Answer to Problem of the Week for 11/18/2002

Answer to November 18, 2002 Problem

by Michael Winckler

The Pentagram Problem

Can you number the five corners and five crosspoints of a pentagram with the numbers 1, 2, 3, 4, 5, 6, 8, 9, 10, and 12 in such a way that the sum of the four numbers on each of the five lines is equal? (See diagram below). I have lettered the ten intersection points of the pentagram to assist you in sending in your solution. Just give the values for A, B, C, D, E, F, G, H, I, and J. Extra Credit: Could you number the ten points with the numbers from 1 to 10 so that the five lines of four numbers add up to the same sum?
Solution to the Problem: Notice first that each pair of lines have exactly one point in common and that each number is in exactly two lines. Counting each number twice, summing up and dividing by the number of lines (5) yields 24 as the sum of each line. There are five ways for 12 to be in a sum: (a) 12, 9, 2, 1 (b) 12, 8, 3, 1 (c) 12, 6, 5, 1 (d) 12, 6, 4, 2 (e) 12, 5, 4, 3 For the number 1, there are six ways to form the sum of 24: (f) 1, 2, 9, 12 (g) 1, 3, 8, 12 (h) 1, 4, 9, 10 (i) 1, 5, 6, 12 (j) 1, 5, 8, 10 (k) 1, 6, 8, 9 We have to choose 2 lines with a 12 in it -- and all the other numbers must appear at most once. This is only possible taking (a, e) or (b, d) above. This shows that 1 and 12 have to be in the same line. For each combination, we now have to choose another line from the second set to find the second line containing 1. I tried all six combinations (f through k) with (a, e) but none of those worked. Then I tried (f through k) with (b, d) and found a solution with (b, d, h). So I placed 12, 8, 3, and 1 on one line; then put 12, 6, 4, and 2 on another line, and made it work with 1, 4, 9, and 10. There was only one position left for the number 5, and everything added up! A = 12 B = 8 C = 3 D = 1 E = 10 F = 4 G = 9 H = 2 I = 5 J = 6 Using similar logic with the numbers 1 to 10, it can be shown that it is NOT possible to solve that problem.

Can you number the five corners and five crosspoints of a pentagram with the numbers 1, 2, 3, 4, 5, 6, 8, 9, 10, and 12 in such a way that the sum of the four numbers on each of the five lines is equal? (See diagram below).

I have lettered the ten intersection points of the pentagram to assist you in sending in your solution. Just give the values for A, B, C, D, E, F, G, H, I, and J.

Extra Credit: Could you number the ten points with the numbers from 1 to 10 so that the five lines of four numbers add up to the same sum?

Solution to the Problem:

Notice first that each pair of lines have exactly one point in common and that each number is in exactly two lines.
Counting each number twice, summing up and dividing by the number of lines (5) yields 24 as the sum of each line.

There are five ways for 12 to be in a sum:
(a) 12, 9, 2, 1
(b) 12, 8, 3, 1
(c) 12, 6, 5, 1
(d) 12, 6, 4, 2
(e) 12, 5, 4, 3

For the number 1, there are six ways to form the sum of 24:
(f) 1, 2, 9, 12
(g) 1, 3, 8, 12
(h) 1, 4, 9, 10
(i) 1, 5, 6, 12
(j) 1, 5, 8, 10
(k) 1, 6, 8, 9

We have to choose 2 lines with a 12 in it -- and all the other numbers must appear at most once. This is only possible taking (a, e) or (b, d) above. This shows that 1 and 12 have to be in the same line. For each combination, we now have to choose another line from the second set to find the second line containing 1.
I tried all six combinations (f through k) with (a, e) but none of those worked. Then I tried (f through k) with (b, d) and found a solution with (b, d, h).
So I placed 12, 8, 3, and 1 on one line; then put 12, 6, 4, and 2 on another line, and made it work with 1, 4, 9, and 10. There was only one position left for the number 5, and everything added up!

A = 12
B = 8
C = 3
D = 1
E = 10
F = 4
G = 9
H = 2
I = 5
J = 6

Using similar logic with the numbers 1 to 10, it can be shown that it is NOT possible to solve that problem.

Correctly solved by:

1. Peter Zhu	Tullängsskolan, Sweden
2. John C. Funk *****	Ventura, California
3. Elizabeth Harp	Columbus, Georgia
4. Gusti Oggenfuss *****	Montet (Broye), Switzerland
5. Jeff Gaither	Winchester, Virginia
6. Josh Feingold	Winchester, Virginia
7. George Gaither	Winchester, Virginia
8. David and Judy Dixon +++++	Bennettsville, South Carolina
9. Kathleen Altemose	Winchester, Virginia

***** Solved the extra credit
+++++ Sent in 10 different solutions