Tends to infinity

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Archive for the ‘Geometry’ Category

Another Geometrical construction…

Posted by tendstoinfinity on February 5, 2011

pichak@scientist.com

I in previous post posed a geometric problem

An angle and a point D lying inside it are given. Construct a line segment with end points on arms of the angle and bisected at D.

let us consider a set of points on a line  BA  and all the segments having one end on these points and mid point at D. then the other end of each line segment will be just the reflection of  the former points. hence the other ends will always lie on a line parallel to BA.

then the above construction can be easily done.

suppose the given angle is <APQ. Image of the vertex P of the angle with respect to D is drawn (draw a line through P and D. cut an arc of length equal to PD from centre D. where it cuts the line is image P’ of P with respect to D

draw a line parallel to PA through P’ . suppose this parallel cuts PQ at R then RS will be required line which has mid- point at D and ends on the arms of the angle

surely you can justify this. I will give a very good  use of this construction

An angle \angle APQ and a point D lying inside it are given. Draw a straight line through D which makes a triangle with the arms of the angle of minimum possible area.

Then obviously the triangle RPS (with reference to the above figure) will have minimum area amongst all the triangles of this type !! got it.

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Posted in Geometry | Leave a Comment »

A Geometrical construction

Posted by tendstoinfinity on January 25, 2011

pichak@scientist.com

An easy challenge  ” Draw a circle of radius r centered at A and another  point C is given. Draw a line through C so that chord cut off by circle along this line is of length d

I request you to try on your own first then see my solution.  you can have better solution than me. if you got another one pl. share.

First this simple geometrical construction ” A circle is given and a point B outside it. Draw a tangent to the circle through B.”   You, I hope will be able to draw the required construction.

Draw the circle “c” with centre A and the point B. Join the points A and B. Bisect the line segment AB at C. draw a circle “d” with centre C and radius BC. let the circle “d” intersects circle “c” at D.
the line through B and D will be required tangent.

Now the original construction

Draw the given circle “c” with radius r centered at A

Draw another circle “d” concentric with the given circle with radius \sqrt{r^{2}-d^{2}/4}

draw the tangent through C to d. this line will cut a chord of length d of circle “c”

hope the construction is justified by you. Again a  challenge to you which I will solve in another post.

An angle and a point D lying inside it are given. Construct a line segment with end points on arms of the angle and bisected at D.

Posted in Geometry | Leave a Comment »