Circle Question

[bugzpodder]

we have an equilateral triangle ABC. we draw circular arc BC of circle that has center A and radius AB, as well as circular arc AC with center B and radius AB. Given that arc BC is 12 units, find the circumference of a circle that is tangent to arc BC, arc AC and base line AB.

[krtxmrtz]

9 * Pi

The derivation is a bit cumbersome to write down so I'll leave it for the coffee break... if I can afford one today...

[bugzpodder]

sorry your answer is incorrect. the answer is an integer (no pi). and plz show your work so maybe we can catch the error. sorry for cuzing the extra trouble

[krtxmrtz]

Here are the triangle (x,y) coordinates according to my setup:

A: (-s/2 , 0)
B: (s/2 , 0)
C: (0 , 0.5*Sqr(3))

where s is the triangle side (s = AB = BC = CA)

Equations of the AC and BC circles:

(AC): (x + s/2)² + y² = s² (1)
(BC): (x - s/2)² + y² = s²

By symmetry the sought circle (call it CX) will have its center at (0,h) h will be its radius and we have to determine h. The equation of CX is:

(CX): x² + (y - h)² = h² (2)

CX is tangent to circle AC and this means they contact in only one point i.e. there is only one pair of values (x,y) that at the same time satisfies both (1) and (2)

From (1): y² = s² - x² - sx - s²/4 (3)

From (2): y² - 2hy + x² = 0 -> x = Sqr(2hy - y²)

and now I substitute this into (3):

3s²/4 - 2hy - s*Sqr(2hy - y²) = 0

This can be finally reduced to this equation for y:

(4h² + s²)y² - 5s²hy + 9s⁴/16 = 0

Its solutions are:

y = {5hs² +/- Sqr[(5hs²)² - 4*(4h² + s²)*9s⁵/16)]} / [1/2(4h² + s²)]

But there must be only one solution, therefore the root must be 0:

(5hs²)² - 4*(4h² + s²)*9s⁴/16 = 0

Simplifying this we get:

h = 3s/8 = Radius

So, finally, the circumference is:

Circ = 2 *Pi * Radius = 2 * Pi * 3s/8 and if s=12, then

Circ = 9*Pi

[kedaman]

large circle circumference=6*12=72

x+y=2z
x^2+z^2=y^2
y=2z-x=sqrt(x^2+z^2)
4z^2-4xz+x^2=x^2+z^2
4z^2-4xz=z^2
4z-4x=z
x=3/4z

small circle circumference=72*(x/2z)^2=72*(9/64)=10.125

[bugzpodder]

krtxmrtz , from your post, i think you screwed up one small detail: the value of s is NOT 12. 12 is the arc length, so we have:

2s*pi/6=12
s=36/pi

Circ = 2 *Pi * Radius = 2 * Pi * 3s/8

so sub in s=36/pi

i get 27


for kedaman,

we have 2*pi*(2z)=72
and we want 2*pi*(x)=2*pi*(3/4)z=36*3/4=27

now my question is, can you prove that in your diagram, x and y is indeed the straight line? in other words, if i draw from A (one of the base corners of the triangle) to the to the point of tangency of the small circle and large circle, I am guarenteed to pass through the center of the smaller circle? your method won't be complete without this proof

[kedaman]

first a silly mistake by me, the circumference is proportional not quadratic proportional to the scale betwen the circles.
72*(x/2z)^2=72*(9/64)=10.125
should be
72*(x/2z)=72*(3/8)=27

the circle is tangent to both the arc ac and bc, which means that they share tangent in that point, as well as normal to that tangent. The normal to a circle's tangent goes trough center.

[bugzpodder]

the circle is tangent to both the arc ac and bc, which means that they share tangent in that point, as well as normal to that tangent. The normal to a circle's tangent goes trough center.

sweet! exactly what I needed! thanks!

[krtxmrtz]

Originally posted by bugzpodder
krtxmrtz , from your post, i think you screwed up one small detail: the value of s is NOT 12. 12 is the arc length...

sigh... this is what happens when you read things too fast...