16

The fraction is Without loss of generality let the radius of the smaller circle equal 1. Edit: Bubbler has observed in a comment that from here it is faster to continue like this We have $JA=JD=r$, and since JH bisects the chord AB, $\angle{JHA}=90^\circ$. We also have $$JA^2=HA^2+(HI^2+IJ^2)$$ $$r^2=1^2+(1^2+(1+\sqrt{3}-r)^2)$$ Solving for r, $$r^2=1^2+1^...


12

My answer is Here is the C code, which took about 3 seconds to run #define _USE_MATH_DEFINES #include <stdio.h> #include <stdlib.h> #include <math.h> #define MAXORDER 20 // max 20 in dima[], dimb[] #define ORDER 16 #define angled dima #define CIRCLE 1024 // 'degrees' in a circle #define CIRCMASK (CIRCLE-1) #...


12

UPDATE: Added more formal proof, see bottom Here is a less rigorous but maybe more intuitive answer: More formal proof:


10

Here's a proof we can't do it with less than circles. Each circle covers at most 8 points. In fact, the max is 6 points except for these five 8-point circles: x O x O x x O O x x x x x x x x x x x x x x O O x O x x x O O x x O x x O O x x x x O O x x O x x O x x x x x O x x O x O x x O x x O x x O x O x x O O x x x O x O O x x O ...


9

I claim the answer is and here's why:


8

Partially inspired by Deusovi's answer, here's another argument. First, Then,


7

Solving this one manually isn't actually that bad. During horizontal and vertical runs they must alternate. So without loss of generality if 1 is vertical, then two is horizontal. The only real choices you get are with the side lengths that are the hypotenuse of a right triangle. Those possibilities are: short leg long leg hypotonuse 3 4 5 6 8 10 5 12 13 ...


7

The following Python 3 code tries to find all paths that start by moving to (0,1) and return back to origin from (16,0). The vector candidates are hardcoded to eliminate unnecessary branches. (I know they can be reordered to eliminate branches earlier, but I'm too lazy.) It took ~5 minutes to run on my PC. import time def cmp(a, b): return (a > b) - (a ...


7

Here's something resembling a solution using mostly just image editing magic and Pythagoras: Here's the result of that operation: Adding up the parts gives the diameter of the bigger circle in small circle radius units: Which gives us the ratio of radiuses Which we can square to get the ratio of areas And as long as we remember that there are two of the ...


5

All these explanation are way too complicated! I propose a simpler one.


5

We can do Things we can't do: This still leaves open the possibility of


4

Yes, as long as you're not too picky about the shape of your page. If you can curl your page so that it forms a cylinder, then you can draw a line forming a ring around the cylinder and put line segments on just that ring so that they all overlap and hit each others' interiors. Then the arguments in the other answers don't make sense, as there is no longer a ...


3

You can do this by painting as follows:


3

The best answer is "time to get a new clock" but the two best times are 9:05:25.452 and 2:54:34.548, with angles HM=119.83°, HS=120.00° and MS=120.17°. Python code: import numpy as np def bestN(a, b, c, N=1): e1 = np.abs(((b-a) % 60) -20) e2 = np.abs(((c-b) % 60) -20) e3 = np.abs(((a-c) % 60) -20) return np.argpartition(e1+e2+e3,N)[:...


3

Continuing Paul's result: Let's check them, using the fact that since only areas are considered we can do arbitrary linear maps to fix points: Therefore,


2

This may be a bit too simple. and not the intended solution, but


2

I believe the answer is Not a proof, but a heuristic argument: EDIT: A "proof" by computer. At least this gives the evidence for the yes/no part, so that a more serious attempt can be made in the right direction.


2

This answer is a bit of a sanity-check for the other answers posted. Note: Depending on how you view the image, its dimensions might be different than my measurements here, but the individual dimensions here don't matter; rather, it's the ratio between them that matter. And since ...we see that the portion of the large semicircle that is shaded is likely


2

Are you looking for something like this?


1

Yes, easy:


1

Answer 1 Answer 2 Answer 3


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