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Constructing a Segment


Date: 09/26/1999 at 22:19:24
From: Alan Medeiros
Subject: Geometry contest question

Greetings,

I am working on a set of problems for a math competition at my school, 
and there is only one problem that I have yet to figure out.

Given a 1" segment and a 2.5" segment, using only a compass and a 
straightedge, find the segment square root of 2.5" .

I have attempted to approach this problem using the Pythagorean 
theorem, and have come very close to the square root of 2.5", but have 
not reached the exact number. Any help or guidance would be greatly 
appreciated.

Sincerely,
A.F. Medeiros


Date: 09/27/1999 at 10:43:19
From: Doctor Floor
Subject: Re: Geometry contest question

Hello Allan,

Thanks for your question.

Let us consider a right triangle with an altitude to the hypotenuse.

     C
     |\        AD is perpendicular to BC.
     | \
     |  D
     A---B

Note that the triangles ABC, DBA and DAC are similar. From that we can 
find:

     CA/AB = AD/DB = CD/AD   ......................................[1]

And thus we can derive:

     CD/DB = CD/AD * AD/DB (first divide and then multiply by AD)
           = CA/AB * CA/AB (using [1])
           = (CA/AB)^2

Or, in other words,

     CA/AB = sqrt(CD/DB)   ........................................[2]

So, let's start with segment AC, with a point B such that AB = 1" and 
AC = 2.5".


     A----B----------C

Draw a circle with AC as the diameter, and a ray from B perpendicular 
to AC. Let the point of intersection of the ray and the circle be D.

          D
         /|   \ 
       /  |       \
     A----B----------C

Triangle ADC must be a right triangle, since it is inscribed in a 
circle and AC is a diameter.

From [2] this means that DC/AD = sqrt(BC/AB) = sqrt(2.5/1) = 
sqrt(2.5). But this is only a ratio, not a segment length.

Now draw a circle with center A through B. Let the point where this 
circle meets segment AD be D'. Construct a line parallel to DC through 
B' and let this line meet AC in point C'.

          D
        D'|   \ 
       /  |       \
     A----B-----C'---C

Now, in the end we have that D'C' is of length sqrt(2.5)", because 
AD' = AB = 1 and D'C'/AD' = CD/AD = sqrt(2.5).

I hope this helped. If you need more help, just write us back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   


Date: 09/27/1999 at 11:28:30
From: Doctor Anthony
Subject: Re: Geometry contest question

Along a straight line measure a distance 2.5" and a further 1" (in the 
same straight line). Let the join of the two segments be the point A. 
Now construct a semicircle standing on this 3.5" line as diameter. 
Draw a perpendicular to the original line from the point A to cut the 
semicircle at B. Then AB has a length sqrt(2.5).

The proof is immediate from the 'Intersecting Chords' theorem.

     AB^2 = 1 x 2.5

       AB = sqrt(2.5)

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Constructions
High School Geometry
High School Triangles and Other Polygons
Middle School Geometry
Middle School Triangles and Other Polygons

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