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Mar 31st, 2002, 07:49 PM
#1
Thread Starter
Member
two lines and a circle construction
Here is something that I came across while solving another problem. I do not know the answer to this question - yet
With only a compass and straight edge, Given two intersecting lines and a circle, construct a circle that is tangent to the three objects. If anyone could save me some time figuring this out I would be grateful.
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Apr 1st, 2002, 11:36 PM
#2
Frenzied Member
I do not know how to solve your problem, but you should start by drawing the bisector of the angle between the two intersecting lines.
A circle tangent to both lines will be on that bisector.
Next, I would experimentally draw a good approximation to a solution. A little fooling around should get you a good diagram.
Analyzing the diagram might give you a clue.
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Apr 2nd, 2002, 06:11 AM
#3
transcendental analytic
(X1-x1)^2+(Y1-y1)^2=r1^2
(X1-x2)^2+(Y1-y2)^2=r2^2
Y2-y3=k3(X2-x3)
(X2-x1)^2+(Y2-y1)^2=r1^2
Y3-y4=k4(X3-x4)
(X3-x1)^2+(Y3-y1)^2=r1^2
solve r1, x1 and y1. r2,x2,y2 is your arbitrary chosen circle, k3,x3,y3 and k4,x4,y4 your lines, k3!=k4.
further on you have the intersections X1,Y1 between the circles, X2,Y2 and X3,Y3 between the drawn circle and each line.
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writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
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Apr 2nd, 2002, 06:24 AM
#4
transcendental analytic
hmm you need to have
(x1-x2)^2+(y1-xy)^2=(r1+r2)^2
k3=(x3-x1)/(y3-y1)
k4=(x4-x1)/(y4-y1)
as well
Use  
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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