*OFFICIAL* AMC 12 2012 Discussion Thread

<p>Will someone please PM me the answers for #1-20?
I’m so nervous; I missed the cutoff by one question last year.
Thanks so much! :)</p>

<p>The answers are posted on their website (main page).</p>

<p><a href=“American Mathematics Competitions | Mathematical Association of America”>American Mathematics Competitions | Mathematical Association of America; (Answers for AMC 12A)</p>

<p>Any chance the cut-off for AIME will be lowered? I got a 99 on the AMC 12A…just 1 point off! Argh!</p>

<p>got a 118.5 on AMC 10A last year and qualified for AIME
Awww I got a 96 on AMC 12 this year… I got 13 questions correct 12 blank… should I go for AMC 12B to make AIME:’(?</p>

<p>i got a 93…what are my chances of qualifying? 14 correct 6 blank 5 incorrect</p>

<p>^ Impossible to say. The general consensus is that the cutoff will be in the mid-90s</p>

<p>Does anyone have the list of questions? My proctor didn’t let us take the pamphlet home.</p>

<p>If you guys can see this, it’s a (rough) solution set
[url=&lt;a href=“http://www.facebook.com/notes/epic-win/amc12a-2012-solutions/10151263892875650]Facebook[/url”&gt;Redirecting...]Facebook[/url</a>]</p>

<p>I got a 96 … answered 13, got all 13 right (1-11, 14, 17). I’m hoping the cutoff is mid-90s this year.</p>

<p>I’m happy though - this is the best I’ve ever done.</p>

<p>does anyone who took the 12A last year think the cutoff will be as low? (i think last year it was the high 80s)</p>

<p>Personally, I think last year’s was a little bit harder. Also, I think the cutoff was a 93 last year, and 88.5 the year before.</p>

<p>Anyone know how to do #21?</p>

<p>@AXanderson. Yeah, so you add them up and you get 2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14. Now, rearrange them like this:</p>

<p>(a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) + (a^2 - 2ac + c^2) = 14.</p>

<p>Yeah. They pulled a perfect squares on us. So now, combine perfect squares, we get:</p>

<p>(a-b)^2 + (b-c)^2 + (a-c)^2 = 14. The only perfect squares that work are 1, 4, and 9, so you get </p>

<p>a-b = 9
b-c = 4
a-c = 1. Solve for a, b, c by plugging in numbers, and then whatever the largest number is will be a (since a>=b>=c). So the answer came out to E: 253.</p>

<p>Thank you so much :)</p>

<p>I got a 96 on this haha (but it’s my senior year, so whatever)</p>

<p>How was #10 done?</p>

<p>The median cuts the triangle into 2 triangle with area 30/2 = 15. The area of a triangle is 1/2 ab sin C so for our desired triangle we have 1/2 (9)(5)sinC=15 sin(theta)= 30/45 = 2/3. (Note that we use 5 as one of the lengths because the median bisects the line of side length 10).</p>

<p>I got a 120 :frowning: down from a 132 last year</p>

<p>And how about #11?</p>

<p>when will the result for amc12a come out?</p>

<p>[2010</a> AMC 8 Statistics](<a href=“American Mathematics Competitions | Mathematical Association of America”>American Mathematics Competitions | Mathematical Association of America)</p>

<p>^i don’t know why the link is named 2010 AMC 8 Statistics but I’m fairly certain these are the 2012 12A stats. the top 5% would be 2457 students, which falls between the 96 and 94.5 score, with 2278 scoring 96 and above and 2740 scoring a 94.5 and above. so i’m not sure exactly what they’ll do about this, but i think it’s safe to say 96 and above works :)</p>