What is your question?
Say is at 1, is at 0 ( ) and we want to be at . (hence we want to be ) Using the intercept theorem we get . If and then !
To construct , one can proceed as follows :
- Draw segment line with .
- Draw segment line perpendicular to the x-axis and such that
- Draw segment line .
- Draw straight line perpendicular to the y-axis which passes through the point . Let be the intersection of this straight line with .
- Draw straight line perpendicular to and passing through F.
- G is the intersection of with the x-axis.
Is there a subject, or branch or part of Geometry that deals about these kind of constructions? It was not taught to us in my school days. Is it something new?
Anyway, as is my way of solving things not taught to us in school, I will use "common sense", or is it intuitive reasoning, here.
From (0,0), or from a, as the center, strike a quarter of a circle whose radius is x units long.
Then draw a line from (1,0) or b to the intersection of the circular curve with the vertical axis at (0,0) or a.
Then subdivide that line into x number of equal parts.
Then from the 1st subdivision, draw a vertical line up or down to the horizontal axis at (0,0) or a.
The horizontal distance from (0,0) or a to the foot of that vertical line is 1/x.
Hello ticbol and flyingsquirrel,
Actually, this guy edited his post because we asked him to post it instead of the pdf... But the true question is that you can only use a compass and draw circles...
(See the first picture) Let and for . Construct the two following circles : Let be the intersection point of with the x-axis such that . Let be an intersection point of the two circles. (it exists since ) As is a diameter of , the triangle is rectangle. Now let be the orthogonal projection of on the x-axis. and are both rectangle triangles. Using the Pythagorean theorem for the three triangles, we get
which gives us . (see (*)) As we want to find one can think that letting is a good idea but it requires us to construct a point such that which is not easy using only a compass. (at least, I don't know how to do it ) A better idea is to let . It 'll give us from which one can easily get . Unfortunately, there is another problem : how can we construct using only a compass ? I don't know... Here is a solution I found to have not to construct this point :
(see the second picture) Let and for . Construct the two circles and as before : (hence and ) and construct the circle of radius 1 and which center is . Let be the intersection point of with which is in the same quadrant as . Let be the orthogonal projection of on the x-axis then, using the three rectangle triangles , one can show that . (same proof as before : see (*)) This is interesting because it gives us : we only need to construct and to find .
(see picture #3) Solution of the problem : Given and for construct the three circles Let and be the intersection points of with and of with , respectively. (choose these two points so that they lie in the same quadrant) Then which can be easily constructed using *only* a compass.
Don't know if it's the "best" solution but that was interesting anyway.
(*) Let's show that :
and give . Substituting and in this equation :