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Investigate polygons with all the vertices on the lattice points of a grid. For each polygon, work out the area A, the number B of points on the boundary and the number of points (I) inside the polygon. Can you find a formula connecting A, B and I?

Great Granddad

Stage: 3 Challenge Level: Challenge Level:2 Challenge Level:2

We received correct solutions from Julia Collins, from Langley Park School for Girls in Bromley, Ross Haines, Ben Young and Philip Bennington from Mosgiel Intermediate School, and Andrei Lazanu from School 205 in Bucharest, Romania. Well done to you all.

Julia Collins said:

"I think I've found a solution to this problem, although I'm sure it's not the best solution, so I will be interested to see what other people have come up with!

We know that A + 100 is an even number, so we know that A is even. Thus we know that one of the primes making up each of the numbers A, B, C and D is 2 (because any other combination of primes is odd).

B = A + 20 = the product of a prime number and a square number

Since A is even, let the prime be 2. Then the only square number that gives a sensible date is

31 2 = 961.

So B = 1922.

So I checked the other numbers to see if the solution was valid, and it was!

A = 1902 = 2 x 3 x 317 (three primes)

B = 1922 = 2 x 961 = 2 x 31 2 (prime and a square)

C = 1982 = 2 x 991 (two primes)

D = 2002 = 2 x 7 x 11 x 13 (four primes and even)

I don't like the solution because it involves a guess, and so doesn't show whether the solution is unique or not. Can it be done purely by logical thinking?"

Andrei Lazanu was able to show that this was a unique solution:

"First I observed that the data known for the problem (A, B, C, D) corresponds to A, A + 20, A + 80 and A + 100.

Because A + 100 is even, all the other terms are also even (because 20, 80 and 100 are even).

Now, I write the data I have:

A = 2 * prime * prime

B = A + 20 = 2 * x 2

C = A + 80 = 2 * prime

D = A + 100 = 2 * prime * prime * prime

I start by considering the condition for B and observe that x must be smaller than 32 (since 2 * 32 * 32 = 2048):

x
B
A = B - 20
C = A + 80
D = A + 100
Satisfies condition
31 2*31 2=1922 1902=2*3*317 1982=2*991 2002=2*7*11*13 YES
30 2*30 2=1800 1780=2 2*5*89 1860=2 2*3*5*31 1880=2 3*5*47 NO
29 2*29 2=1682 1662=2*3*277 1742=2*13*67 1762=2*881 NO
28 2*28 2=1568 1548=2 2*3 2*43 1628=2 2*11*37 1648=2 4*103 NO
27 2*27 2=1458 1438=2*719 1518=2*3*11*23 1538=2*769 NO
26 2*26 2=1352 1332=2 2*3 2*37 1412=2 2*353 1432=2 3*179 NO
25 2*25 2=1250 1230=2*3*5*41 1310=2*5*131 1330=2*5*7*19 NO
24 2*24 2=1152 1132=2 2*283 1212=2 2*3*101 1232=2 4*7*11 NO
23 2*23 2=1058 1038=2*3*173 1118=2*13*43 1138=2*569 NO
22 2*22 2=968 948=2 2*3*79 1028=2 2*257 1048=2 3*131 NO
21 2*21 2=882 862=2*431 942=2*3*157 962=2*13*37 NO
20 2*20 2=800 780=2 2*3*5*13 860=2 2*5*43 880=2 4*5*11 NO
19 2*19 2=722 702=3 4*13 782=2*17*23 802=2*401 NO
18 2*18 2=648 628=2*2*157 708=2 2*3*59 728=2 3*7*13 NO
17 2*17 2=578 558=2*3 2*31 638=2*11*29 658=2*7*47 NO
16 2*16 2=512 492=2 2*3*41 572=2 2*11*13 592=2 4*37 NO
15 2*15 2=450 430=2*5*43 510=2*3*5*17 530=2*5*53 NO
14 2*14 2=392 372=2 2*3*31 452=2 2*113 472=2 3*59 NO
13 2*13 2=338 318=2*3*53 398=2*199 418=2*11*19 NO
12 2*12 2=288 268=2 2*67 348=2 2*3*29 368=2 4*23 NO
11 2*11 2=242 222=2*3*37 302=2*151 322=2*7*23 NO
10 2*10 2=200 180=2 2*3 2*5 260=2 2*5*13 280=2 3*5*7 NO
9 2*9 2=162 142=2*71 222=2*3*37 242=2*11 2 NO
8 2*8 2=128 108=2 2*3 3 188=2 2*47 208=2 4*13 NO
7 2*7 2=98 78=2*3*13 158=2*79 178=2*89 NO
6 2*6 2=72 52=2 2*13 132=2 2*3*11 152=2 3*19 NO
5 2*5 2=50 30=2*3*5 110=2*5*11 130=2*5*13 NO
4 2*4 2=32 12=2 2*3 92=2 2*23 112=2 4*7 NO

I stop here because from now on A is below 0.
So, 1902 is the unique year that matches all conditions for Great Granddad's birth ."

PS. Julia Collins wrote to us pointing out a slight flaw in Andrei's argument. They discussed this on the web-board, and their discussion can now be read at
http://www.nrich.maths.org/askedNRICH/edited/4087.html