George Rebane
For a little relief from the travails of the election season, I offer a little bagatelle for your intellectual entertainment – something provided by neither candidates nor commentators during these last couple of years. This jumped from my fevered brain one night as I was noodling over some candidate problems for Techtest 2025 coming up next spring.
From the figure we see a green semicircle of radius R with center at O in which resides the smaller blue semicircle of radius r with center at E anchored, as shown, on the right end D of the large semicircle’s diameter. From the left end A of the diameter issues the red line AC which is tangent to the small semicircle at B and terminates on the circumference of the large semicircle at C. The only data available is the value of R. The task at hand is to find the value of r and the length L of the line which is equal to the perimeter length of the small semicircle, or demonstrate that no such solution is feasible.
At an appropriate time in the future I will append this post with the solution, and also give you immediate feedback if you post your solution as a comment or email it to me.
[4nov24 addendum] Well that didn’t take long; we have some very astute readers led by Efox who submitted the correct solution in his 232pm comment. What made me smile was the description of his approach using Excel™. We would call it the direct bigger hammer method. He correctly argued that a solution is indeed feasible and immediately recognized the scalability of the problem which allowed him to deduce that r must be a fixed fraction k of the given radius R – i.e. r = kR. He then extracted a relevant algebraic equation from the geometry that contained both R and r – see the first equation below the figure. Realizing that an analytical solution for r would be too tedious to pursue, he got out Excel and used his bigger (“old fashioned”) hammer by inserting a numerical value for R and then kR for r, and then started incrementally varying k until his equation was satisfied with the k interpolated to yield at least a five-place decimal accuracy. Bravo.
The numerical optimization algorithm I used is one of the tools in the Solver add-in of the Excel spreadsheet. Solver is a very valuable part of that spreadsheet, and when installed is accessible under the ‘Data’ tab. The tool can be used for everything from solving equations (like above) to finding ‘best’ values for certain input parameters such as corporate ‘operating points’ with complex cashflow models. It has been a lifesaver for me over the years, and I could wax eloquent over its manifest benefits. But alas, such accolades are out of scope here.
Do I correctly understand that R is not the only available data because it is also specified that L is equal to the perimeter length of the small semicircle?
Posted by: Michael R. Kesti | 03 November 2024 at 02:50 PM
MichaelK 250pm - Good clarifying question. R is indeed the only *data* that is provided. The rest describes the constraining geometry of the problem, such as r and the line's length L which must be computed from the solution once given the numerical value of R.
Posted by: George Rebane | 03 November 2024 at 03:07 PM
Let's take the minimum & maximum values of r to see if there is a solution.
rmin = 0, in which case L=2R. Perimeter P = 0.
Let rmax = R (a coincident semicircle to the green one), then L=0. P = (2+pi)*R (Add the semicircle + line segment AD.) Both P and L are continuous functions, so they will intersect if you plot them together. So there is a solution.
Look at the diagram. Angle ABE is 90-deg; angle ACD is also 90-deg, since it is inscribed and intersects the diameter. So these 2 triangles are similar right triangles. So we want to solve for AC (which is L) in the equation (AD/AE) = (AC/AB).
Let r = k*R. Solve ABE rt triangle for AB. AE= R(2-k), EB=k*R, AB^2=[R(2-k)]^2 - k^2R^2.
AB=2R√(1-k); by inspection, AD = 2R; AE = R(2-k). AC=(AD*AB)/AE.
So AC = L= 2R* 2R√(1-k) /(R(2-k)) = 4R*√(1-k)/(2-k). Need to set this equal to k*R(2+pi) (the perimeter).
So 4*√(1-k) / (2-k) = k(2+pi).
Way too complicated for an analytical solution, so I used Excel.
k = 0.37826, then r = 0.37826R and L = 1.9448R
Posted by: The Estonian Fox | 03 November 2024 at 03:23 PM
EFox -323pm - Correctamundo! And your use of Excel's gradient search Solver was also the most efficient way to go. Analytically solving the resulting quartic polynomial is like getting into a land war in Asia.
Posted by: George Rebane | 03 November 2024 at 06:40 PM
Actually George, I did it the old-fashioned way. Stepped k-column from 0.1 to 0.5 by 0.01. Narrowed easily to 0.38, stepped 0.0001 to get to 4 places, then I eye-balled the final 5th-place digit. Took less than a minute. Since I don't use Solver much, it would have taken me longer to remember how to use it. So, efficient enough for me.
I still use Excel 2007, because I have the ACTUAL DISK for Office 2007. I don't need no stinkin' cloud storage or yearly fee. Office 2007 has been re-loaded at least 7 additional times onto my & my daughter's new & used computers over the years.
Posted by: The Estonian Fox | 04 November 2024 at 05:26 AM
And besides, what country would be stupid enough to get into a land war in Asia?
OK, OK, more than twice?
Posted by: The Estonian Fox | 04 November 2024 at 05:40 AM
Efox 540am - What country? How about the US? Let us count the ways - Spanish-American (1898+), anti-Bolshevik expedition (Arkhangelsk in September 1918), WW2, Korea, Vietnam, Iraq, Afghanistan,... ;-)
Posted by: George Rebane | 04 November 2024 at 02:01 PM