Stumped by (simple) geometry problem

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Discussion Overview

The discussion revolves around a geometry problem involving the construction of two points M and N on a line segment AB, based on specific ratios related to the segments created. The problem is approached without the use of a compass, relying solely on a pen and ruler.

Discussion Character

  • Homework-related, Exploratory, Technical explanation

Main Points Raised

  • One participant presents the geometry problem and expresses difficulty in solving it.
  • Another participant provides a detailed method for constructing points M and N, including calculations of segment lengths and a step-by-step construction process.
  • A later reply acknowledges the simplicity of the solution and expresses gratitude for the help received.
  • Another participant reiterates the construction method and questions whether the solution can be derived from established theorems or if it relies on trial and error.
  • A response suggests that the solution is primarily based on logic and practice, indicating that prior knowledge of line division techniques could be beneficial but is not strictly necessary.

Areas of Agreement / Disagreement

Participants generally agree on the method of construction provided, but there is some uncertainty regarding the theoretical underpinnings of the solution and whether it can be derived from known theorems.

Contextual Notes

The discussion includes references to specific segment ratios and the construction process, but does not resolve the question of whether established theorems can be applied to the problem.

dextercioby
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I've been quite under desperate lately. I got no clue on how to solve this (apparently easy) geometry problem:

Consider a circle C of center O in a plane and a line segment AB not necessarily outside the circle C.
Having no compass, just a pen and a simple ruler, construct 2 points M and N inside the segment AB so that

$$ \frac{AM}{2,333...} = \frac{MN}{1,666...} = \frac{NB}{2} $$

Any ideas?
 
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Note that:- ##MN=\frac{5}{6}NB\\
AM=\frac{7}{6}NB\\
AB=3NB##
Now draw a long line ##BY## from ##B##
Now match one of the endpoints of the scale with the center of the given circle, and using the pen mark the radius of the circle onto the scale.
Now use this distance and mark 18 equidistant points on the line ##BY##. Now join the 18th point with ##A##. Slide the scale along ##BY## and join the remaining points with ##AB## as well. (sliding will make sure all the lines are parallel.)
So now we have divided ##AB## into 18 equal parts. The 6th point will be ##N## and the 11th point along the same direction will be ##M##.
fig-4-6.png
 
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Wow, amazing. It was so simple in the end. So yes, my imagination definitely needs a reset.
Thank you.
 
certainly said:
Note that:- ##MN=\frac{5}{6}NB\\
AM=\frac{7}{6}NB\\
AB=3NB##
Now draw a long line ##BY## from ##B##
Now match one of the endpoints of the scale with the center of the given circle, and using the pen mark the radius of the circle onto the scale.
Now use this distance and mark 18 equidistant points on the line ##BY##. Now join the 18th point with ##A##. Slide the scale along ##BY## and join the remaining points with ##AB## as well. (sliding will make sure all the lines are parallel.)
So now we have divided ##AB## into 18 equal parts. The 6th point will be ##N## and the 11th point along the same direction will be ##M##.
View attachment 83272
Is there a way to figure that out from some theorems? Or its something you either should read in some old paper or try and fail until you find out?
 
Shyan said:
Is there a way to figure that out from some theorems? Or its something you either should read in some old paper
Not really. It's mostly logic and practice :-)
[But you could know before-hand the way to divide a line into equal parts. But it's simple enough that it's not necessary.]
 

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