<p>To those who are knowledgeable about the AMC tests, have you heard about the changes for 2010? Aka the introduction of the USAJMO?</p>
<p>I, personally, think this is silly, especially due to the ridiculous 270/230 split. </p>
<p>My question is, though, will "USAJMO qualifier" even be appealing on a college app?
This new method is unlikely to attract the smartest 9th and 10th graders to USAJMO, because those very gifted students will take the AMC 12 in an attempt to qualify for USAMO, resulting in above-average-but-not-amazing 9th and 10th graders landing in the USAJMO category.</p>
<p>Additionally, it makes 9th and 10th graders have to guess which AMC to take. Should I take the AMC 10 and do great and potentially qualify for USAJMO? Or should I take the AMC 12 and have to do REALLY great and potentially qualify for USAMO, without the possibility of qualifying for USAJMO?
This is exacerbated by the fact that many school districts only offer one administration of the AMC, so 9th and 10th graders who would like to take the AMC 10A AND the AMC 12B can not do so. (Some will say, if your school only offers one test, take the second test date at another high school or university. But what if the closest high school/university that offers the second test date is nowhere near where you live? Then you're stuck with choosing which test to take.)</p>
<p>I guess a "USAJMO qualifier" is a step up from being only an "AIME qualifier," but overall, I think this new change is stupid. Do you think colleges will care about USAJMO? </p>
<p>(For anyone who is not aware of this change, I've pasted the description below.)</p>
<p>
[quote]
On May 18, 2009, the MAA Executive COmmittee formally approved the attached proposal from the Committee on the American Mathematics Competitions. </p>
<p>In short, the changes are: </p>
<p>The current USA Mathematical Olympiad administered to about 500
students in all grades will be split into two flights: </p>
<ol>
<li><p>The USA Mathematical Olympiad administered to about 270 students
who qualify from the AMC 12 with a top USAMO index based on their AMC
12 score. (along with a few students who took only the AMC 10
and scored 11 or more on the AIME.) </p></li>
<li><p>The USA Junior Mathematical Olympiad administered to about 230
students who qualify from the AMC 10 with a top USAMO index based on
their AMC 10 score. </p></li>
<li><p>The current "floor score" system will no longer be used. </p></li>
</ol>
<p>In practical terms this means:
1. For 11th and 12th graders, there is essentially no change. </p>
<ol>
<li><p>Students who want to qualify for the USAMO should take the AMC 12,
since the AMC 12 is the gateway to the USAMO. (The exception is
students who take the AMC 10 only but score 11 or more on the AIME.
Evidence from the last few years indicates that this is typically
about 5 students. These students later excelled at all levels of the
AMC program, and so based on this evidence we have built in this
exception.) </p></li>
<li><p>Students in 10th grade and below who want a chance at the USAMO
should take the AMC 12, but may also want to hedge their chances by
taking the AMC 10. Students in 10th grade and below who take the AMC
12 will have their AMC 12-based USAMO index considered without
consideration of age or grade or AIME score. Of course this means
they are considered with 11th and 12th graders and compete for the
approximately 250-270 USAMO spots on AMC 12 index alone. Students in
10th grade and below who qualify for the USAMO are eliminated from the
pool of AMC 10 takers competing for the 230 invitations to the USA Junior
Math Olympiad. </p></li>
<li><p>The gateway for the USA Junior Math Olympiad is the AMC 10. This
automatically limits participants in the Junior Math Olympiad to
grades 10 and below. </p></li>
<li><p>The USA Junior Mathematical Olympiad would have 6 problems and be
administered over 2 days, the same as the USAMO. Problems J1,J2 on
Day 1, and Problems J4, J5 would be different from the USAMO, and
would be close in level and content to problems 13-15 on the AIME.
Problem J3 would be the same as Problem 1 (Day 1) on the USAMO,
Problem J6 would be the same as Problem 4 (Day 2) on the USAMO. All
six problems on the USA Junior Mathematical Olympiad would require
written answers, perhaps a detailed algebraic, number theoretic,
combinatoric or geometric solution rather than a proof. Problems
would be graded on the same 7 point scale (42 points total) as the
USAMO, and have the same rigorous grading. </p></li>
<li><p>Selection to "Red MOSP" will work in the following way (assuming
funding for Red MOSP continues.) Among all 9th graders taking the
USAMO and JMO, the scores on problems J3 = 1 and J6 = 4 will be
totalled and ordered. Then we will select the students with the top
25 scores on these two problems. Ties among the top scores in each
category will be broken by using the scores from the other 4 problems
on each Olympiad.
[/quote]
</p></li>
</ol>