AMC 10/12/A/B 2013 Discussion Thread

Pre-College Issues High School Life
181

<p>Did you guys get your results through your school already? I am still waiting for mine.
Fingers crossed. I found 12b to be much easier than 12a, but AIME practice problems look tough.</p>

182

<p>I am taking the AIME for the first time this year.</p>

183

<p>Just took the AIME I.</p>

184

<p>Aren’t you supposed to wait until March 14th to take it…?</p>

185

<p>you can only take it march 14th or the alternate AIME II date…</p>

186

<p>Different time zones, perhaps?</p>

187

<p>does anyone want to compare for aime or have the answers for it.
pm me</p>

188

<p>The instructions that my teacher got in the mail with the test said that it could be administered before the 14th, but not after under any circumstances. I was the only one in my school to qualify What did you guys think? Anyone want to compare answers?</p>

<p>I felt that the first 8 or so were pretty straightforward. 14 really threw me through a loop until I remembered Taylor Series’. I didnt get anywhere on 15.</p>

189

<p>Pretty sure you didn’t need Taylor Series. All Aime questions can be solved using precalculus concepts.</p>

190

<p>Of course you did’t need it, but it sped up the process greatly.</p>

191

<p>@charlietidmarsh you want to compare?</p>

192

<p>what did you get for number fourteen, the question about the series for trig, and the question about folding the equilateral triangle</p>

193

<p>I took it two days ago, so I dont remember my answers for each one, but what did you get? It will jog my memory. I was not allowed to keep my test booklet with the problems because I took it early.</p>

<p>the ones I do remember my answers for are:</p>

<p>Number 1 (AIME Triathlon) - 150 minutes
Number 2 (5 digit number, sm divisible by 5 ect…) - 105 possible solutions</p>

<p>sound familiar?</p>

194

<p>wasn’t #2 like 200 possible?</p>

195

<p>What was your approach?</p>

196

<p>yeah i got 200 too, what did u guys get for number four about the box arrays where you rotate it and it asks what is the probability the colors would be the same, and what did u guys get for the equilateral triangle being folded and it gives you the length of the fold</p>

197

<p>I got 200 as well. Multiple of five (end in 0 or 5) and and first and last digit are the same (can’t have a leading 0), so it is 5ABC5. So for the sum to be a multiple of five, A+B+C=0 (mod 5). A+B can be any number between 0 and 18 inclusive. For each of those possible sums, there are 2 numbers for C that satisfy the condition. Therefore, for every possible A+C, there are 2 solutions. A can be 0 to 9, B can be 0 to 9, so 10<em>10=100=>100</em>2=200.</p>

<p>For the triangle fold one, I got 459. From left to right, let the points created by folding be D and E. Then I set AD=12-BD and AE=12-CE. Angle C and B are both 60, so you can find AD and AE in terms of BD and CE. Then set the two equations for AD equal to each other and the same for AE. Then I used law of cosines again on angle A (60) and AD and AE. Sorry, I don’t remember the exact numbers, but they were kinda ugly. Anyway, they added to 459. I might have made an arithmetic error though, so correct me if I’m wrong.</p>

198

<p>I also got 459. Same approach.</p>

<p>The approach on number 2 looks sound. I may have gotten 200, not sure. I remember answering 105 for one of them.</p>

<p>How about 14?</p>

199

<p>the triangle fold one is 113: 39(sqrt 39)/25…look on aops</p>

<h1>2 is 200(i missed it since i used stars and bars and got a bunch of extra cases :frowning: )</h1>

<h1>14 I didn’t get but according to aops is 036.</h1>

<p>Other answers:

  1. (time) 150
  2. (square 9/10 area one) 018
  3. (rotate 90 degrees) 429
  4. (cubic equation) 098
  5. (counting question) 047
  6. (rectangular prism) 041
  7. (arcsin log question) 371
  8. (complex roots 65 one) 080
  9. (16, 15, 14 mod arithmetic one) 148
  10. (equilateral hexagon) 021
  11. (similar triangle recursion) 961
  12. 272
    overall I got 9 correct. do you think 126/9 is good enough for JMO???</p>
200

<p>Dear math people on CC,</p>

<p>As an international student currently attending a high school in Dallas, Texas, I look forward to taking the AMC 12 next spring. I understand very well that there’re certain eligibility restrictions for me as an international student (F1-visa holder), >.< but I’m still curious if, with high performance on AMC and subsequently AIME, I would be qualified to take the USAMO. </p>

<p>Any idea?</p>