AIME II 2014 #12: Implications for competitive mathematics

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CreateAloha

7 posts

CreateAloha

#1 h Mar 29, 2014, 4:43 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Now a senior in high school, I have participated in competitive math for 6 years. After reading forum posts and discussions about the similarity between #12 on the AIME II 2014 and a question on one of this year's AMC Advantage tests, I am extremely disappointed in the MAA. I will no longer promote the AMC competitions in the state of Hawaii, nor ever recommend this discipline to future generations. I have no malice toward AMC Advantage users, bur rather toward the greed and deception of the MAA. You can attribute my feelings to a shortcoming of USAMO (9 on AIME 1, so close!), and while this may be partially true, you cannot hold the MAA innocent for this incident. Nearly verbatim problems that occur within such a short time period? Please. What happened to the supposedly innovative, thought-provoking questions that the AMC tests are commended for? From their website, "the AMC program identifies, recognizes, and rewards excellence in mathematics through a series of national contests." They do not reward excellence; they reward expenses toward their corrupt AMC Advantage. I fear for the future of competitive math. I have been deceived for 6 years.

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Wolstenholme

543 posts

Wolstenholme

#2 h Mar 29, 2014, 4:56 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

While your anger is somewhat justified, please understand that unlike in proof based contests like the USA(J)MO in which problems can be innovative everytime, problems of this level are often repeated or at least their ideas are repeated. I assure you that if you look at every amc10, amc12, aime, hmmt, pumac, Stanford math contest, first rounds from other countries, etcetera you will find TONS of repeats. It just happened to be unlucky that these were so close in time span and in "provider."

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kwansun

20 posts

kwansun

#3 h Mar 29, 2014, 5:45 AM • 21 Y

Y by lucylai, droid347, jubjubdaboss, sicilianfan, henrikjb, GeneralCobra19, tigerzhang, CyclicISLscelesTrapezoid, Toinfinity, mathtiger6, ike.chen, Adventure10, aidan0626, and 8 other users

Yeah, the one month old program AMC Advantage has deceived you for 6 years

((((( AMC must be really evil to come up with a program that can do something like that. It doesn't even matter that the AMC's motivated countless numbers of kids to deeply study mathematics, that one aime problem is ridiculous!!!! I had a feeling they've been plotting something like this for six years...

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Tan

494 posts

Tan

#4 h Mar 29, 2014, 11:59 AM • 1 Y

Y by Adventure10

I believe that your justification is invalid. You can find all AMC, AIME and maybe even USAMO and USAJMO problems (or at least similar in concept) in many textbooks, etc. I really don't understand why you are really mad at one problem being on AMC Advantage. Then again, this is only my opinion.

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Konigsberg

2228 posts

Konigsberg

#5 h Mar 29, 2014, 1:38 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

That's because AMC Advantage is very closely related to the MAA, and I think that MAA, the ones who run the AIME, earns money from AMC Advantage.

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droid347

2679 posts

droid347

#6 h Mar 29, 2014, 1:41 PM • 2 Y

Y by Adventure10, Mango247

Konigsberg wrote:

That's because AMC Advantage is very closely related to the MAA, and I think that MAA, the ones who run the AIME, earns money from AMC Advantage.

I'm pretty sure the MAA runs AMC advantage.

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dantx5

1464 posts

dantx5

#7 h Mar 29, 2014, 5:46 PM • 1 Y

Y by Adventure10

^slippery slope there

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borntobeweild

331 posts

borntobeweild

#8 h Mar 30, 2014, 1:50 AM • 2 Y

Y by Adventure10 and 1 other user

Has anyone considered the possibility that perhaps they just had the problem lying around somewhere, and it accidentally got used for both contests? (Something very similar occurred to let this problem get used for both USAMO and BMO.) The people who run math contests are human too, and I find it much more likely that this seeming corruption was the result of an honest mistake.

Though I'm sure others as well as myself would like to see an apology, formal or just on AoPS, given by one of the people from AMC headquarters. So far, I haven't heard any word from them, but maybe I just haven't searched carefully enough?

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v_Enhance

6877 posts

v_Enhance

#9 h Mar 31, 2014, 7:32 AM • 33 Y

Y by geoishard, csmath, TheMaskedMagician, henrikjb, pedronr, mathwizard888, MSTang, 62861, champion999, anantmudgal09, GeneralCobra19, 277546, HamstPan38825, tigerzhang, CyclicISLscelesTrapezoid, Toinfinity, Mogmog8, sehgalsh, Geometry285, Adventure10, Mango247, aidan0626, and 11 other users

This topic is a really perfect example of a bias called prior neglect! Let's explain this in detail.

Suppose I introduce you to my friend Tom and you find out that he's shy. Is he more likely to be a business major or a math major?

As an estimate, let's say 70% of math majors are shy and 10% of business majors are shy. Of course, $70\% \gg 10\%$ , so most people would quite naturally assume that Tom is probably a math major. Guess what? Wrong answer!

Why? There are about 20 times more business majors than math majors. If you meet $105$ random people, you would find maybe $100$ business majors and $5$ math majors. Of these, maybe 3-4 of the math majors would be shy; 7 of the business majors would be shy. So in fact, it's about twice as likely that Tom is a business major.

Oops. This is prior neglect -- people tend to forget their initial perceptions after seeing evidence. It is very dangerous because it leads you to wrong conclusions. (Suppose a test is 99% accurate and you test positive for a rare disease which requires immediate treatment. What's the chance you actually have it? Better hope your doctor gets that right.)

---
Anyways, time to actually respond to the original post. Aside from the prior neglect I mentioned, you are also vastly underestimating the chance that two similar problems appear.

CreateAloha wrote:

What happened to the supposedly innovative, thought-provoking questions that the AMC tests are commended for?

They are hard to write.

CreateAloha wrote:

Nearly verbatim problems that occur within such a short time period? Please.

Oh, it happens. In fact, it happens a lot.

WOOT AIME 3 2011 #2 wrote:

In a convex polygon with 18 sides, the angles are all positive integers (when measured in degrees) and form a nonconstant arithmetic sequence. Find the measure of the smallest angle, in degrees.

AIME II 2011 #3 wrote:

The degree measures of the angles of a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle.

WOOT Problem of the Day 11/1/2010 wrote:

Choose three random distinct vertices $A$ , $B$ , $C$ from a regular 123-gon. The probability that triangle $ABC$ is obtuse can be expressed as $\frac{m}{n}$ for relatively prime positive integers m and n. Find m+n.

AIME I 2011 #10 wrote:

The probability that a set of three distinct vertices chosen at random from among the vertices of a regular $n$ -gon determine an obtuse triangle is $\frac{93}{125}$ . Find the sum of all possible values of $n$ .

This happens when lots and lots of problems exist. We don't do it intentionally, but it's basically impossible to verify that a new idea has never been seen before. (In fact I basically assume that any idea for a problem I come up with has appeared before and just try to make sure it hasn't frequently appeared before.)

Wolstenholme wrote:

While your anger is somewhat justified, please understand that unlike in proof based contests like the USA(J)MO in which problems can be innovative everytime...

Nah, the same thing happens in olympiads too. Especially geometry.

So before you all accuse the MAA of intentionally placing two similar problems on close exams, please consider the possibility that the MAA is not evil. Seems pretty likely to me.

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Konigsberg

2228 posts

Konigsberg

#10 h Mar 31, 2014, 10:02 AM • 1 Y

Y by Adventure10

If this appeared somewhere where MAA has no direct relation to then there would be not that much of an uproar. However, this came out ALMOST verbatim from the AMC advantage, wherein MAA earns money. I believe that for most people, date is not the issue. It is the involvement of the same organization for both tests and the possibility of corruption, etc.

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v_Enhance

6877 posts

v_Enhance

#11 h Mar 31, 2014, 12:21 PM • 7 Y

Y by geoishard, csmath, HamstPan38825, Adventure10, and 3 other users

I mean you can't expect the AMC Advantage to go out of their way to try and fare better than random chance. In fact, my guess would be that whoever prepares the actual exams for AMC Advantage does not actually have access to the current year's AIME problems (which, by the way, are prepared two years in advance).

As I read it, the OP is suggesting that someone decided to intentionally put a current AIME problem in the Advantage's mock AIME in order to artificially and systematically inflate the scores of Advantage students. I don't think this claim has much merit to it.

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ksun48

1514 posts

ksun48

#12 h Mar 31, 2014, 1:59 PM • 4 Y

Y by mentalgenius, Adventure10, and 2 other users

@v_Enhance: I'm not necessarily disagreeing with your point, but it appears to me that AMC Advantage is a program sponsored by the MAA, by looking at the AMC Advantage site and some pages on the AMC website. If this is true, then your argument is not exactly valid, as it assumes that AMC Advantage and the AMC are completely separate, which may not be the case. (If this isn't true, then you can ignore this post).

EDIT: From the AMC Advantage: "The AMC Advantage test courses have been designed and created under the leadership of Dr. Dave Wells who is the chair of the Mathematical Association of America’s Committee on the AMC competitions. AMC Advantage is a joint initiative of the Mathematical Association of America (MAA) and Edfinity."

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tmathman

2923 posts

tmathman

#13 h Mar 31, 2014, 4:08 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

v_Enhance wrote:

This topic is a really perfect example of a bias called prior neglect! Let's explain this in detail.

Suppose I introduce you to my friend Tom and you find out that he's shy. Is he more likely to be a business major or a math major?

As an estimate, let's say 70% of math majors are shy and 10% of business majors are shy. Of course, $70\% \gg 10\%$ , so most people would quite naturally assume that Tom is probably a math major. Guess what? Wrong answer!

Why? There are about 20 times more business majors than math majors. If you meet 105 random people, you would find maybe 100 business majors and 5 math majors. Of these, maybe 3-4 of the math majors would be shy; 7 of the business majors would be shy. So in fact, it's about twice as likely that Tom is a business major.

Oops. This is prior neglect -- people tend to forget their initial perceptions after seeing evidence. It is very dangerous because it leads you to wrong conclusions. (Suppose a test is 99% accurate and you test positive for a rare disease which requires immediate treatment. What's the chance you actually have it? Better hope your doctor gets that right.)

Is this "prior neglect" what takes place in the birthday problem?

Birthday Problem wrote:

What is the smallest $n$ such that in a group of $n$ people, the probability that at least one pair of people share the same birthday is greater than $0.50$ ?

Funny how psychology takes so much effect in math, huh?

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MSTang

6012 posts

MSTang

#14 h Mar 31, 2014, 7:40 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

In the birthday problem, the main thing that people fail to understand is that increasing the number of people will increase the probability by gradually larger amounts. They think, "Really? Only 23 people? But this one person among the 23 is very unlikely to share a birthday with somebody else in the group." This is true, but there are other pairs of people in the group not including that one person.

In fact, there are $\dbinom{n}{2} = \dfrac{n(n-1)}{2}$ such pairs; this is a quadratic function, so if we add people gradually, the number of pairs will increase and increase by more each time! Once you hit $23$ people, which is the answer to the problem, there are $\dbinom{23}{2} = 253$ pairs, which is a lot of possible matches!

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happiface

1300 posts

happiface

#15 h Mar 31, 2014, 9:15 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

v_Enhance wrote:

A Really Detailed Post

This topic is a really perfect example of a bias called prior neglect! Let's explain this in detail.

Suppose I introduce you to my friend Tom and you find out that he's shy. Is he more likely to be a business major or a math major?

As an estimate, let's say 70% of math majors are shy and 10% of business majors are shy. Of course, $70\% \gg 10\%$ , so most people would quite naturally assume that Tom is probably a math major. Guess what? Wrong answer!

Why? There are about 20 times more business majors than math majors. If you meet $105$ random people, you would find maybe $100$ business majors and $5$ math majors. Of these, maybe 3-4 of the math majors would be shy; 7 of the business majors would be shy. So in fact, it's about twice as likely that Tom is a business major.

Oops. This is prior neglect -- people tend to forget their initial perceptions after seeing evidence. It is very dangerous because it leads you to wrong conclusions. (Suppose a test is 99% accurate and you test positive for a rare disease which requires immediate treatment. What's the chance you actually have it? Better hope your doctor gets that right.)

---
Anyways, time to actually respond to the original post. Aside from the prior neglect I mentioned, you are also vastly underestimating the chance that two similar problems appear.

CreateAloha wrote:

What happened to the supposedly innovative, thought-provoking questions that the AMC tests are commended for?

They are hard to write.

CreateAloha wrote:

Nearly verbatim problems that occur within such a short time period? Please.

Oh, it happens. In fact, it happens a lot.

WOOT AIME 3 2011 #2 wrote:

In a convex polygon with 18 sides, the angles are all positive integers (when measured in degrees) and form a nonconstant arithmetic sequence. Find the measure of the smallest angle, in degrees.

AIME II 2011 #3 wrote:

The degree measures of the angles of a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle.

WOOT Problem of the Day 11/1/2010 wrote:

Choose three random distinct vertices $A$ , $B$ , $C$ from a regular 123-gon. The probability that triangle $ABC$ is obtuse can be expressed as $\frac{m}{n}$ for relatively prime positive integers m and n. Find m+n.

AIME I 2011 #10 wrote:

The probability that a set of three distinct vertices chosen at random from among the vertices of a regular $n$ -gon determine an obtuse triangle is $\frac{93}{125}$ . Find the sum of all possible values of $n$ .

This happens when lots and lots of problems exist. We don't do it intentionally, but it's basically impossible to verify that a new idea has never been seen before. (In fact I basically assume that any idea for a problem I come up with has appeared before and just try to make sure it hasn't frequently appeared before.)

Wolstenholme wrote:

While your anger is somewhat justified, please understand that unlike in proof based contests like the USA(J)MO in which problems can be innovative everytime...

Nah, the same thing happens in olympiads too. Especially geometry.

So before you all accuse the MAA of intentionally placing two similar problems on close exams, please consider the possibility that the MAA is not evil. Seems pretty likely to me.

Well, I'm not sure how you guys do it, but the problems from NIMO/OMO all feature pretty innovative use of techniques--I don't think it would be a longshot to call most NIMO/OMO problems "new." I would expect the MAA to have problem standards at a level just as high as a group of high school students (very creative and devoted ones, but still).

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Fry

2 posts

Fry

#16 h Apr 1, 2014, 12:45 AM • 4 Y

Y by Adventure10 and 3 other users

Yeah guys, be thankful that the MAA holds such great contests.. We have all seen reused problems before..

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tmathman

2923 posts

tmathman

#17 h Apr 1, 2014, 3:26 PM • 2 Y

Y by Adventure10, Mango247

I'll put it this way: in the Bulgarian (I think it was Bulgarian? it might have been somewhere else) lottery, the sets of winning numbers came up twice in a row. O.o is it rigged? No, simply a freak event occurred. Actually, for the American Powerball lottery, the same set of numbers is expected to appear every 40-50 years. That's pretty often, a lot more often than people think. Same thing here, except that maybe problems repeat 5-10 years. Back to AIME #12, the repeated question was, yes, a more unlikely than likely event, but still, like the Bulgarian lottery, it could happen. The Bulgarian company was investigated, and no fraud was found. Thus, all this was just an occurrence of what is also called the "Improbability Principle", where the improbable is not so improbable as people think.

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sugar_rush

1341 posts

sugar_rush

#18 h Aug 9, 2022, 11:39 PM

Y by

v_Enhance wrote:

CreateAloha wrote:

Nearly verbatim problems that occur within such a short time period? Please.

Oh, it happens. In fact, it happens a lot.

WOOT AIME 3 2011 #2 wrote:

In a convex polygon with 18 sides, the angles are all positive integers (when measured in degrees) and form a nonconstant arithmetic sequence. Find the measure of the smallest angle, in degrees.

AIME II 2011 #3 wrote:

The degree measures of the angles of a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle.

WOOT Problem of the Day 11/1/2010 wrote:

Choose three random distinct vertices $A$ , $B$ , $C$ from a regular 123-gon. The probability that triangle $ABC$ is obtuse can be expressed as $\frac{m}{n}$ for relatively prime positive integers m and n. Find m+n.

AIME I 2011 #10 wrote:

The probability that a set of three distinct vertices chosen at random from among the vertices of a regular $n$ -gon determine an obtuse triangle is $\frac{93}{125}$ . Find the sum of all possible values of $n$ .

This too

WOOT AIME 3 2016 #14 wrote:

Find the number of permutations $(a_1, a_2, \dots, a_6)$ of the numbers $(1, 2, \dots, 6)$ , with the property that for any integer $k$ , $1 \le k \le 5$ , $(a_1, a_2, \dots , a_k)$ is not a permutation of the numbers $(1, 2, \dots, k)$ .

AIME II 2018 #11 wrote:

Find the number of permutations of $1,2,3,4,5,6$ such that for each $k$ with $1\leq k\leq 5$ , at least one of the first $k$ terms of the permutation is greater than $k$ .

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programmeruser

2455 posts

programmeruser

#19 h Aug 10, 2022, 12:06 AM

Y by

sugar_rush wrote:

v_Enhance wrote:

CreateAloha wrote:

Nearly verbatim problems that occur within such a short time period? Please.

Oh, it happens. In fact, it happens a lot.

WOOT AIME 3 2011 #2 wrote:

In a convex polygon with 18 sides, the angles are all positive integers (when measured in degrees) and form a nonconstant arithmetic sequence. Find the measure of the smallest angle, in degrees.

AIME II 2011 #3 wrote:

The degree measures of the angles of a convex 18-sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle.

WOOT Problem of the Day 11/1/2010 wrote:

Choose three random distinct vertices $A$ , $B$ , $C$ from a regular 123-gon. The probability that triangle $ABC$ is obtuse can be expressed as $\frac{m}{n}$ for relatively prime positive integers m and n. Find m+n.

AIME I 2011 #10 wrote:

The probability that a set of three distinct vertices chosen at random from among the vertices of a regular $n$ -gon determine an obtuse triangle is $\frac{93}{125}$ . Find the sum of all possible values of $n$ .

This too

WOOT AIME 3 2016 #14 wrote:

Find the number of permutations $(a_1, a_2, \dots, a_6)$ of the numbers $(1, 2, \dots, 6)$ , with the property that for any integer $k$ , $1 \le k \le 5$ , $(a_1, a_2, \dots , a_k)$ is not a permutation of the numbers $(1, 2, \dots, k)$ .

AIME II 2018 #11 wrote:

Find the number of permutations of $1,2,3,4,5,6$ such that for each $k$ with $1\leq k\leq 5$ , at least one of the first $k$ terms of the permutation is greater than $k$ .

How did you find this topic

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sugar_rush

1341 posts

sugar_rush

#20 h Aug 10, 2022, 12:15 AM

Y by

I was doing 2014 AIME II/12 for AIME practice, then I found this thread which was linked on the original topic.

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