A seemingly easy logical question

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Discussion Overview

The discussion revolves around a logical puzzle involving the connection of three circles to three triangles without crossing lines. Participants explore the feasibility of this task and share their attempts and insights.

Discussion Character

  • Exploratory, Debate/contested

Main Points Raised

  • One participant expresses difficulty in solving the puzzle, noting they have managed eight connections but not the required nine.
  • Another participant questions the clarity of the problem by asking for the picture referenced in the initial post.
  • A different participant argues that the task is impossible, providing a detailed explanation involving the separation of areas created by the connections between the triangles and circles.
  • Some participants share visual aids (gifs) to illustrate the problem, suggesting alternative approaches, such as considering the connections on the surface of a torus.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the solvability of the puzzle, with some asserting it is impossible while others explore different representations and methods.

Contextual Notes

The discussion lacks a visual reference for the problem, which may affect participants' understanding and contributions. The assumptions regarding the nature of the connections and the definitions of the areas are not fully clarified.

maverick_bih
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This looks like an easy one, but I haven't found anyone yet who has answered it correctly, including me...
So, the goal is (like shown in the picture), to connect all three circles to each of the three triangles (nine connections in total, i have managed eight). Give it a try. You can go around the triangles and circles with the lines, just as long as they DON'T cross.

∆ ∆ ∆

O O O
 
Last edited:
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What picture?
 
It's impossible. Let's call the triangles T1, T2, T3 and the circles C1, C2, C3. There will be a separated area on each side of the lines going from T1 to C1 to T2 to C2. Let's call one of these areas Ž and the other Đ. There will also be a line going from C1 to T3 to C2, subdividing one of the areas, let's say Đ, into two new areas, one touching T1 which we will call Æ and one touching T2 which we will call Ø, giving a total of three separated areas. T3 now neighbours areas Æ and Ø, T1 neighbours Ž and Æ and T2 neighbours Ž and Ø. No matter which of these areas C3 is in, only one of the points T1, T2 and T3 will be on the edge of this area.
 
See attached gif.
 

Attachments

  • Connect6.gif
    Connect6.gif
    7.6 KB · Views: 525
Martin Rattigan said:
See attached gif.

You could also do it in "two dimensions" on the surface of a torus (donut).
 

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