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As a fun diversion, here’s your Easter Egg problem albeit with a round egg. Given three mutually perpendicular line segments of lengths a, b, c, as shown in the figure, Problem: (A) derive the expression for the radius r of the circle, then (B) for (a, b, c) = (13,6,18) calculate the value of r.
[23apr25 update] My solution is shown below.
lol. Let’s see Dr Rebane. My skills are rusty.
X is the distance from line C to the center of the circle.
X = (c squared minus b squared minus ac) divided by 2b
R = the square root of (((b plus x) squared) plus (one half a squared))
If a = 13 and b = 6 and c = 18. Then r= about 12.349 and some change.
How did I do professor?
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There are 3 points of interest that intersect the circle – point A, at the bottom of the “a” line; point B at the intersection of the “a” and “b” line; point C at the bottom of the “c” line. Any 3 points determine a circle, or lie on the circumference of a circle.
Can use the equation of a circle (x-xi)^2 + (y-yi)^2 = R^2. So need 3 equations, using all 3 points, to find x, y, R, where (xi, yi) are the 3 coordinates of A, B, C. I don’t like to use this method. Personal preference.
Trig functions can be used also. But that’s cheating, when algebra can do.
You have 2 chords whose end-points lie on the circle – AB, AC. If you use the equation of the perpendicular bisector of the mid-point of both chords, then find their point of intersection, you have just found the center of the circle. Then can use Pythagoras theorem to find the radius.
Take point C as the origin of our coordinate system. C=(0,0); A = (-b,c-a); B=(-b,c). Mid-point of AB = (-b, c-a/2). Slope = infinite. So perp-bisec has slope m=0, so y1= mx1+b = c-a/2. That’s our 1st of the 2 lines.
2nd mid-pt = (-b/2, (c-a)/2). Slope of AC = (y1-y2)/(x1-x2) = (c-a)/(-b). So slope of perp-bisec = b/(c-a). So line bisector y2 = bx/(c-a) + b. This intersects line y1 at the center of circle.
Use point-slope form to find equation of 2nd line.
y2-(c-a)/2 = b/(c-a) * (x+b/2). Set y2 = y1 = c – a/2. Solve for x. End result:
x=c(c-a)/2b – b/2. This is the X-coordinate of center of circle. To find the long leg, add b to this, so x=c(c-a)/2b + b/2. And y=c-a/2, the Y-coordinate of center, but short leg subtracts (c-a) from this, so y=a/2.
Right triangle has short leg y = 6.5; long leg x = 10.5. Center of our circle is (4.5, 11.5).
R^2 = 6.5^2 + 10.5^2 = 42.25 + 110.25 = 152.5. R = 12.35.
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Had to use algebra because it is been 35 years since I did a trig problem. I tried with trig but gave up and went back to old Pythagoras.
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BarryP and EF – You both nailed paths to the numerical answer and gave the correct answer. But you didn’t do part A of the problem which required you to derive and express an elegant form of R = f(a,b,c). Back to the drawing boards 😉
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That’s the best I can do Professor. I’ll take a C+. Lol
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R^2 = x^2 + y^2.
R^2 = [c(c-a)/2b + b/2]^2 + [a/2]^2
R= [[[c^2 – ac + b^2]/2b]^2 + [a^2/4]]^0.5, so R = f(a,b,c)
R= [[(18^2 -13*18 + 36)/12]^2 + 169/4]^0.5 =12.35
Not sure what else you’re looking for George; maybe our definitions of ‘elegant’ are different.
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The function can be reduced much more but I get lost in the math relatively quickly.
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R^2 = x^2 + y^2
R^2 = [[(c(c-a)/2b + b/2]^2 + a^2/4]
R = [[c^2 – ac + b^2]/2b]^2 + a^2/4]^0.5 so this is R=f(a,b,c)
R = [[[324 – 13(18) +36]/12]^2 + 169/4]]^0.5
R = [10.5^2 + 42.25]^0.5 = 12.35
Not sure what you’re looking for George. Maybe our definitions of ‘elegant’ are different.
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Figured it out.
X = (c squared minus b squared minus ac) divided by 2b
R = the square root of (((b plus (c squared minus b squared minus ac) divided by 2b squared) plus (one half a squared)) squared) plus (one half a squared))
R = square of (2b2/2b plus (c squared minus b squared minus ac) divided by 2b) plus (one half a2)
R = square of ((c2 + b2 -ac)/2b) squared + (one half a2))
R = square of (((c2 + b2 -ac) squared) divided by 4b2) + one half a2
R = square of (((c2 + b2 – ac) squared) divided by 4b2) + 2a2b2/4b2)
R = square of ((((c2 + b2 -ac) squared) + 2a2b2) divided by 4b2)
R = square of (((c2 + b2 – ac) squared + 2a2b2)) divided by 2b?
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Typo removed.
Figured it out.
X = (c squared minus b squared minus ac) divided by 2b
R = the square root of (((b plus (c squared minus b squared minus ac) divided by 2b squared) plus (one half a squared))
R = square of (2b2/2b plus (c squared minus b squared minus ac) divided by 2b) plus (one half a2)
R = square of ((c2 + b2 -ac)/2b) squared + (one half a2))
R = square of (((c2 + b2 -ac) squared) divided by 4b2) + one half a2
R = square of (((c2 + b2 – ac) squared) divided by 4b2) + 2a2b2/4b2)
R = square of ((((c2 + b2 -ac) squared) + 2a2b2) divided by 4b2)
R = square of (((c2 + b2 – ac) squared + 2a2b2)) divided by 2b?
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EF 359pm – Our definitions of elegance concur.
Barry 623pm – not sure of your translation of algebra, but if your formula can be reduced to the Fox’s R = [[c^2 – ac + b^2]/2b]^2 + a^2/4]^0.5, then you did express R = f(a,b,c). In any case, thanks for the valiant effort.
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That was fun. Thank you George. Happy Easter!
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