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Hi all...
This is a nice Geometry Problem proposed by one of my close friends!
I'm really stuck and need some help....
Thanks!
If two or more thoughts intersect, there has to be a point!
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Let P, Q be the points where the tangent touches the bigger and smaller arcs.
Let AM intersect XY at Z.
Compute length AM.
AZ = 2(MZ)...... (ratio of the 2 similar rt-angled triangles); so AZ = 2/3 AM.
Solve the triangles APZ and MQZ.
Compute angles BAZ and AZX: gives the slope of XY.
We know XY goes through the point P which we can compute.
So we can get the equation of line XY.
Solve against the equations for FE and CD to get the coordinates of X and Y.
From these compute length XY.
It's the activity of the intelligence above all that gives charm to existence.
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I showed him your reply and...
mmmm... Can we try solving it using Euclidean Geometry only?
Sort of... NO USE of co-ordinate system and vectors!
(he says he does 'em using Similar Triangles and Trigonometry )
Thanks for looking into it!
If two or more thoughts intersect, there has to be a point!
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Here is another (and perhaps Tougher) version of the same question...
Try calculating the answer for this and see!!
If two or more thoughts intersect, there has to be a point!
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He is looking for a General Solution for the lengths of all such XY's for all Regular Polygons (no. of sides > 4)!
I don't know if thats possible or not...
May be, he'll come up with a General Approach instead!?
If two or more thoughts intersect, there has to be a point!
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I used the idea from gurthbruins to solve this problem.
This is a very hard problem. I had printed out the problem on a sheet of
paper and decided to try to solve it. First, we draw a straight line connecting
points A and M. The point where this line intersects XY we'll call point Z.
We'll make another point called G which is the mid-point of AB. The point of
tangency for the big circle we'll call W. The center of the hexagon is the origin.
Notice that
Last edited by Fruityloop (2012-05-23 20:50:23)
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