Six-point two-color triangle puzzle: why five is different

A six-point two-color triangle puzzle gives a small gathering something definite to investigate together. Draw six dots, connect every pair, and assign each connection one of two labels. The challenge is to find three dots whose three connecting lines all have the same label. Such a triangle must exist, however the labels were chosen. A five-dot drawing can avoid one, so the extra dot changes what is guaranteed.

The paper activity uses stipulated inputs. The dots need no relationship to actual guests, and the labels need no meaning beyond R and B. Orange and navy ink can help distinguish the lines, but letters beside the lines work too. Keep the rule visible so participants can inspect the same problem. The activity concerns a finite diagram, without claiming anything about friendships, people's preferences or how a real gathering will unfold.

Define every pair before starting

Name the dots A, B, C, D, E and F. For every pair of different dots, draw one undirected connection. AB and BA therefore refer to the same line. There are five connections from A, four more from B to the later letters, then three, two and one: 5 + 4 + 3 + 2 + 1 = 15. No pair may be left blank. Each of those fifteen connections receives exactly one label, R or B.

A triangle here means three distinct named dots together with all three pairwise connections. For instance, ABC uses AB, AC and BC. It does not mean any triangular-looking region formed where unrelated lines cross. A crossing creates no extra dot unless the rules explicitly name one, which this exercise does not. You can redraw the diagram to reduce clutter while preserving the pair list and labels; the conclusion depends on that list rather than the shape on the page.

Write a short checklist below the sketch: all six dots present, all fifteen pairs assigned, no pair given two labels. A missing edge or an extra label changes the question. Someone finding no uniform triangle in an incomplete drawing has not disproved the claim. Complete the inputs first, then let one person trace candidate triples while another checks the labels against the recorded pair list. Both readers should be able to recover the same three entries.

Look at five lines from one dot

Start at A and inspect AB, AC, AD, AE and AF. These are five connections, each with one of two labels. At least three must share a label. To check that statement directly, suppose each label appeared at most twice. That would account for at most four connections, leaving the fifth unassigned. Since all five have labels, that supposition fails. Three agreeing lines are therefore available without assuming anything about the remaining ten connections.

Call their common label R for the proof. If the three actually carry B, exchange the names R and B throughout the argument; that consistent relabeling preserves which connections agree. Name three neighbors reached by those lines B, C and D, renaming the vertex letters for the argument if necessary. The recorded facts are now AB = R, AC = R and AD = R. Write them down rather than relying on a quickly drawn patch of color.

There might be four or five matching lines from A. Selecting any three is sufficient; the argument does not require exactly three. The unused vertices remain part of the original complete drawing. They simply are not needed to finish this proof. This distinction helps when someone objects that a fourth matching line was ignored: leaving it unused in the reasoning does not erase it from the defined graph or alter its assigned label.

Finish with two exhaustive cases

Inspect the three connections among B, C and D: BC, BD and CD. First ask whether at least one carries R. If BC is R, then AB, AC and BC are all R, so ABC is the required triangle. If BD is R instead, ABD works. If CD is R, ACD works. One matching neighbor connection is enough because the two connections back to A were already recorded as R.

The remaining case is that none of BC, BD and CD carries R. Every connection must have exactly one of the two labels, so all three then carry B. Those three connections form the triangle BCD, entirely labeled B. This completes the argument. The first case supplies a triangle containing A; the second supplies one among its three selected neighbors. They cover every allowed assignment to the three inspected connections.

Notice what has been proved and what has been left open. Every valid six-dot assignment contains at least one same-label triangle. The proof has not counted all such triangles or identified a single triple that works for every assignment. A different labeling may require a different triple and may contain many successful triples. The guarantee concerns existence somewhere in each complete drawing, with a short procedure for finding a witness from the chosen starting dot.

Make the smaller counterexample together

Draw a separate page with only A, B, C, D and E arranged around a pentagon. Label the five outer connections AB, BC, CD, DE and EA with R. Label the other five pairs AC, AD, BD, BE and CE with B. Every pair now has one label, and the total is ten. Keep this five-dot page separate from the six-dot page, so that removing F cannot silently remove required connections from the first exercise.

The R connections form a five-step loop. Each vertex has exactly two R neighbors, and those two neighbors are connected to each other by B. For example, A's R neighbors are B and E, while BE is B. B's R neighbors are A and C, while AC is B. The same check works at C, D and E. Any all-R triangle would need two R neighbors joined by R, so none exists.

The B connections also form a five-step loop, in the order A, C, E, B, D, A. This uses AC, CE, EB, BD and DA, precisely the listed B pairs. Each vertex's two B neighbors are joined by an R connection: A has C and D with CD = R, for instance. The same obstruction excludes an all-B triangle. The two labels therefore both fail to produce a uniform triangle on this particular five-dot page.

If the loop explanation feels too compressed, inspect all ten triples. ABC has R, B, R; ABD has R, B, B; ABE has R, R, B; ACD has B, B, R; ACE has B, R, B. Continue with ADE as B, R, R; BCD as R, B, R; BCE as R, B, B; BDE as B, B, R; and CDE as R, B, R. Each list follows the alphabetical pair order within its triple. Every triple contains both labels.

Keep the threshold claim precise

The five-dot page shows that five dots do not force the desired triangle. It does not say that a five-dot page can never contain one. Labeling every pair R would create plenty. The conclusion is that an avoiding assignment exists at five, while no avoiding assignment exists at six under the complete-pair two-label rules. Those two statements together establish the exact threshold for this specific question, without requiring claims about larger patterns or additional labels.

For a relaxed group activity, let each participant label one complete six-dot page and exchange it with a neighbor. Ask the checker to name a successful triple and its three pair labels. Someone who prefers a written role can maintain the pair list while another person draws. These are optional ways to share the task, not predictions about what guests will enjoy. Keep participation separate from the proof: the mathematics remains the same if one reader works through it alone.

When a proposed answer seems wrong, identify the exact triple rather than telling someone the whole picture is confusing. A claim that ABC is all R can be checked against AB, AC and BC. If AC was copied differently, resolve that single input before checking the claim again. Retain the original and revised pages as different versions when someone deliberately changes a label. A new assignment deserves a new witness; it does not rewrite the earlier result.

Keep the supporting references separate

The DollarDays receptacle pencil sharpener account supplies optional writing-tool context. Its sole gallery shows a Creative Colors package with two complete two-hole manual sharpeners. Catalog and pictured labels establish no maker equivalence here. The image does not independently count the stated case quantity or verify materials, fit, safety or sharpening results.

The beige electric pencil sharpener account shows one whole tool with an integral cutter and selector. A DELI mark appears while the catalog says NOBRAND; no maker relationship is authenticated. Three pictured pencils are unconfirmed supplied props. No diameter fit, automatic motion, charging behavior, blade life or tip performance is verified.

The guide on graph coloring on paper assigns labels to vertices and requires connected endpoints to differ. This puzzle assigns two labels to edges freely and seeks a uniform triangle. The five-cycle serves a different role in each proof; vertex-label constraints do not establish the edge-label guarantee.

The minimum connection cost guide selects a connected subset of weighted links. Here all pairs remain present and no costs are assigned. A cheap spanning path would answer neither whether all fifteen pairs were labeled nor whether a same-label triangle exists.

Keep any tool purchase outside the puzzle

If the paper activity prompts a separate writing-tool purchase, keep the order evidence apart from the diagram. Completing a triangle proof establishes no package quantity, supplied brand, pencil compatibility or blade performance. Before ordering through republeq.com, inspect the exact current item and resolve any product question that affects the purchase. A photograph of two sharpeners cannot independently count a larger shipping case, and pencils pictured beside a tool do not necessarily belong to the supplied contents. If cryptocurrency is being considered, check the actual payment choices presented for that particular order. Mentioning USDT in an article does not establish that the token, a particular network or a particular wallet is supported at checkout. Where an appropriate cryptocurrency option is actually offered, verify the requested asset and network, destination, amount and order reference against the current transaction instructions before authorizing a transfer. Include shipping, taxes and any transaction charges when evaluating the complete amount. Keep the item reference, applicable delivery and return terms, and any written clarification with the order record. A changed item or changed payment instruction needs a fresh check rather than an assumption that an earlier answer still applies. Save the resulting order confirmation and payment record together so that a later inquiry can identify the exact transaction. If the payment or order status remains uncertain, reconcile that evidence before considering another payment. None of these purchasing steps changes which triples appear in the drawing. You can complete and share the puzzle without buying either optional tool, while any later order remains dependent on its own current product and transaction information.