# STEM quiz #2

It appears that @Ether is no longer posting, so I’ll try to take up the slack of his math and physics quizzes with my STEM quizzes. OBTW, if you haven’t seen them and are interested, search on @ether and quiz for a cornocopia of interesting problems.

Draw the following circles on a Cartesian coordinate system (x, y, radius): (0,0,1), (3,0,2), (0,4,3)*. Each of the circles will be externally tangent to the other two. [It may be helpful to prove this to yourself.] Draw a fourth circle externally tangent to the three original circles, centered at “D” and inside the “talon” shape they define. What are the radius and center coordinates of this smaller circle?
* Fixed; thanks @Jon_Stratis !

(0.913043,0.869565,0.2608695)

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I feel like solving this with CAD is cheating… but that’s what I did at first

Summary

Now, working on a purely mathematical solution… that might end up being tomorrow morning!

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Did your answer come out to be a rational number? Because it turns out that the radius and coordinates are all rational., and I’m only going to approve rational answers.

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Awww. I’ll probably give it another go tomorrow morning then, have a lot of work due tonight to deal with at the moment.

Unrelated, but who was/is ether? I know they made a lot of popular swerve math posts but don’t know much else. Was it just someone who was good at math (and explaining it, the hard part) ?

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Ether is/was a person who liked going into the math of omni-directional drives, AND any other math he could get his hands on. I know a few CDers know his real name, but I can’t say for sure if anybody’s talked to him recently.

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I had some communications with Ether back around 2015.
We were both from South Bend and went to Purdue.
Ether = Physics 1973. Me = Engineering 1984.

He lived in SW Michigan and health issues prevented him from traveling far, but I had hoped we could meet in Detroit at some point.

He was an iconic presence in many discussions here. His avatar of Ernest Rutherford may give you a hint to his inclinations. His last post that I can find is from 2018.

Hoping he is lurking here and will chime in.
Maybe with an elegant solution.

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When revisiting a mathematical solution, I discovered something…

These are not the circles @GeeTwo drew. Inferring dimensions by comparison, he drew (0,0,1), (0,4,3), and (3,0,2). Both my CAD drawing and @SiddheshDash made that same mistake!

On to the solution:
r=6/23, x=20/23, y=21/23

Summary

I created a system of 3 equations using the pythagorean theorem and the 3 triangles highlighted below:

For each triangle, our X and Y values are based on the centers of the known circles and the center of the unknown circle. Likewise, the hypotenuse is based on the radius of the known and unknown circles.

x^2 + y^2 = (r+1)^2
x^2 + (3-y)^2 = (r+2)^2
(4-x)^2 + y^2 = (r+3)^2

As for solving that system of equations… Wolfram Alpha to the rescue!

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We had the honor to be visited by him in the summer of 2017. He just stopped in to our teams open house event. Hope he’s still watching.

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Exactly what I did, I just thought about it a bit differently.

Solution

Rather than thinking about tangent circles, I “extended” the radius of each circle by some value r and drew three new circles from the centers of the original circles with radii r larger than the originals.

The point where all three circles meet is the center of the middle circle. It’s then easy to solve for the point (x,y) where all three circles meet by solving the three equations simultaneously for x, y, and r.

(x-0)^2 + (y-0)^2 = (1+r)^2 \\ (x-3)^2 + (y-0)^2 = (2+r)^2 \\ (x-0)^2 + (y-4)^2 = (3+r)^2

I used Mathematica to solve it because I’m lazy (but it could be done by hand).

The first solution isn’t valid because r can’t be negative, so the solution is

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My Solution, not as simple as @Jon_Stratis' or @AriMB's

Strategy: Use the law of cosines to get cosine and sine of the three angles at one of the large circle’s center and use an angle sum formula. Here we’ll do A.

BC²=AC² + AB² - 2 AC AB cos(CAB)

As the line segment connecting the centers of externally tangent circles is the sum of the radii, we can easily see that AB=3, AC=4, BC=5. Letting the radius of the small circle be r, we also have: AD=1+r, BD=2+r, CD=3+r.

Solving for the cosines, we have:

cos(CAB) =-(5²-4²-3²) / (2·4·3) = 0/24 = 0
cos(DAB) = -((r+2)² - (r+1)² - 3²) / (2(r+1)3) = (3-r)/(3r+3)
cos(CAD) =-((r+3) ² - 4² - (r+1)²) / (2·4(r+1)) =(2-r)/(2r+2)

When cosφ = x/y , sinφ = √(1-x²/y²) = √(y²-x²)/y, so

sin(DAB) = √((3r+3)² - (r-3)²) / (3r+3)= √(8r²+24r)/(3r+3)
sin(CAD) = √((2r+2)² - (r-2)²) / (2r+2)= √(3r²+12r)/(2r+2)

But cos(α+β)=cosαcosβ – sinαsinβ, so

cos(CAB) = cos(CAB) cos(DAB) - sin(DAB) sin(CAD)

As the left hand side is zero and the denominators of the two terms on the right match and are clearly positive, we’ll multiply by that denominator and just look at the numerators:

0= (r-3)(r-2) - √(8r²+24r) √(3r²+12r)
√(24r⁴+168r³+288 r²) = r² - 5r + 6
24r⁴ + 168r³ + 288r² = r⁴ -10r³ +37r² - 60r + 36
23r⁴ + 178r³ +251r² +60r - 36 = 0

Inspired by the rational root theorem and that the central circle is about a quarter the radius of the smallest outer circle (radius 1), we try factoring this by 23r-6, and succeed! This factors as (23r-6)(r+6)(r+1)². The negative roots are not interesting as a solution to this problem, so r=6/23.

Given that r is rational, let’s see if the coordinates of the inscribed circle are, as well : 1+6/23=29/23. It is well known that 29² = 20² + 21², and using the Pythagorean theorem, it is easy to prove that (21/23,20/23) is 29/23 from A, 52/23 from B, and 75/23 from C.

But then, when I did this solution, I hadn’t assigned coordinates for the circle centers, I was just working with the radii of 1, 2, and 3 (the simplest case I could think of where none of the three radii were equal). I didn’t realize that the centers made a right triangle until I found that cos(CAB)=0.

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This is such an elegant solution. Very nicely done.

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Based on some papers I found in technical journals, Ether is his actual surname.

I learned his name from @Richard_Wallace after supposing that he wasn’t actually a real person (taking the word “Ether” a bit too literally).

I do hope he’s doing well still. I still owe him a beverage of his choice for his help many years ago.

Are we allowed to say it? Lest Mr. Ether’s first name achieve Voldemort status?

Just a bit of follow up:
Using @Jon_Stratis (or @AriMB)'s equations, note that all have identical second order terms. Equation 1 only has linear term in r, 2 in y & r, 3 in x & r. Therefore, subtracting Eq 1 from Eq 2 and from Eq 3 will give linear relationships between y and r and between x & r. In particular, x=1-r/3, y=1-r/2. Plugging these into equation 1 gives a quadratic in r which can be solved by factoring or the quadratic equation to give the two solutions for r, 6/23 and -6.
6/23 is the desired solution.
Note that using -6 to calculate x & y, and drawing a circle of radius 6 defines a circle which circumscribes the three original circles.

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