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## 19 December 2017 I can solve it, George... but since you already know the answer I doubt there's a pot of gold waiting for me if I work it down to the end, and my kid is busy working problems that are paying his bills. :) Gregory 507pm - The puzzle came without an answer, so the only thing that I and my mentee got out of it was just the sheer joy of solving it (as is often the case). The solution set for R in [0, 0.5] is pretty straightforward. It gets more interesting in the (0.5, Rmax] domain. For R<= 0.5: score area = [0.25 - πr^2/4 -(0.5-r)^2]
I used 6 drawings (in EXCEL) to illustrate various values of R = 0+; 0.2; 0.5; 0.5858. Needed a very detailed drawing for R>0.5 for understanding what is happening. My drawings used gold color for the scoring area for R<0.5, and green for R>0.5

To calculate values for R>0.5 required creating a kite shape in the unit square/circle combination; then calculating the lengths of the 3 sides of a half-kite; then using Heron's formula to find the area of the half-kite; doubling this area to find the area of the whole kite; then subtract the area of the sector of the circle.

..........Score
0, 0, 0.0%
0.05, 0.0455, 18.2%
0.10, 0.0821, 32.9%
0.15, 0.1098, 43.9%
0.20, 0.1286, 51.4%
0.25, 0.1384, 55.4%
0.26, 0.1393, 55.7%
0.27, 0.1398, 55.9%
0.28, 0.1400, 56.0%
0.29, 0.1398, 55.9%
0.30, 0.1393, 55.7%
0.35, 0.1313, 52.5%
0.40, 0.1143, 45.7%
0.45, 0.0885, 35.4%
0.50, 0.05365, 21.5%
0.5258, 0.0242, 9.7%
0.5508, 0.0079, 3.2%
0.5758, 0.0007, 0.3%
0.5858, 0.0000, 0.0%

Since the Snipping Tool doesn't copy onto this area, understanding the formulas below will be tough without seeing the drawing.
A pix is really worth 1000+ words sometimes.

For finding values R>= 0.5 to 0.5858 Rmax took some work.
Area of kite (ABCD)- area of circle sector (BCD) = green scoring area
CE = 1-R
ED = x = sqrt (2R-1)
DA = 1 - x - R
AC = (sqrt (2))/2 - [(R-0.5)*sqrt(2)]
CD = BC = R
Use Heron's formula for area of triangle ACD (half the kite):
s = [(DA + AC + CD)/2]
then AREA = sqrt [s*(s-DA)*(s-AC)* (s-CD)]
So green area = 2*AREA - (f/360*pi*R^2)
where f = 90-2q; q = arccos [(1-R)/R].
q=angle of base triangle inside the circle; f = small angle of kite from the center of the circle with radius R.

..R,.....CE,.....ED=x,....DA,.......AC,..........CD,.........s,
0.500, 0.500, 0.000, 0.500, 0.707106781, 0.500, 0.853553391,
0.525, 0.475, 0.224, 0.251, 0.671751442, 0.525, 0.724072322,
0.550, 0.450, 0.316, 0.134, 0.636396103, 0.550, 0.660084169,
0.575, 0.425, 0.387, 0.038, 0.601040764, 0.575, 0.606871215,
0.5858, 0.4142, 0.414, 0.000, 0.5857814, 0.5858, 0.585779619,

AREA,.....q,......f,....f/360*pi*R^2,.....green,....Prob.,
0.1250, 00.00, 90.00, 0.196349541, 0.05365045, 21.5%,
0.0597, 25.21, 39.58, 0.095206996, 0.02420477, 9.7%,
0.0301, 35.10, 19.81, 0.052285136, 0.00791236, 3.2%,
0.0080, 42.34, 5.31, 0.015334482, 0.000688726, 0.3%,
0.0000, 45.00, 0.00, -5.03737E-06, 1.00758E-05, 0.0%

(f/360*pi*R^2) is the formula for the sector drawn in the circle.

Merry Christmas! Well done sir and a Merry Christmas to you too! ;-) DonC 607pm - Excellent Mr Carrera. A simple way to solve for R in [0,0.5] is to attack the quarter square in which the forbidden areas are the quarter circle of radius R in the outside corner and the quarter square measuring (1-2R)/2 on the inside corner. Taking the derivative wrt R of the remaining area quickly gives the optimal R = 2/(pi + 4) and Pmax = 4/(pi + 4). Good work on ferreting our the probs for R in (0.5, Rmax] where Rmax is seen to be the circle centered on the square's diagonal, passing through one corner, and tangent to the two opposite sides from that corner. This requires that Rmax + Rmax*cos(pi/4) = 1 which yields the correct Rmax = 0.5858. And you're, of course, right about making the whole explication simpler with appropriate graphics. Merry Christmas! George,

I'm a retired EE, and I've been volunteering one day a week for the last 9 years in a local Middle School, in their math & science classes. I am currently reviewing some math work with three 6th graders. I've already covered Carl Friedrich Gauss's way to add an infinite number of integers, and how to prove the Pythagorean Theorem. It's within their grasp, and they've used that info to solve several problems.

But I will show them what to learn from this problem. Not what you think though. Sad to say, I spent 30 minutes wondering why using a 'modified' Pythagorean theorem (a^2 + b^2 = c) did not work. Fortunately, I only wasted a half hour before the idiot part of my brain went to sleep, and I found the missing "c^2". And it wasn't a fatal error. So good fortune all around.

Yes, I set the problem up the same way you described, with a smaller quarter-square. And used calculus to find the max value. But what's the fun in that? I emphasize to the kids that there is usually more than one way solve a problem - sometimes clever, sometimes not. So for middle schoolers, using geometry, trig & algebra can often times be more understandable.
Don C. DonC 628am - I too was taught that there is more than one way to climb a mountain. DonC 628am - Upon closer inspection, our answers for R in (0.5, Rmax] differ a bit. Could you email me your graphic so I can append it to this post, and I will also do the same for my solution in that range. You can reach me at gjrebane@gmail.com. Thanks. re my 135pm - In a subsequent communication Mr Carrera demonstrated that in fact his solution and mine produce identical numerical results. My error for not picking this up in the first place.

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