geometry construction problem
|Sep24-11, 01:08 PM||#18|
geometry construction problem
The problem is a bog standard one in surveying for railway, mining and road engineering as is the solution I offered.
|Sep24-11, 02:57 PM||#19|
Secondly, I misstated why I feel you didn't give an adequate solution:
The OP is looking for a geometric solution that doesn't involve calculations that are subject to rounding errors. (Think straightedge and compass). For example, it is easy to construct the bisector of an angle using only those tools. Granted, we could also measure the angle with a protractor, divide that measurement in half and draw in the bisector by measuring this new angle with the protractor, but this method is much less accurate (even in AutoCAD).
|Sep24-11, 04:30 PM||#20|
I also stated that I didn't have any sort of drawn solution only a calculated one, which I though was better than none.
I would, however be interested in a drawing solutuin since it must be possible to draw what can be calculated. I am just not a first class draughtsman (or even second class really).
So please post if you come up with anything.
If it helps I can let you have the surveyors screed on the subject.
|Sep25-11, 11:59 PM||#21|
This problem seems so easy on the surface but difficult in practice. I have started an approach to the problem by solving it with constructions if the lines were parallel. Of course I know they are not. I will post my approach in hopes of it sparking someone else while I continue working on it.
|Oct1-11, 08:59 PM||#22|
Here is a solution, it is not a solution with both circles tangent to each other, but they intersect in equal arcs with equal radii. This seems to me to be adequate for any construction job. All of the following steps correspond to the attached drawing:
1) Draw line segment AB
2) Draw perpendicular from point A
3) Draw perpendicular from point B
4) Extend line A
5) Extend line B
6) The measure of angle CAB = a
7) The measure of angle ABD = a-θ
8) Compute angle from AC and BD to be equal solve: a-x = a-θ+x, x = θ/2
9) Add θ/2 to angle DBA, subtract θ/2 from CAB
10) Draw the new angle line segments to intersect at point F, This forms isosceles triangle AFB.
11) Bisect segments BF and AF
12) Draw perpendiculars from the midpoints of the segments to intersect BD and AC
13) Points H and J are the center of the two circles that are tangent to line A and line B
14) The circles are not tangent to each other but intersect at a point to create equal arcs
Yes, the above requires some cleanup and formalization, however, you get the point.
|Oct21-11, 05:33 PM||#23|
I have discovered and proven the following, which reveals one point that lies on the line tangent to both circles (also know as the perpendicular bisector of the line segment connecting the circles' centers). I can do this using ONLY drafting techniques. The problem is, I haven't found a way to locate a 2nd point on that line.
But, it's a start; maybe it will help.
Here are the steps:
1) extend lines A and B so that they intersect at point P
2) draw the angle bisector of angle APB
3) draw perpendicular bisector of line segment AB
4) the intersection of these 2 lines lies on the common tangent line to the 2 circles
If we can find a 2nd point on that line, then we can find the common tangent (obviously), which means we can find the angle of the line connecting the circles' centers. Using this information, I would be able to solve the problem.
|Oct21-11, 08:19 PM||#24|
I think that the radii have a minimum and maximum that can make this true. That is why this becomes such a difficult problem, as 3 simultaneous events must occur.
|Oct22-11, 01:26 PM||#25|
Here is the proof (albeit, not a completely formal one):
To show this is true, I start with 2 circles and the tangent lines which intersect at point P.
As per the problem requirements, the circles have the same radius and intersect at only one point.
A couple of definitions:
- Let the distance from a point to a line be defined as the length of the segment drawn from the point perpendicularly to the line.
- Any point lying on the bisector of an angle is equidistant from each ray of the angle
- Any point lying on the perpendicular bisector of a line segment is equidistant to each of the segmentís endpoints.
Line AP is tangent to the upper circle (circle C) at point A.
Likewise, line BP is tangent to the lower circle (circle D) at point B.
- Draw the line segment AB
- Draw the perpendicular bisector of segment AB
- Draw the line segment CD
- Draw the perpendicular bisector of CD. This line is tangent to both circles
- Label the intersection of the 2 bisectors as point E
Now, all I have to do is show that point E lies on the angle bisector of angle APB.
- Draw segment EF perpendicular to AP with point F on AP
- Draw segment EG perpendicular to BP with point G on BP
(for clarity, since I have located point E, I will remove segment AB, circles C and D, and the perpendicular bisectors from the drawing)
So, if I can show that EF = EG then Iíve shown that it lies on the bisector of angle APB.
- Draw segments AC and BD
- We know that AC = BD since they represent the radius of the circles
- We know that AC is perpendicular to AP and BD is perpendicular to BP (since AP is tangent to circle C and BP is tangent to circle D)
- Draw segments EA and EB
- Since E lies on the perpendicular bisector of AB, then EA = EB
-Draw segments EC and ED
- Since E lies on the perpendicular bisector of CD, then EC = ED
- Therefore, by SSS, we know that triangle ACE = triangle BDE
- EF is parallel to AC since both are perpendicular to AP
- EG is parallel to BD since both are perpendicular to BP
- Angle AEF = angle CAE by alternate interior angles
- Angle BEG = angle DBE by alternate interior angles
- Angle CAE = angle DBE by congruent triangles
- Therefore, angle AEF = angle BEG by transitivity
- Angle EFA = angle EGB = 90 degrees, by definition of perpendicular
- Triangle AEF = triangle BEG by AAS
- Therefore EF = EG by congruent triangles
|Oct24-11, 06:35 AM||#27|
BTW, Point E lies on the circumscribed circle of triangle ABP, for what that's worth.
|Oct24-11, 08:19 AM||#28|
That may be helpful! What's the proof for that?
|Oct24-11, 08:58 AM||#29|
Referring to your diagram above:
AB is a chord of the circumscribed circle for triangle ABP
A perpendicular bisector of a chord AB, bisects the arc at E
angle P is an inscribed angle, intercepting the arc AB
The bisector for angle P, bisects the intercepted arc AB at its midpoint E
|Oct25-11, 02:26 PM||#30|
Here is another proof for coolul007's claim.
In the quad AEBP:
A = EBG -- (congruent triangles)
EBG + B = 180 degrees -- (angles on a straight line)
A+B = 180
E+P = 360 - (A+B) = 180
AEBP is a cyclic quad -- (opposite angles form straight angles).
|Jul15-12, 11:24 AM||#31|
Im a 2nd year student and really really confusing for us this things...
|Jul15-12, 11:50 AM||#32|
Hello physs and welcome to Physics Forums.
Perhaps if you were to explain your difficulty someone could help.
Did you really want to post in this old thread?
|Similar discussions for: geometry construction problem|
|Geometry Problem in Context of Image-Charge Problem (Griffiths,p. 125, [3.15],[3.16])||Advanced Physics Homework||1|
|Simple geometry problem but only using geometry||Calculus & Beyond Homework||7|
|Calculus Highway Construction Problem||Calculus & Beyond Homework||1|
|Geometry problem||Calculus & Beyond Homework||1|
|Geometry: Construction||Introductory Physics Homework||2|