Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)
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The four point condition implies slim triangles

Statement

In a geodesic space satisfying the four-point condition with constant κ0, every geodesic triangle is 4κ-slim.

Facts & Assumptions

Given: Such a space, specified sides of a triangle (a,b,c), and p[a,b].

[F1]

The four-point hypothesis gives the product inequality with the same constant at every basepoint, by The gromov product inequality implies the four point condition.

Proof

1.1

Write =d(a,b), t=d(a,p) and α=(bc)a. Suppose first that tα. Since αd(a,c) there is q[a,c] with d(a,q)=t. Products along a radial geodesic give (pb)a=t and (cq)a=t. Applying the product inequality first through b, then through c, gives (pc)atκ and (pq)at2κ. Hence d(p,q)=2t2(pq)a4κ.

F1givenalgebra
2.1

If tα, use b as basepoint. Indeed (ac)b=α by expansion, and d(b,p)=tα. The argument of step 1.1 with a and b interchanged gives a point on [b,c] at distance at most 4κ from p. At t=α either construction works.

step 1.1algebra
3.1

Thus every point of [a,b] is within 4κ of the other two sides; relabeling vertices proves this for all sides and all specified triangles. Zero side lengths require only t=0, and when κ=0 the produced point equals p. There were only finitely many segment choices.

step 1.1step 2.1

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