Gadfly

avatar
BenutzernameGadfly
Punkte121
Membership
Stats
Fragen 0
Antworten 50

 #1
avatar+121 
0

 

 

I have redrawn and recolored the image to make the following claims:

1) All regions of the same color in this image have the same area., and

2) the measure of the area of each green region is twice that of the area of any blue region.

 

If these claims are proved, the ratio of the smallest area to the largest area would of course be 1/2.

The proofs I give will be mostly based on the image, something that math people normally discourage for good reasons, but in this case should be justified since the facts are well known and the details could become overwhelming.

 

First we notice that regions 1 and 18 together make up half of the area of the rectangle ABGF, which in turn has area equal to half of the square CHFG that has area 1 square unit; so

Area of 1 +Area of 18 = 1/4.

Next consider regions 18 and 19, which again together make up a triangle with area 1/4 of the square. So

Area of 19 + Area of 18 = 1/4.

This forces not only  regions 1 and 19 to have the same area, but all the othe other green regions other than 16, 24, 21 and 7 due to obvious congruence.

 

To show that the latter 4 regions have the same area, note that 16, 24, 15, and 17 together make up a triangle that has area 1/2, and that 24 and 15 together make up a triangle with area 1/4, which implies that 16 and 17 also add up to 1/4. Since 15 and 17 are congruent and therefore have the same area, we are forced to conclude that 16 and 24 (and therefore 21 and 7 by congruence) also have the same area.

 

Similar reasoning shows that all blue regions also have the same area.

 

To prove that each green region has area twice as large as each blue region, consider regions 24 and 18 and refer to the image. Region 18 is divided into two congruent triangles by means of the line segment CE, each of which is congruent to region 18 triangle.

Of course this proof consists of mainly 'hand-waving' and is mainly heuristic, but the reader may provide the details!!

12.11.2019
 #1
avatar+121 
+4

I have a solution to this that is very involved. But I will try to outline it even though it may not be the most elegant solution. Here it goes:

Here I assume that x is the number of people stranded on the island, n is the number of coconuts gathered and piled before dark, and k1, k2, k3, k4, and k5 are the number of coconuts, respectively, in each of the x piles after the additional coconut is given to the monkey and before the night watches 1 through 5 hide their own portion. We can write a system of 5 equations involving the 7 variables(n, x,  k1, k2, k3, k4, and k5) as follows:

\(k_1x= n -1\)

\(k_2x=n-k_1-2 \)

\(k_3x=n-(k_1+k_2)-3\)

\(k_4x=n-(k_1+k_2+k_3)-4\)

\(k_5x=n-(k_1+k_2+k_3+k_4)-5\)

 

This system evidently has an infinite number of solution since the number of variables is larger than the number of equations and can be approached in multiple ways, but all methods would involve a reduction of the number of variables. Here we are asked to find the minimum value of n for a given value of x, so we will reduce the number of equations to only one, a quintic polynomial equation in the variable x, the number of stranded people, and containing two other variables, n and k5. I will only do step one of this process since the rest can be accomplished similarly.

Multiply the 5th equation by x to get:

\(k_5x^2=nx-(k_1+k_2+k_3+k_4)x-5x\)

Replace k1x, k2x, k3x, and k4x by their values given by the 1st through 4th equation in the system to get :

\(k_5x^2=nx-[(n-1)+(n-k_2-2)+(n-k_1-k_2-3)+(n-k_1-k_2-k_3-4)]-5x\)

\(=nx-4n+10+3k_1+2k_2+k_3-5x\)

Note that k4 has been eliminated from the equation and by repeating this process 3 more times , we can also eliminate k1, k2, and k3 to get the quintic we were aiming at:

 

\(k_5x^5=(n-5)x^4+(10-4n)x^3+(6n-10)x^2+(5-4n)x+(n-1)\)

 

Moving all terms to the left side we get:

 

\(k_5x^5+(5-n)x^4+(4n-10)x^3+(10-6n)x^2+(4n-5)x+(1-n)=0\)

 

Well this is the formula we were looking for. The author of this question claims that if the there are only six people stranded on the island, the initial number of coconuts would have to be a minimum of 7,771. Let's see if this holds true. Substituting 6 for x in the above equation we get:

\(0=7,776k_5+1,296(5-n)+216(4n-10)+36(10-6n)+6(4n-5)+(1-n)\)

\(7,776k_5+4,651-625n\), which implies that

 

\(n=\frac{7,776k_5+4,651} {625}\)

 

Here we need to find the smallest value of k5 that results in an integer value for n. I gave the problem to EXCEL to solve. EXCEL did a search for k5=1, 2, ..., 10,000, we decided to limit the search, and came up with 16 values for k5 that resulted in integer values for n.The smallest of these values was k5=624 and the largest was k5=9,999. Plugging these for k5 in the formula for n, gave us, what do you know, n= 7,771 and n= 124,411.

11.11.2019