Draw a Line of Length √3: Tips from an Old Man

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SUMMARY

The discussion centers on constructing a line of length √3 using a set square. Two primary methods are proposed: the first involves creating a 30-60-90 triangle with a hypotenuse of 2 and a leg of 1, while the second method utilizes an isosceles right triangle with legs of 1. Participants highlight the impossibility of achieving an exact length of √3 with drawing instruments, emphasizing that such precision exists only in theoretical mathematics. The conversation also references the historical figure Pythagoras and the concept of irrational numbers.

PREREQUISITES
  • Understanding of right triangle properties and Pythagorean theorem
  • Familiarity with geometric constructions using a set square
  • Knowledge of irrational numbers and their implications in geometry
  • Basic skills in drawing geometric figures accurately
NEXT STEPS
  • Explore the properties of 30-60-90 triangles in geometric constructions
  • Study the implications of irrational numbers in mathematics
  • Learn advanced geometric construction techniques using only a compass and straightedge
  • Investigate the historical context and contributions of Pythagoras and Hippasus
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Mathematicians, geometry enthusiasts, educators teaching geometric constructions, and anyone interested in the historical aspects of mathematics.

Edgardo
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An old man gives you a set square (http://www.buchhandlung-umbach.de/pbs/geodreick.gif ) and then asks you to draw a line of exactly the length \sqrt{3}.

How would you do it?
 
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In white: Draw a right triangle of lengths 1 and 2. The hypotenuse will have length \sqrt{3}.
 
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first you what units are we using?

second, as far as i can see there are two ways:
first you can straight away construct a 30-60-90 triangle with the hypotenuse (2) and the other leg of 1 (the one which is adjacent to the 60 degree) and then the other leg is sqrt(3).
the second approach is you build an isosceles right triangle with the legs of 1, and then on the hypotenuse you bulid another 1 leg (when the degree between sqrt(2) and the third leg is 90), and then the second hypotenuse is sqrt(3).

this was way too easy, so perhaps i totally misinterpratated your question.
 
@jimmy: Your length would be \sqrt{5}

@ Loop:
I thought about the second approach. You're right.


Next question: What was the name of the old man :biggrin:
 
Edgardo said:
@jimmy: Your length would be \sqrt{5}
Man, talk about a blind spot! I have no explanation for what I was thinking when I wrote that.
 
think quite easy !

1- first you draw a squareroot of 2 by using P=1 AND b =1

then measure the length of the squareroot of 2, take it as p , and take 1 as b, then you can join the end of the these two sides.

correct !

hope so !
 
the word "exactly" means none of these answers are correct, nor is there any correct answer. this is impossible with drawing instruments, but only in the mind, or in the fantasy world of perfect mathematics can this be done exactly.
 
I figure the old man must be Pythagoras.

To get the answer “Exactly” we can’t be drawing so many lines as has been suggested.
Observing the “Square” provided by the old man we see all three sides are neatly inscribe with halfway points. Thus we shall define it as 2 units by 2 units with a H of 2*sqrt{2} units.
After inscribing an exact right angle of 1 by 2, we place a mark on the long length of exactly sqrt{2} by using the midpoint of the H side of the provided square.
Now, again only using the square provided, we see it is long enough to draw a straight line from that marked point to the end of the short line. Giving us an exact sqrt{3} .

But we will never be able to provide an exact measure of that line based on any fractional measure of the units we have selected, ie that measure is irrational.
Note: The old man would appreciate it if you kept this part a secret – he fears the idea that a number can be irrational could cause a public panic! :wink:
 
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