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 ...................................... And thus we can derive: CD/DB = CD/AD * AD/DB (first divide and then multiply by AD) = CA/AB * CA/AB (using ) = (CA/AB)^2 Or, in other words, CA/AB = sqrt(CD/DB) ........................................ 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  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/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.