supseteq (1 + 1/m')S = S_{m'}$ since $S$ is star-shaped about $0$. Hence if $\alpha \in S_{m_0}$ for some $m_0$ and $\alpha \in S_{m'}$ for some $m' > m_0$, and more, $\alpha \in S_m$ for all $m \leq m_0$ by nestedness; combined with infinitely many witnesses ensures the intersection is over all $m$.)[/guided]

[step:Derive the quantitative bound $|\alpha|^2 \leq 4 A(\Lambda)/\pi$ by optimizing over disc size]It remains to prove the second statement: there exists $\alpha \in \Lambda \setminus \{0\}$ with \begin{align*} |\alpha|^2 \leq \frac{4 A(\Lambda)}{\pi}. \end{align*} Choose $r > 0$ so that the closed disc $S = \overline{B}(0, r)$ satisfies the area hypothesis with equality: \begin{align*} \operatorname{area}(S) = \pi r^2 = 4 A(\Lambda) \quad \iff \quad r^2 = \frac{4 A(\Lambda)}{\pi}. \end{align*} By the first statement (just established in Step 4), $S$ contains a non-zero $\alpha \in \Lambda$. Since $\alpha \in \overline{B}(0, r)$, $|\alpha| \leq r$, hence \begin{align*} |\alpha|^2 \leq r^2 = \frac{4 A(\Lambda)}{\pi}. \end{align*} This gives the desired $\alpha$ and completes the proof.[/step]

[guided]The quantitative bound comes from **applying the first statement to the smallest disc permitted by the area hypothesis**. Any disc $S$ with $\operatorname{area}(S) \geq 4 A(\Lambda)$ works for the first statement, and smaller discs give tighter bounds on $|\alpha|$. **Choosing the disc.** We want $\operatorname{area}(S) = 4 A(\Lambda)$ exactly — any smaller and the first statement might fail; any larger and we waste room. A closed disc of radius $r$ has area $\pi r^2$, so we set \begin{align*} \pi r^2 = 4 A(\Lambda), \qquad r = \sqrt{\frac{4 A(\Lambda)}{\pi}} = 2 \sqrt{\frac{A(\Lambda)}{\pi}}. \end{align*} Let $S = \overline{B}(0, r)$ for this specific $r$. **Applying the first statement.** We have just proved (Steps 1-4) that any closed disc $S$ centred at $0$ with $\operatorname{area}(S) \geq 4 A(\Lambda)$ contains a non-zero $\alpha \in \Lambda$. Here $\operatorname{area}(S) = 4 A(\Lambda)$ satisfies the hypothesis with equality, so there exists $\alpha \in S \cap \Lambda$, $\alpha \neq 0$. **Reading off the bound.** By definition $\alpha \in \overline{B}(0, r)$ means $|\alpha| \leq r$, hence \begin{align*} |\alpha|^2 \leq r^2 = \frac{4 A(\Lambda)}{\pi}, \end{align*} which is the desired inequality. Squaring both sides is valid because $|\alpha| \geq 0$ and $r \geq 0$, so the squaring function is monotone on the non-negative reals. **Sanity check on the constant.** The constant $4/\pi \approx 1.273$ is the area of a square of side $2$ divided by the area of the inscribed disc of radius $1$; equivalently, it is the area of the smallest disc covering the fundamental parallelogram $[0,1]^2$ divided by the area of that parallelogram. This matches the geometric intuition that circumscribing a parallelogram by a disc introduces a factor of roughly $4/\pi$.[/guided]