**Contest Registration and Books of Past Contests
**Please mail next year's registration (enclosed) soon. *You may ask us to bill
you in the Fall.* We also sponsor an *Algebra Course 1 Contest* and contests
for grades 4, 5, 6, 7, & 8. Use our Secure
Internet Order Form to register for contests or to order books of past contests.

**Future Contest (& Alternate) Dates-Ends In March **The contests (and alternate) dates are these Tuesdays: Oct. 17 (Oct. 10), Nov. 14 (Nov. 7), Dec. 12 (Dec. 5), Jan. 16 (Jan. 9), Feb. 13 (Feb. 6), Mar. 20 (Mar. 13). Do you have a conflict (such as the Feb. 13 AMC-formerly the AHSME)? If so, you may put the alternate date on your calendar. We have moved the calendar forward because of a number of requests that we have the final contest in March instead of in April. If you have an opinion, please write to us.

**End-of-Year Awards and Certificates **Symbols identify winners (we shipped plaques to the advisor). Errors? Write to: *Math Plaques, P.O. Box 17, Tenafly, NJ 07670-0017.* Identify the award, the contest level, your name, and the school's name and address. The Contest #5 envelope contained Certificates of Merit for the highest scoring student overall, and on each grade, for the year. Do you need extra certificates for ties? If so, send a self-addressed stamped envelope large enough to hold certificates (you need to use * *DOUBLE* * postage) to: *Certificates, P.O. Box 2117, Miller Place, NY 11764-8896.* (Allow 2-4 weeks.)

**Comments About This Year's Contests **Sharon Wilson said "Great final contest. Thanks." Because Susan Wong and Tricia Rothenberg independently asked for an alternate contest date to avoid a conflict with the AMC, we now offer an alternate date for each contest date **(see the "contest dates" paragraph above)** in case your school calendar cannot accommodate the preferred contest date. Gerald Lamb wrote "I cannot thank you enough for a wonderful set of contests this year. You have a great blend of problems in terms of difficulty and content area." Terry John wrote "Thanks for another year of great contests. I thought that the questions were particularly interesting and accessible this year. I rarely had students stop working prior to the end of the 30 minutes." Ed Imgrund "thought this was a good year. The kids see where in the curriculum most of these topics are covered or where in math team practice we discussed them before. Keep up the good work." Eric Bromaugh said "I especially enjoy going over the problems the next day and watching lightbulbs turn on as students understand how to reach a solution. Thanks!" Bob Smith wrote "The contests this year were very well done. The students enjoy them." Dave Farber wrote "It's time for me to plan for the Math Contest Awards Party for our participants. I look forward to next year which will be my last year before I retire and start a new career." Brian Balsdon wrote "Thanks for another great year of contests. Our students look forward to them every month." Richard Perrins said "Thanks for another good year. The last two problems totally stumped all our students, even though #5 shouldn't have."

**Problem 6-2: A Comment **Bob Smith noted that #6-2 "makes students read carefully and decide what the question asks, after which it is very straight forward."

**Problem 6-3: An Appeal (Denied) & An Alt. Solution **One appeal claimed it was unclear which prices should be compared. The appeal was denied without comment by Professor Brian Conrad, Math Department, Harvard University. Dave Farber used two variables in his solution.

**Problem 6-5: Comments & Great Alternate Solution **Bob Smith wrote that 6-5 was "a question that most students can do with some help, but they have trouble deciding on a plan of attack." Susan Osterkamph used trigonometry. Let the angle at the top of the triangle be 2*q*. Draw a radius *r* from that vertex to the center of the circle. Finally draw a perpendicular from the center to the midpoint of either leg. Since sin*q* = 0.6 = 3/5, we know that cos*q* = 4/5 = 0.8. In the small right triangle, 0.8 = cos*q* = 5/*r *; so *r* = 6.25. Thus, *C* = 2pi(6.25) = 12.5pi.

**Problem 6-6: Inquiry, Alternate Solution, Comments **Kay Nelson said "The last problem was a killer, but one student recognized the *u*-substitution from calculus and figured it out." One student asked "Graphically, *y* = (*x*-3)/(1-3*x*) and *y* = 8/(9*x*) do not coincide. If they are transformed into each other, it seems they should coincide." They should not coincide until *after* one is transformed into the other. An unnamed advisor said "It seems that Question 6 was too difficult, as not even our best students were able to come up with much more than a guess. The other problems were great." The simplest solution was sent independently by Mike Sikora, Dr. Robert Wollkind, and Brian Balsdon. As explained in our Method I, to shift the intersection of the asymptotes to (0,0), use the translation (*x* - 1/3, *y* + 1/3). Now, *y* = (*x*-3)/(1-3*x*) has an *x*-intercept of (3,0). Since (3,0) shifts to (8/3,1/3), (8/3,1/3) is on *xy* = *k*, and *k* = 8/9.

Prob #, % Correct (top 5 each school)

6-1 97% |
6-4 62% |

6-2 93% |
6-5 41% |

6-3 90% |
6-6 10% |