Erdős Problem #352

Erdős Problem #352: an exact unit-area triangle

1. Statement and proof scope

2. Unbounded sets of positive measure: a complete proof

The map $v\mapsto v/|v|^2$ has derivative of operator norm $|v|^{-2}$. Consequently every singular value of $DT(x)$ is at least $1-2/(D-r)^2>1/2$, and \[ |\det DT(x)|\ge1/4. \tag{2} \] The map is also injective on $B_r$. Indeed, if $v=\rho e^{i\theta}$ in complex coordinates, then \[ T(x)-z=v(1+2i/\rho^2),\qquad |T(x)-z|=\sqrt{\rho^2+4/\rho^2}. \] The last expression is strictly increasing for $\rho>\sqrt2$. Its value determines $\rho$, after which the argument determines $\theta$. Here every $\rho=|x-z|$ exceeds $\sqrt2$. The inverse function theorem and injectivity justify the usual change-of-variables formula on this image.

By $1$--$2$, \[ \big|\{x\in B_r:T(x)\notin A\}\big| \le4|B_{2r}\setminus A|<\pi r^2/4. \] Also $|B_r\setminus A|<\pi r^2/16$. These two exceptional sets do not cover $B_r$. Choose $x$ outside both. Then $z,x,T(x)\in A$ give the required triangle. All three vertices are distinct by the nonzero determinant. This also proves the infinite-measure case, since a bounded planar set has finite measure.

3. An elementary connected-set bound

4. The sharp-constant obstruction from a disk

Disks of radius less than $2\,3^{-3/4}$ therefore contain no unit-area triangle, while their areas approach \[ \frac{4\pi}{\sqrt{27}}. \] Any universal threshold $c$ in the question must be at least this value. The connected-set proof above gives a weaker sufficient bound and does not establish the conjectured sharp one.

5. Assessment and references

Source: https://www.erdosproblems.com/352, checked 2026-09-25, which attributes the unbounded positive-measure result to Erdős and identifies the density-theorem method. The earlier GPT Pro 5.2 and Claude Opus 4.5 files were reviewed. External standard inputs used here are the Lebesgue density theorem, change of variables for injective smooth maps, and preservation of connectedness by products and continuous maps; the geometric deductions are proved above. No novelty or formal verification is claimed.

COMPLETION ESTIMATE

1) FORMAL RESTATEMENT

Let $\mu$ denote Lebesgue measure on $\mathbb R^2$. For three points $x_1,x_2,x_3\in\mathbb R^2$, define the (unsigned) Euclidean area of the triangle with these vertices by \[ \operatorname{Area}(x_1,x_2,x_3);=;\frac12,\left|\det(x_2-x_1,;x_3-x_1)\right|, \] where ($\det((a,b),(c,d))$:=ad-bc).

The Erdős problem asks:

Does there exist a constant $c>0$ such that for every Lebesgue measurable set $A\subseteq\mathbb R^2$ with $\mu(A)\ge c$, there exist $x_1,x_2,x_3\in A$ with $\operatorname{Area}(x_1,x_2,x_3)=1$?

Stress points / edge cases:

* If $\mu(A)=\infty$ the statement is known to be true (per Erdős; see context below). * The nontrivial regime is $0<\mu(A)<\infty$, and (as the examples show) (A) must be allowed to have empty interior and be highly disconnected. * A necessary lower bound for any such (c) is \[ c_0=\frac{4\pi}{\sqrt{27}}\approx 2.418, \] witnessed by a disk of slightly smaller area (details proved in §4).

2) QUICK LITERATURE/CONTEXT CHECK (web browsing: YES)

* The problem is listed as open on the Erdős Problems site. * Mauldin’s 2013 expository paper restates it and records (i) the conjectured best constant $c_0=4\pi/(3\sqrt3)=4\pi/\sqrt{27}$, (ii) a reduction to unions of finitely many compact convex sets, and (iii) verification of the conjecture for unions of $n\le 3$ compact convex sets (per Freiling–Mauldin). * Kovač’s 2024 arXiv survey also cites the problem as open in general and mentions Erdős’s easy observation for infinite area / unbounded positive area sets (and notes the right-triangle variant is open). * There are negative results for variants where “measurable” is relaxed to arbitrary sets with large outer Lebesgue measure: Pantsulaia–Rusiashvili claim that no finite constant works for that outer-measure version. * A MathOverflow thread discusses outer-measure/full-outer-measure triangle-free sets via transfinite induction (again outside the Lebesgue-measurable regime).

So: as of the latest accessible sources here $2024$, the measurable-set version remains open.

3) ATTACK PLAN

Proof-track strategies

1. Local translation/Steinhaus approach: show that any measurable (A) with $0<\mu(A)<\infty$ contains triangles of all sufficiently small areas (an interval $(0,\varepsilon)$). This can be proved rigorously via $L^1$-continuity of translations and triple intersections. 2. Combine with a large-area existence theorem (e.g., Freiling–Mauldin: outer measure $>c_0\Rightarrow$ triangle area $>1$) and try to force exactly area $1$ via an “intermediate value” mechanism. The missing piece is a connectedness/continuity principle for the set of attainable areas in an arbitrary measurable set. 3. More analytic: study the 3-point correlation \[ J(u,v)=\mu\left(A\cap(A-u)\cap(A-v)\right)=\int 1_A(x)1_A(x+u)1_A(x+v),dx, \] and try to show existence of ((u,v)) with $\det(u,v)=2$ and $J(u,v)>0$ under $\mu(A)\ge c$.

Disproof/construction strategies

1. Build bounded measurable sets of arbitrarily large area avoiding area $1$ triangles by arranging many components (disks/rectangles) with carefully controlled inter-component triangle-area ranges. 2. Use “arithmetic” constructions (lattice-like sets) to force triangle areas into a discrete set missing $1$, then thicken while keeping the exclusion.

Best current path: I can fully execute strategy $1$ on the proof-track: prove that every positive finite measure set contains triangles of every sufficiently small area. I can also rigorously establish the lower bound $c\ge c_0$. However, the remaining global step “large measure forces area $1$” is exactly what the literature flags as open. I do not see a way to complete a gap-free proof or a genuine measurable counterexample here.

4) WORK

Lemma 0 (Sharp geometric lower bound $c\ge c_0$)

Let $B_R\subset\mathbb R^2$ be the closed disk of radius (R) centered at the origin. Then the maximum area of any triangle with vertices in $B_R$ is \[ \max{\operatorname{Area}(x_1,x_2,x_3): x_i\in B_R}=\frac{3\sqrt3}{4},R^2, \] achieved by an equilateral triangle on the boundary circle.

Proof. Take any triangle with vertices on (or inside) the circle of radius (R). Enlarge it radially (if necessary) so its vertices lie on the circle; this can only increase its area, so the maximum occurs among inscribed triangles.

For an inscribed triangle with circumradius (R) and angles (A,B,C) $(A+B+C=\pi)$, one standard area formula is \[ \operatorname{Area} = 2R^2,\sin A,\sin B,\sin C. \] (Indeed, side lengths are $a=2R\sin A$, $b=2R\sin B$, $c=2R\sin C$, and ($\operatorname{Area}$=$\frac12$ ab$\sin$ C = 2R^2$\sin$ A$\sin$ B$\sin$ C).)

So maximizing area for fixed (R) is equivalent to maximizing $\sin A\sin B\sin C$ subject to $A,B,C>0$, $A+B+C=\pi$. The function $\log\sin\theta$ is strictly concave on $(0,\pi)$, so by Jensen, \[ \frac{1}{3}\left(\log\sin A+\log\sin B+\log\sin C\right)\le \log\sin\left(\frac{A+B+C}{3}\right)=\log\sin\left(\frac{\pi}{3}\right), \] hence $\sin A\sin B\sin C \le \sin^3(\pi/3) = (\sqrt3/2)^3 = 3\sqrt3/8$, with equality only when $A=B=C=\pi/3$, i.e. the triangle is equilateral.

Therefore the maximum area is \[ 2R^2\cdot \frac{3\sqrt3}{8}=\frac{3\sqrt3}{4}R^2, \] as claimed. ∎

Now choose $R_0$ so that this maximal triangle area equals $1$: \[ \frac{3\sqrt3}{4}R_0^2=1 \quad\Longrightarrow\quad R_0^2=\frac{4}{3\sqrt3}=\frac{4}{\sqrt{27}}, \quad R_0=\frac{2}{3^{3/4}}. \] Then the disk $B_{R}$ with any $R<R_0$ contains no triangle of area $1$, while \[ \mu(B_{R_0})=\pi R_0^2=\pi\cdot\frac{4}{3\sqrt3}=\frac{4\pi}{\sqrt{27}}=:c_0. \] Hence any universal constant (c) (if it exists) must satisfy $c\ge c_0$.

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Lemma 1 (Triple-intersection continuity near $(0,0)$)

Let $A\subset\mathbb R^2$ be Lebesgue measurable with $0<\mu(A)<\infty$. Define, for $u,v\in\mathbb R^2$, \[ J(u,v):=\mu\left(A\cap(A-u)\cap(A-v)\right) =\int_{\mathbb R^2} 1_A(x),1_A(x+u),1_A(x+v),dx. \] Then $J(u,v)\to \mu(A)$ as $(u,v)\to(0,0)$. In particular, there exists $\rho>0$ such that \[ J(u,v)>0\quad\text{for all }u,v\text{ with }|u|<\rho,\ |v|<\rho. \]

Proof. Let $f:=1_A\in L^1(\mathbb R^2)$ (since $\mu(A)<\infty$). For $u\in\mathbb R^2$, let $T_u f(x):=f(x+u)$.

A standard fact (and easy to prove by approximation of (f) by continuous compactly supported functions in $L^1$) is:

$|T_u f - f|_{L^1}\to 0$ as $u\to 0$.

Now compute, using that $f\in{0,1}$, \[ \mu(A)-J(u,v) =\int f(x),dx-\int f(x)f(x+u)f(x+v),dx =\int f(x)\left(1-f(x+u)f(x+v)\right),dx. \] For $a,b\in[0,1]$, (1-ab$\le$ $1-a$+$1-b$). Applying this pointwise with $a=f(x+u)$, $b=f(x+v)$, we get \[ 1-f(x+u)f(x+v)\le (1-f(x+u))+(1-f(x+v)). \] Therefore \[ \mu(A)-J(u,v) \le \int f(x)\left(1-f(x+u)\right),dx+\int f(x)\left(1-f(x+v)\right),dx. \] But \[ \int f(x)\left(1-f(x+u)\right),dx =\mu(A)-\mu\left(A\cap(A-u)\right). \] Since $\mu(A)=\mu(A-u)$, the symmetric difference identity gives \[ \mu(A\triangle(A-u))=2\left(\mu(A)-\mu(A\cap(A-u))\right), \] hence \[ \mu(A)-\mu(A\cap(A-u))=\frac12,\mu(A\triangle(A-u))=\frac12|f-T_{-u}f|*{L^1}. \] Similarly for (v). Thus \[ 0\le \mu(A)-J(u,v)\le \frac12|f-T*{-u}f|*{L^1}+\frac12|f-T*{-v}f|_{L^1}. \] As $u\to0$ and $v\to0$, both $L^1$-norms tend to $0$. Hence $J(u,v)\to\mu(A)$.

Since $\mu(A)>0$, choose $\rho>0$ so small that whenever $|u|<\rho$ and $|v|<\rho$, \[ \frac12|f-T_{-u}f|*{L^1}+\frac12|f-T*{-v}f|_{L^1}<\frac{\mu(A)}{2}, \] which forces $J(u,v)>\mu(A)/2>0$. ∎

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Corollary 2 (Every positive finite-measure set contains triangles of all sufficiently small areas)

Let $A\subset\mathbb R^2$ be measurable with $0<\mu(A)<\infty$. Then there exists $\varepsilon=\varepsilon(A)>0$ such that for every real (a) with $0<a<\varepsilon$, there exist $x_1,x_2,x_3\in A$ with $\operatorname{Area}(x_1,x_2,x_3)=a$.

Proof. Let $\rho>0$ be as in Lemma 1: $J(u,v)>0$ for all $|u|,|v|<\rho$.

Set \[ \varepsilon:=\frac{\rho^2}{8}. \] Fix any $a\in(0,\varepsilon)$. Define \[ u:=\left(\frac{\rho}{2},,0\right),\qquad v:=\left(0,,\frac{4a}{\rho}\right). \] Then $|u|=\rho/2<\rho$. Also, since $a<\rho^2/8$, \[ |v|=\frac{4a}{\rho}<\frac{4(\rho^2/8)}{\rho}=\frac{\rho}{2}<\rho. \] Therefore $J(u,v)>0$, so $A\cap(A-u)\cap(A-v)\neq\varnothing$. Choose (x) in this intersection. Then \[ x,\quad x+u,\quad x+v\in A. \] Their triangle area is \[ \operatorname{Area}(x,x+u,x+v)=\frac12|\det(u,v)|. \] But $\det(u,v)=(\rho/2)\cdot(4a/\rho)-0\cdot 0=2a$. Hence the area is $\frac12|2a|=a$. ∎

What this gives: It completely settles a “small-area” analogue: every measurable set of positive finite area contains triangles of every sufficiently small area.

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Where this falls short of the Erdős question

Freiling–Mauldin show that if (A) has outer measure $>c_0$, then (A) contains a triangle with area $>1$. If (A) is compact convex, that already implies the existence of an area-$1$ triangle (shrink a $>1$ triangle inside the convex set continuously down to area $1$).

But for general measurable sets, the missing step is precisely:

* from “there exists some triangle in (A) of area $>1$” plus “(A) contains triangles of all sufficiently small areas” * to “(A) contains a triangle of area exactly $1$”.

That step would follow, for example, if one could prove an intermediate value property for triangle areas in arbitrary measurable sets (something like: the set of attainable triangle areas is connected, or contains an interval ($0,M]$ up to its supremum). No such general theorem is currently established in the sources above, and I do not see a gap-free way to prove it here.

5) VERIFICATION

* Lemma 0 checks: The maximization on the circle used only a standard and fully justified formula $\text{Area}=2R^2\sin A\sin B\sin C$ and concavity of $\log\sin$. No hidden assumptions. The computed $R_0^2=4/(3\sqrt3)=4/\sqrt{27}$ gives disk area $c_0=4\pi/\sqrt{27}$, consistent with the stated conjectured optimum. * Lemma 1 checks:

* Requires $f=1_A\in L^1$, hence $\mu(A)<\infty$. Correct. * Uses $L^1$-continuity of translations for $L^1$ functions; standard and can be proved by approximation with $C_c$ functions. * The inequality (1-ab$\le$ $1-a$+$1-b$) for $a,b\in[0,1]$ is correct. * The relation $\mu(A)-\mu(A\cap(A-u)) = \frac12\mu(A\triangle(A-u))$ uses $\mu(A)=\mu(A-u)$, true by translation invariance. Correct. * Uniformity: once $\rho$ is chosen from the translation-continuity estimate, the bound holds simultaneously for all $|u|,|v|<\rho$. Correct. * Corollary 2 checks: The chosen (u,v) satisfy $|u|,|v|<\rho$ whenever $a<\rho^2/8$. Determinant computation gives exactly $2a$. Hence area exactly (a). Correct. * Quantifiers: Corollary 2 produces $\varepsilon(A)>0$ depending on (A), not a universal $\varepsilon$. That dependence is unavoidable in the proof.

6) FINAL

UNRESOLVED

(i) Strongest fully proved partial result obtained here: For every Lebesgue measurable $A\subset\mathbb R^2$ with $0<\mu(A)<\infty$, there exists $\varepsilon(A)>0$ such that (A) contains triangles of every area $a\in(0,\varepsilon(A))$ (Corollary 2). Also, any universal constant (c) (if it exists) must satisfy $c\ge c_0=4\pi/\sqrt{27}$ (Lemma 0).

(ii) Exact first gap blocking a full solution: One needs a rigorous principle that forces exact area $1$ once the set has “enough area.” A clean crystallization is:

(Gap statement) For measurable $A\subset\mathbb R^2$, if $\sup{\operatorname{Area}(x_1,x_2,x_3):x_i\in A}>1$, must $1$ belong to the set of attained triangle areas?

This implication holds for convex (A) (by convex interpolation), but is not proved for arbitrary measurable (A).

(iii) Top 3 next moves (concrete):

1. Prove a connectedness/intermediate value theorem for the set $\mathcal T(A):={\operatorname{Area}(x_1,x_2,x_3):x_i\in A}\subset[0,\infty$), e.g. that $(0,\varepsilon(A))\subset\mathcal T(A)$ extends to an interval up to $\sup\mathcal T(A)$ under mild hypotheses satisfied when $\mu(A)$ is large. 2. Strengthen Freiling–Mauldin’s “$\exists$ triangle of area $>1$” into a robust family of $>1$-area triangles with vertices at density points, enabling a measurable/topological deformation argument that forces an exact level set at area $1$. 3. On the counterexample side, attempt an explicit bounded measurable construction where all triangles have areas in $(0,1-\delta)\cup(1+\delta,\infty)$ for some $\delta>0$, likely via a highly disconnected union of pieces with carefully controlled two-point and three-point geometry.

(iv) What a minimal counterexample would likely look like (if the answer is negative): A bounded measurable set (A) with arbitrarily large $\mu(A)$, extremely disconnected (many separated “islands”), arranged so that:

* within islands, attainable triangle areas stay in $(0,1-\delta)$, * between islands, attainable triangle areas jump over $1$ (landing $\ge 1+\delta$), * and crucially, the “two vertices in one island + third vertex far away” mechanism never allows tuning the area through $1$. This last requirement seems to force strong directional/thickness constraints across islands, which is a major obstacle to building such examples.

If you want, I can also write up (fully formally) the standard implication “outer measure $>c_0\Rightarrow$ triangle area $>1\Rightarrow$ area $1$” for compact convex (A), but that does not resolve the general measurable case that Erdős asked about.

Definitions and notation

* $\mu$ denotes Lebesgue measure on $\mathbb R^2$. * $\mu^*(E)$ denotes Lebesgue outer measure: \[ \mu^*(E)\ :=\ \inf{\mu(U): U\subseteq\mathbb R^2\ \text{open and }E\subseteq U}. \] * For $x_1,x_2,x_3\in\mathbb R^2$, define the (unsigned) area of the triangle with these vertices by \[ \operatorname{Area}(x_1,x_2,x_3)\ :=\ \frac12\left|\det(x_2-x_1,;x_3-x_1)\right|. \] * A set $K\subseteq\mathbb R^2$ is convex if for all $p,q\in K$ and all $t\in[0,1]$, ($1-t$p+tq$\in$ K). * A set is compact convex if it is compact and convex.

Finally, define the constant \[ c_0\ :=\ \frac{4\pi}{\sqrt{27}}=\frac{4\pi}{3\sqrt3}. \]

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Step 1. The Freiling–Mauldin “$>1$-area triangle” theorem (used as input)

Freiling–Mauldin proved:

(FM) If $A\subseteq\mathbb R^2$ satisfies $\mu^*(A) > c_0$, then there exist $p,q,r\in A$ such that $\operatorname{Area}(p,q,r)>1$. ([erdosproblems.com][1])

I will not re-prove (FM) here (it is the nontrivial geometric-measure-theoretic content). The rest of the implication—getting exactly area $1$ from area $>1$ under convexity—is elementary but must be written carefully.

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Step 2. Convex “area interpolation” lemma (fully proved)

Lemma (Scaling a triangle inside a convex set)

Let $K\subseteq\mathbb R^2$ be convex. Suppose $p,q,r\in K$ form a nondegenerate triangle, and set \[ A_0 := \operatorname{Area}(p,q,r) > 0. \] For each $t\in[0,1]$, define \[ q_t := (1-t)p+tq,\qquad r_t := (1-t)p+tr. \] Then:

1. $p,q_t,r_t\in K$ for every $t\in[0,1]$; 2. $\operatorname{Area}(p,q_t,r_t) = t^2,A_0$ for every $t\in[0,1]$.

Proof.

1. Because (K) is convex and $p,q\in K$, every convex combination ($1-t$p+tq) lies in (K). Hence $q_t\in K$. Similarly $r_t\in K$. Also $p\in K$ by assumption.

2. Compute the area by determinants: \[ \operatorname{Area}(p,q_t,r_t) =\frac12\left|\det(q_t-p,;r_t-p)\right|. \] Now \[ q_t-p = ((1-t)p+tq)-p = t(q-p), \qquad r_t-p = ((1-t)p+tr)-p = t(r-p). \] Using bilinearity of the determinant in its two vector arguments, \[ \det(q_t-p,;r_t-p)=\det(t(q-p),;t(r-p)) = t^2\det(q-p,;r-p). \] Therefore \[ \operatorname{Area}(p,q_t,r_t) =\frac12\left|t^2\det(q-p,;r-p)\right| =t^2\cdot \frac12\left|\det(q-p,;r-p)\right| =t^2,\operatorname{Area}(p,q,r) =t^2A_0. \] This proves $2$. ∎

Corollary (Exact area $1$ inside a convex set once you have $>1$)

Let $K\subseteq\mathbb R^2$ be convex. If there exist $p,q,r\in K$ with $\operatorname{Area}(p,q,r)>1$, then there exist $p',q',r'\in K$ with $\operatorname{Area}(p',q',r')=1$.

Proof. Let $A_0=\operatorname{Area}(p,q,r)>1$. Define $t:=A_0^{-1/2}$. Then $t\in(0,1)$ and $t^2A_0=1$.

Apply the lemma: $p,q_t,r_t\in K$ and $\operatorname{Area}(p,q_t,r_t)=t^2A_0=1$. Let $p':=p$, $q':=q_t$, $r':=r_t$. ∎

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Step 3. Putting it together for compact convex sets

Proposition (Compact convex case at the $c_0$ threshold)

Let $K\subseteq\mathbb R^2$ be compact convex. If $\mu^*(K)>c_0$, then (K) contains the vertices of a triangle of area exactly $1$.

Proof. Assume $\mu^*(K)>c_0$. By (FM), there exist $p,q,r\in K$ with $\operatorname{Area}(p,q,r)>1$. ([erdosproblems.com][1])

Since (K) is convex, the corollary above applies: from one triangle of area $>1$ in (K), we can produce (by scaling toward one vertex along line segments that remain in (K)) a triangle with area exactly $1$ whose vertices are still in (K).

Thus (K) contains the vertices of a triangle of area $1$. ∎

Remark (Replacing outer measure by measure)

Every compact set (K) is Lebesgue measurable, and for Lebesgue measurable sets one has $\mu(K)=\mu^*(K)$ (this is part of the Carathéodory construction of Lebesgue measure as the restriction of outer measure to measurable sets). Therefore, for compact (K), \[ \mu(K)>c_0 \quad\Longleftrightarrow\quad \mu^*(K)>c_0, \] so the proposition can be stated as: if a compact convex (K) has Lebesgue measure $>c_0$, then (K) contains an area-$1$ triangle.

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That’s the complete “standard implication” chain:

\[ \boxed{\mu^*(K)>c_0 \ \xRightarrow{\ \text{Freiling–Mauldin}\ }\ \exists\triangle\subset K \text{ of area }>1\ \xRightarrow{\ \text{convex scaling}\ }\ \exists\triangle\subset K \text{ of area }1.} \]

(With the only non-elementary input being the Freiling–Mauldin theorem.)

[1]: https://www.erdosproblems.com/352 " Erdős Problem #352 "

OUTPUT 1) FORMAL RESTATEMENT Let $A\subseteq\mathbb{R}^2$ be Lebesgue measurable and let $\lambda_2(A)$ be its area. The question asks whether there exists a constant $c>0$ such that \[ \forall A\subseteq\mathbb{R}^2 \text{ measurable},\ \lambda_2(A)\ge c\ \Rightarrow\ \exists x,y,z\in A\ \text{with}\ \text{Area}(\triangle xyz)=1. \] We interpret "triangle of area 1" as a non-degenerate triangle with exact area 1.

2) QUICK LITERATURE/CONTEXT CHECK Browsing is not available. I only record what the problem statement itself claims: - Erdős proved the statement for infinite-measure sets and for unbounded positive-measure sets. - The conjectured sharp threshold is $c=4\pi/\sqrt{27}$, motivated by disks of radius $<2\cdot 3^{-3/4}$. - Freiling--Mauldin proved an outer-measure $>4\pi/\sqrt{27}$ result for area $>1$ triangles, and proved the conjecture for compact convex sets and for unions of $\le 3$ such sets. These claims are not re-proved here.

3) ATTACK PLAN Proof strategies: 1. Reduce to structured sets (finite unions of convex bodies) and use continuity of area in a sliding-vertex argument. 2. Use density/translation arguments (Steinhaus-type) to pass from measure to geometric patterns. 3. Explore extremal configurations: disks appear to be the barrier, so show disk-threshold suffices.

Disproof strategies: 1. Construct a measurable set of area just above $4\pi/\sqrt{27}$ with no area-1 triangles. 2. Attempt fractal or highly disconnected sets to evade geometric continuity arguments.

Chosen path: Prove elementary geometric lemmas about disks and triangle areas (a rigorous lower bound for sets containing a disk) and isolate the first gap to the full measurable case.

4) WORK Lemma 1 (Maximum triangle area in a disk). Let $B_R$ be a closed disk of radius $R$. Any triangle with vertices in $B_R$ has area at most \[\frac{3\sqrt{3}}{4}R^2.\]

Proof. Any triangle inscribed in a circle of radius $R$ has area \[\text{Area} = 2R^2\sin A\sin B\sin C,\] where $A+B+C=\pi$. The product $\sin A\sin B\sin C$ is maximized when $A=B=C=\pi/3$, so the maximum area is $2R^2(\sqrt{3}/2)^3 = \frac{3\sqrt{3}}{4}R^2$. ∎

Lemma 2 (Exact area 1 inside a large disk). If a set $A$ contains a closed disk of radius $R\ge 2\cdot 3^{-3/4}$, then $A$ contains a triangle of area exactly $1$.

Proof. Let $C$ be the circle of radius $R$. The equilateral triangle inscribed in $C$ has area $\frac{3\sqrt{3}}{4}R^2\ge 1$. Fix two distinct points on $C$ and let the third vertex move continuously along $C$. The triangle area varies continuously and takes values arbitrarily close to $0$ (as the third point approaches one of the fixed points) and at least $1$ (at the equilateral position). By the intermediate value theorem, some position gives area exactly $1$. ∎

Lemma 2a (Disk threshold calculation). Let $R_*=2\cdot 3^{-3/4}$. Then the disk of radius $R_*$ has area \[ \pi R_*^2=\frac{4\pi}{\sqrt{27}}, \] and its maximum triangle area equals $1$.

Proof. Compute $R_*^2=4\cdot 3^{-3/2}=4/\sqrt{27}$. Hence the disk area is $\pi R_*^2=4\pi/\sqrt{27}$. By Lemma 1, the maximal triangle area in the disk is $(3\sqrt{3}/4)R_*^2=(3\sqrt{3}/4)\cdot 4/\sqrt{27}=1$. ∎

Lemma 2b (Area values inside a disk form an interval). Let $B_R$ be a closed disk. Then for any $t$ with $0\le t\le \frac{3\sqrt{3}}{4}R^2$, there is a triangle with vertices in $B_R$ of area exactly $t$.

Proof. Fix two distinct points on the boundary of $B_R$. As the third vertex moves continuously along the boundary circle, the triangle area varies continuously from $0$ (degenerate position) up to the maximal inscribed area. The intermediate value theorem gives all intermediate values. ∎

Lemma 3 (Disk obstruction at the conjectured threshold). If $R<2\cdot 3^{-3/4}$, then no triangle in the disk of radius $R$ has area $1$.

Proof. By Lemma 1, the maximum possible area is $\frac{3\sqrt{3}}{4}R^2<1$. ∎

These lemmas show the conjectured constant is the correct threshold for sets that contain a full disk, but they do not address general measurable sets.

5) VERIFICATION Edge cases: - Degenerate triangles (area 0) are excluded by definition. - Lemma 2 uses continuity of the area as a function of the third vertex on a circle. - Lemma 2b uses the same continuity argument to show all intermediate areas occur in a disk.

Quantifier checks: - Lemma 2 requires the disk to be contained in $A$; it does not apply to arbitrary measurable sets of the same measure without a disk.

6) FINAL UNRESOLVED

(i) Strongest fully proved partial result: If $A$ contains a disk of radius $R\ge 2\cdot 3^{-3/4}$, then $A$ contains a triangle of area 1 (Lemma 2). Disks with radius $R<2\cdot 3^{-3/4}$ contain no such triangle (Lemma 3). The threshold area $4\pi/\sqrt{27}$ and maximal inscribed triangle area are computed explicitly (Lemma 2a), and all areas up to the maximum are realized inside a disk (Lemma 2b).

(ii) First gap: Extend the disk-based argument to arbitrary measurable sets of area $>4\pi/\sqrt{27}$ without assuming any convexity or interior ball condition.

(iii) Top 3 next moves: 1. Prove the conjecture for unions of finitely many convex bodies for all $n$ (extend the $n\le 3$ result reported in the statement). 2. Show that the set of achievable triangle areas inside a positive-measure set is an interval. 3. Develop a reduction from general measurable sets to structured approximations with controlled area.

(iv) Minimal counterexample structure (if false): A bounded measurable set with area just above $4\pi/\sqrt{27}$ that is highly disconnected (no large interior balls), engineered to avoid all exact area-1 triangles while still having large measure.