Answer to February 10, 2003 Problem of the Week

Each year, both the National League and the American League pick a Cy Young Award Winner in baseball for the outstanding pitcher in each league.

Shown below is only a portion of the information about the five National League players who received votes for the award in 2001. Each voter casts one first place ballot, one second place ballot, and one third place ballot.

Examine the incomplete chart and use mathematics to complete it.

Pitcher	1st	2nd	3rd	Total Points
R. Johnson	30	2	0	156
C. Shilling	2	29	1	98
M. Morris	0			31
J. Lieber	0			2
R. Oswalt	0			1

Solution to the Problem:

From column 1, it is possible to determine that there were 32 voters. That means that there are only 32 2nd place ballots and 32 3rd place ballots (Many problem solvers sent in Morris with ten 2nd place votes, but that wouldn't add up). Adding up the total points gives you 288 points. Each voter contributed 288 / 32 or 9 points to the total. In other words, each ballot is worth 9 points.

Randy Johnson earned 30 first-place votes, 2 second-place votes, and a total of 156 points. Curt Shilling earned 2 first-place votes, 29 second-place votes, and 1 third-place vote for a total of 98 points. If x, y, and z represent the number of points awarded for each first-place, second-place, and third-place vote, respectively, the preceding information indicates that
30x + 2y = 156 and 2x + 29y + 1z = 98. In addition, we know that each ballot is worth 9 points, so x + y + z = 9.

Solving these three equations in three unknowns yields x = 5 points for a first-place vote, y = 3 points for a second-place vote, and z = 1 point for a third-place vote.

Once we know the point values for the different votes, it is easy to fill in the table. Oswalt received 1 point, so he received a single third-place vote. Lieber received two points, so he must have received two third place votes and no other votes. This, Morris received the remaining 28 third-place votes and must have received the one remaining second-place vote. Here is the completed table:

Pitcher	1st	2nd	3rd	Total Points
R. Johnson	30	2	0	156
C. Shilling	2	29	1	98
M. Morris	0	1	28	31
J. Lieber	0	0	2	2
R. Oswalt	0	0	1	1

Correctly solved by:

1. Rick Jones	Kennett Square, Pennsylvania
2. Richard Johnson	La Jolla, California
3. Kathleen Altemose	Winchester, Virginia
4. Adam Carter	Fort Wayne, Indiana
5. Tina Zahel	Winchester, Virginia
6. Matt Stewart	Columbus, Georgia
7. David and Judy Dixon	Bennettsville, South Carolina
8. Walt Arrison	Philadelphia, Pennsylvania
9. Jeff Gaither	Winchester, Virginia
10. Erik Svahnström	Tullängsskolan, Örebro, Sweden
11. Johan Bernström	Tullagen, Sweden
12. Bobby Martin	Columbus, Georgia
13. Filip Åkerling	Tullängsskolan, Sweden
14. Johan Bernberg	Tullängsskolan, Örebro, Sweden
15. Kelsey Tomblin	Columbus, Georgia
16. Matt Stillwagon	Winchester, Virginia
17. Rich Murray	Ridgetown, Ontario, Canada
18. Court Pifer	James Madison University Harrisonburg, Virginia

February 10, 2003 Problem of the Month

Correctly solved by:

February 10, 2003
Problem of the Month