circleseuclidean-geometrygeometric-constructiongeometry
I found this problem in Hawthorne's Geometry: Euclid and Beyond exercise $2.11$.
Given a line $l$, a line segment $d$, and a point $O$, construct a circle with center $O$ that cuts off a segment congruent to $d$ on the line $l$.
I think it's possible to move $d$ onto $l$ such that it is perpendicularly bisected by a line from $O$, then the circle would cut the $d$ from the line. I'm curious if anyone knows an elegant solution to an interesting construction problem.
1. Pick an arbitrary point C on the circle. Draw an arc centered at C and radius AB, intersecting the original circle at point D.
If there is no intersection, stop: AB is too long for a solution to exist.
2. Draw line segment CD.
We now have a chord which is congruent to AB. We need to find a congruent chord which passes through P, so the first thing we need to do is figure out what point on CD will correspond to P.
3. Draw an arc with center O and radius OP. Let this intersect CD in point E.
If there is no intersection, stop: AB is too short for a solution to exist. Otherwise, E is the point we needed. OE is congruent to OP, and CED is congruent to the desired chord. Now if we find the point Q that C corresponds to, we'll have the chord we want.
4. Draw an arc with center P and radius EC to intersect the original circle at point Q.
The (undrawn) triangles OPQ and OEC are congruent, so PQ and EC make the same angle with the radius, and that determines the length of the chord.
5. Draw line segment PQ, extended beyond P to meet the circle on the other side.
This is the desired chord. It is congruent to CD, which is congruent to AB.
Two circles + five lines = 7 steps:
Select an arbitrary point $P_0$ on a given circle $C_0$ and draw a circle $C_1$ with an arbitrary radius $r$, which is small enough to intersect $C_0$.
Draw a circle $C_2$ centered at the intersection point $P_1$ with the same radius $r$ to get the intersection points $P_2$ and $P_3$.
Draw a line through points $P_2,P_1$, intersecting $C_0$ at point $P_4$.
Draw a line through points $P_1,P_3$, intersecting $C_0$ at point $P_5$.
Draw a side $P_5 P_4$ of the inscribed equilateral triangle.
Draw a side $P_4 P_0$ of the inscribed equilateral triangle.
Draw a side $P_0 P_5$ of the inscribed equilateral triangle.
Best Answer
Follow these steps:
1- Mark A and B(A left and B at right) the ends of the segment d. Mark it's midpoint as D.Draw a circle center on A and radius AB.
2- draw a line from A parallel with line l, it meets the circle at C.
3- A perpendicular from O on l intersect it at O'. O' is orthogonal projection of O on l.
4- Connect D to O'.
5- Draw two lines from A and C parallel with DO', they meet l at point N and M respectively.
Clearly $MN=AC=AB$. The circle center on O passing points points M and N is what you are looking for.
Update; I made a figure but I can not attach it due to problem I have with the site. I can send you if you give me an email address.