Below is the general case submitted by Roland Sampy.
There seems to be at least 2 expressions for r.
We had a solution, but it looked a little clumsy. Moreover, we lost it :)
There seems to be at least 2 expressions for r.
We had a solution, but it looked a little clumsy. Moreover, we lost it :)
So, do help us to recover... Oh yes, do enjoy as well...
ReplyDeleteEach value of a < 2 and b <2 we have 2 positions of C and 2 positions of B.
So r is not only depend on a and b but also depend on position C relative to B .
Peter Tran
Sounds correct. In that case the final expression for r
ReplyDeleteshould give two separate roots. (Most likely quadratic)
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The problem appeared as problem “fox 347” at 8foxes.blogspot.com in 2012 and solution is not available.
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I tried to solve it by taking a large circle of radius R with center at (0,0) and two chords of length a>b at point(0,-R). Assuming that small circle center lies at bisector chord, the question is to find the radius of inscribed circle?
Updated on 19th October 2021
Today, I have switched to trigonometric representation of equations for chords a>b and bisector chord. However, all chords start at (0,-R). Further, I have chosen the small circle center on the bisector at (x,y). I have imposed two conditions on the center (x,y) of small inscribed circle:
(1) the expression for distance of point (x,y) on bisector chord, from chord b is made equal to small circle radius “r”,
(2) the distance between large circle center (0,0) and small circle center (x,y) is taken equal to difference of their radii “(R-r)”.
The equations for chords and bisector are:
(1) y=tan(α)x-R, sin(α)=1/2(a/R), for chord “a”
(2) y=tan(β)x-R , sin(β)=(1/2)(b/R) for chord “b”
(3) y= tan[1/2(α+β)]x-R for bisector chord.
Distance of (x,y) on bisector from chord “b”,
(4) r=-xsin(β)+ycos(β)+Rcos(β)
and lastly, the distance between two circle centers,
(5) √[x²+y²]=(R-r) and r>0.
These equations have been solved on WolframAlpha to determine x, y, r and results are consistent with geometric display.
Typical script on WolframAlpha is as follows:
solve y= tan[(θ+ϕ)/2]x-R, r=[-xsin(ϕ)+ycos(ϕ)+Rcos(ϕ)],√(x²+y²)=(R-r), r>0, θ=asin[a/(2R)], ϕ=asin[b/(2R)], R=10, a=16,b=10
However, the closed form of expression for radius “r” of small inscribed circle is the aim of this question on this forum?
Remark
All chords are covered by choosing (0,-R) as the start point. Shorter chord “b” is always in the right part of the large circle. Long chord “a” can be easily switched in left part by converting α to π−α.
Update 21st October 2021,
A Geometric Solution
Case-I
Both chords are on two sides of circle center a>b
Draw perpendicular from large circle center and small circle center on both chords:
(1) the tangent chords give X+a/2 = Y+b/2. ——-(1)
(2) Further two following relations are:
X² = (R-r)² - [r - √[R² - (a/2)²]]². ———————-(2)
Y² = (R-r)² - [√[R² - (b/2)²] - r]². ————————(3)
The equations (1) to (3) were solved for radius r of inscribed circle.
Case-II
Both chords are on one side of large circle center,
In this case, the equation (2) is modified as below:
X² = (R-r)² - [r + √[R² - (a/2)²]]². ——————-(2)
Once again equations (1) to (3) are solved for r radius of inscribed circle.
In fact closed form of solution for both Case-I and Case-II are available from above solutions and will be posted in future.
**Update 23rd October 2021**
ReplyDelete**A closed form of solution to problem “fox 347” is presented below**
**A further simple and unified expression is based on chords angle for pair of chords (a,b) with a>b>0 in a circle of radius R with center at(0,0). Both chords start at (0,-R).**
**Chord "a" has angle "α" and chord "b" has angle "β" with X-axis at (0,-R). One uses the following cosine relation in triangle with sides (R-r), R and bisector chord between (0,-R) and inscribed circle center with length r/sin((α−β)/2). The angle opposite (R-r) side is [π/2-(α+β)/2]. The cosine law is written as**
**(R-r)²= R² + [r/{sin((α−β)/2)]² - 2Rr[sin((α+β)/2)/sin((α−β)/2)]**
**The radius of inscribed circle "r" is written after algebraic simplification as:**
**r/R = 2x(y-x)/(1-x^2)**
Where,
**x=[sin(α−β)/2]**
**y=[sin(α+β)/2]**
**This , then, is the final answer to problem “fox 347”.**
For Readers interested in solution of problem “fox 347”, I recommend to view the details of solution, examples and discussion at web site math.stackexchange.com using “fox 347” as search string. Enjoy the problem and it’s solution.
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