## Monday, April 20, 2015

### Learn How to Learn by BOGTRO from AoPS forum -- Thanks a bunch !!

I love the following quotes :

Insanity: doing the same thing over and over again and expecting different results.
I previously thought it's from Albert Einstein, but it's not. I love it anyway.

You can practice shooting eight hours a day, but if your technique is wrong, then all you become is very good at shooting the wrong way. Get the fundamentals down and the level of everything you do will rise.”
from Michael Jordan

"When you first start off trying to solve a problem, the first solutions you come up with are very complex, and most people stop there. But if you keep going, and live with the problem and peel more layer of the onion off, you can often times arrive at some very elegant and simple solutions."
- Steve Jobs, 2006

Below, BOGTRO from AoPS has graciously allow me to post his well-thoughtout article on "Learn How to Learn".

I wish more students will read it , and don't just read it once, but many times at different intervals and really internalize the method. It will help you not just with problem solving/competition math, but learning in general.

Learn How to Learn

About a month ago I was PMed by a member, asking for advice as to how to prepare for MATHCOUNTS. I (strangely) get a lot of these types of PMs, but this one was slightly different. Whereas normally I could answer something along the lines of "read Volume 1, do practice tests, profit", this user was complaining that despite having rigorously worked through Volume 1 and CMMS (I still don't know what this is, but it's implied to be a book), he was still scoring only in the low 20s on sprints.
To some extent, I was able to relate. Back in my MATHCOUNTS days, I was doing loads of practice tests, learning new techniques to shave off precious seconds, and even practicing hitting a buzzer quickly. But my results only marginally improved. Gradually I understood that he was facing the exact same problem I was - although we were doing plenty of work, we were doing it in the wrong way.

After some thought, I formulated a long but fairly detailed response. Given that state-national season is rolling around, and with it the usual abundance of "how do I prepare" threads, I'm reproducing it below (with some minor edits). I referenced sprint several times because that was the specific complaint by the user, but obviously you can replace "sprint" with "target" or even "countdown", or any combination thereof.
________________________________________________________________________________________________________________________
You first need to determine why it is that you're getting low scores on sprint. Are you running out of time? Making stupid mistakes? Bad at computation? Or do you honestly not know how to do the problems? The former three are rectified with simply a lot of (effective) practice, where I say "effective" because simply blazing through problems, checking your score, and moving on is not going to help you very much. You need to be critically analyzing almost every problem - not just the ones you got wrong. Sure, you don't need to think too hard about your process on #2, but questions that take you longer than you would like, you get wrong, or you do in a "bashy" way need to be reviewed.

Essentially, you should be following something similar to the following process. Of course, this is not something that is going to work for 100% of people. The point here is not that you should be following these guidelines like a bible, but that you need to think about how to get the maximum benefit out of each practice test you take. You may very well find that the below system doesn't work for you (though you should at least give it a chance - it may seem "boring" at first, but after some time you'll be going through it like it's second nature and learning excellent habits along the way), in which case you should come up with an alteration that works for you. If (or more likely when) you choose to develop your own preparation system, keep in mind that the basic elements should be present - rigorous review of problems you got wrong, self-reflection on why you got them wrong, and so on.
• Take any MATHCOUNTS sprint round under contest conditions. It doesn't really matter which one you take, though it should be fairly recent for best results. When you're done, score with a simple checkmark or X system - don't look through the solutions immediately. Make a note of the problems that took you a long time, even if you got them correct.
• Without timing yourself (though you shouldn't spend more than 15 minutes or so), solve the problems that you either got wrong or didn't answer during the test. This will partially tell you if you're getting questions wrong because of time constraints or because you don't know the material.
• At this point you should have 4 separate categories of problems:
• Completely correct - don't worry about these at all. Though there is some benefit to looking these over, they are significantly less important than all the other questions.
• Correct, but took you a long time. Identify why it took you a long time - and if it matters. A problem taking you 2-3 minutes may sound like a killer, but in general if you only have a couple of these questions that's completely fine. Even if there's only one "timesink", you should be looking through alternate solutions to doing these problems. I find that problems that usually cause timesinks are either geometry problems that are semi-direct applications of similar triangles (which are naturally fairly easy to coordinate bash or something similarly slow, but this may take a while) or counting problems where you just listed out the possibilities and counted them up. Unfortunately, many MATHCOUNTS problems have this as their intended solution, so there's not a great deal you can do about those. However, even though there may not be a cleaner solution, minute steps during your bashing may prove important. And in the event that even with optimizations the problem will still take 2-3 minutes, you may want to just skip it altogether even if you know exactly how to do it.
• Incorrect (or blank), but you solved it after the test. These are questions that you know how to do, but you ran out of time doing. Important is to determine how long it took you to solve these questions. If you solved 2 questions in 30 seconds each after the test, clearly that's worse than solving one problem in the second category. These second and third categories are quite similar and should be evaluated against each other (a quite reasonable rule of thumb is to save any counting question that you don't see how to do within ~10 seconds for later).
• Incorrect, and you couldn't solve it after the test. Look up the solution, searching (or even posting) on AoPS if necessary (which you should likely do anyway, as MATHCOUNTS official solutions are often horrendous). If it's a situation where you just forgot something that you really knew, it's easy to pass this off as a fluke and move on. However, this is a grave mistake. Perhaps if it happens once or twice in an otherwise good practice, you can kind of gloss over it. But make a note of it anyway. Whenever you hit two problems in the same general category that you didn't solve (keep your categories broad, but not too broad. "Geometry" is too broad a category, while "trignometric relations in geometric models of algebraic inequalities" is too specific to be helpful. Something like "similar triangles" or "factoring" is a much better type of category), you should immediately stop your practicing and look up the relevant sections in whatever book you have (e.g. Volume 1, or whatever CMMS is, or even just an internet search, etc.). Don't move on until you are confident in that area. By "confident", I don't mean that you can approach these kinds of problems once in a while. I mean that once you identify a question as being in your category, you should be able to solve it relatively quickly at least 75% of the time.
• File away every single problem that you got wrong. Categorize these as either "I solved this afterwards" (include the time it took you to solve it - approximate is fine) or "I didn't solve this afterwards". You will need these later. Take a break - read a book, play some FTW, go outside, play League of Legends, whatever floats your boat. There's not much value in overloading yourself, especially so close to chapter. If you're feeling particularly ambitious, review a chapter on a topic that you have trouble with.There is no point to reviewing topics you already can solve problems in regularly.
• At the end of the week, collect every single problem on your "incorrect problems list". If you're going through a test a day, these shouldn't number more than 50. Do these like you would a test under contest conditions. Compare your results to your incorrect problems paper (how long it took you to solve the problems, and whether you got them correct). The fact that you've seen the problems already should compensate for the fact that you need to work quicker. If you get a problem wrong, do the same process - don't time yourself while solving all of the remaining problems.
• If you got the same problem wrong twice, there are 3 scenarios:
• You got it wrong both times, but finished it after the test both times. This speaks to your (lack of?) time management, something that comes much more naturally with practice. Keep in mind that MATHCOUNTS really only tests a very small amount of concepts (relatively speaking), so working through old problems virtually guarantees that almost all MATHCOUNTS problems will already be more or less familiar to you on test day.
• You couldn't solve it at all the first time, but solved it after the test the second time. This is improvement, so it's perfectly fine.
• You didn't solve it the second time around. This means that you don't understand the concept - back to the books.
• Take all the problems you got correct (during the test) off your "incorrect problems" sheet, and continue to repeat the process from the top.

This may seem like quite a bit of work when typed up here, but in reality it's not. Instead of perpetuating the cycle of "do a practice test, score it, move on, read some books in some disorganized fashion, take another practice test, hope for improvement" (not even necessarily in that order, which is even more problematic), instead we optimize this routine by taking a single practice test a day and making sure that we get everything possible out of it. There are only so many tests, and a frequent complaint is that people have run out of old contests to do. While this may be true, this most likely means that they're not doing the tests properly. A single test with the time taken to reflect, organize, and perform a targeted review is significantly more beneficial than 5 tests taken without a goal in mind.

All in all, this should take at most a little over an hour per day (a little more at the end of the week). You are, of course, welcome to do more, but there's a sort of diminishing returns law past a certain point. Devoting a great deal of time to MATHCOUNTS is going to seem like a serious mistake in hindsight (I was among the most guilty of this), especially if you realize you were spending time incredibly inefficiently. I won't give an exact quote here (simply because I don't remember it and a quick search doesn't turn it up), but one MATHCOUNTS winner (Albert Ni?) said something along the lines of
Quote:
I knew that I wouldn't be the smartest mathlete competing. But I could, quite realistically, be the hardest working one [...]

In MATHCOUNTS, that's all that's required. But quantifying the term "hard work" is necessary - someone who is pushing a boulder from point A 25% of the way to point B is doing a lot of work for very little benefit, while someone who uses a truck to carry the same boulder to point B is doing significantly less work for significantly more benefit. Perhaps as a more accurate analogy, take two people in a shooting contest. As soon as the whistle blows, person A starts shooting haphazardly at his target, hitting it once in a while but constantly having to reload. Person B, on the other hand, takes his time, lines up his shots, and hits the target with deadly accuracy. This is very similar to MATHCOUNTS. Person A is blowing through his material quickly, getting little benefit overall, but naturally with the experience of shooting comes some slight improvement. On the other side of things, person B is taking the time to think about how best to use his limited resources to improve as best he can. Sure, he starts off a bit slower, and at the end of the day he might still have some ammunition left unused, but overall he hits the target more. The first approach is popular because it's very easy to feel like you're doing something - after all, if you're spending 4 hours a day on practice MATHCOUNTS tests, you're outworking everyone else, right? Don't fall into this trap. Line up your shots.

### Hints/links or Solutions to 2014 Harder Mathcounts State Sprint and Target question

Links, notes, Hints or/and solutions to 2014 Mathcounts state harder problems.

Sprint round:

#14 :
Solution I :
(7 + 8 + 9)  + (x + y + z)  is divisible by 9, so the sum of the three variables could be 3, 12, or 21.
789120 (sum of 3 for the last three digits) works for 8 but not for 7.
21 is too big to distribute among x, y and z (all numbers are district),
thus only x + y + z = 12 works and z is an even number
__ __ 0 does't work (can't have 6 6 0 and the other pairs all have 7, 8 or 9)
264 works (789264 is the number)

Solution II :
789000 divided by the LCM of 7, 8 and 9, which is 504 = 1565.47...
Try 504 * 1566 = 789264 (it works)
The answer is 264.

#18:
Watch this video from Mathcounts mini and use the same method for the first question,
you'll be able to get the answer. It's still tricky, though.

#23 : Drop the heights of the two isosceles triangles and use similar triangles to get the length of FC.
Then solve.

#24:
The key is to see 210 is 1024 or about 103

230 = ( 210 ) or about (103  )3about 109 so the answer is 10 digit.

#25:
As you can see, there are two Pythagorean Triples : 9-12-15 and 9-40-41.
Base (40-12) = 28 gives you the smallest area.
The answer is 28 * 18 = 504

#26 : Let there be A, B, C three winners. There are 4 cases to distribute the prizes.
A     B    C
1      1     5    There are 7C1 * 6C1 * $$\dfrac {3!} {2!}$$ = 126 ways -- [you can skip the last part for C
because it's 5C5 = 1]

1       2    4    There are 7C1* 6C2 * 3! = 630

1      3     3    There are 7C1 * 6C3 * $$\dfrac {3!} {2!}$$ = 420

2      2    3     There are 7C2 * 5C2 * 3 (same as above)

Add them up and the answer is 1806.

If you can't see why it's $$\dfrac {3!} {2!}$$ when there is one repeat, try using easier case to help you understand.

What about A, B two winners and 4 prizes ?
There are 2 cases, 1 3 or 2 2, and you'll see how it's done.

#27 : Read this and you'll be able to solve this question at ease, just be careful with the sign change.
Vieta's Formula and the Identity Theory

#28: There are various methods to solve this question.
I use binomial expansion :
$$11^{12}=\left( 13-2\right)^{12}=12C0*13^{12}$$+ $$12C1*13^{11}*2^{1}$$+... $$12C11*13^{1}*2^{11}$$+ $$12C12*2^{12}$$ Most of the terms will be evenly divided by 13 except the last term, which is $$2^{12}$$ or 4096, which, when divided by 13, leaves a remainder of 1.

Solution II :
$$11\equiv -2\left ( mod13\right)$$ ; $$(-2)^{12}\equiv 4096\equiv 1\left ( mod13\right)$$

Solution III :
Or use Fermat's Little Theorem (Thanks, Spencer !!)
$$11^{13-1}\equiv 11^{12}\equiv1 (mod 13)$$

Target Round :

#3: This question is very similar to 2012 chapter target #8.
See this link for similar question.

#6: This question is very similar to this Mathcounts Mini.
My students should get a virtual bump if they got this question wrong.

#8: Solution I : by TMM (Thanks a bunch !!)
Using similar triangles and Pythagorean Theorem.

The height of the cone, which can be found usinthe Pythagorean  is $\sqrt{10^2-5^2}=5\sqrt{3}$.
Usingthediagram below, let $r$ be the radius of the top cone and let $h$ be the height of the topcone.
Let $s=\sqrt{r^2+h^2}$ be the slant height of the top cone.

Drawing the radius as shown in the diagram, we have two right triangles. Since the bases of the top cone and the original cone are parallel, the two right triangles are similar. So we have the proportion$$\dfrac{r}{5}=\dfrac{s}{10}=\dfrac{\sqrt{r^2+h^2}}{10}.$$Cross multiplying yields $$10r=5\sqrt{r^2+h^2}\implies 100r^2=25r^2+25h^2\implies 75r^2=25h^2\implies 3r^2=h^2\implies h=r\sqrt{3}.$$This is what we need.

Next, the volume of the original cone is simply $\dfrac{\pi\times 25\times 5\sqrt{3}}{3}=\dfrac{125\sqrt{3}}{3}$.

The volume of the top cone is $\dfrac{\pi\times r^2h}{3}$.
From the given information, we know that $$\dfrac{125\sqrt{3}}{3}-\dfrac{\pi\times r^2h}{3}=\dfrac{125\sqrt{3}}{9}\implies 125\sqrt{3}-r^2h=\dfrac{125\sqrt{3}}{3}\implies r^2h=\dfrac{250\sqrt{3}}{3}.$$We simply substitute the value of $h=r\sqrt{3}$ from above to yield $$r^3\sqrt{3}=\dfrac{250\sqrt{3}}{3}\implies r=\sqrt[3]{\frac{250}{3}}.$$We will leave it as is for now so the decimals don't get messy.

We get $h=r\sqrt{3}\approx 7.56543$ and $s=\sqrt{r^2+h^2}\approx 8.7358$.

The lateral surface area of the frustum is equal to the lateral surface area of the original cone minus the lateral surface area of the top cone. The surface area of the original cone is simply
$5\times 10\times \pi=50\pi$.
The surface area of the top cone is $\pi\times r\times s\approx 119.874$.
So our lateral surface area is

All we have left is to add the two bases. The total area of thebases is $25\pi+\pi\cdot r^2\approx 138.477$. So our final answer is $$37.207+138.477=175.684\approx\boxed{176}.$$
Solution II
Using dimensional change and ratio, proportion.

Cut the cone and observe the shape.

The circumference of the larger circle is 20pi (10 is the radius) and the base of

the cone circle circumference is 10pi (5 is the radius), which means that the cut-off cone shape is a half circle because it's $$\dfrac {10\pi } {20\pi }$$ or $$\dfrac {1 } {2 }$$ of the larger circle. (180 degrees)

To find the part that is the area of the frustum not including the top and bottom circles,

you use the area of the half circle minus the area of the smaller half circle.

Since the volume ratio of the smaller cone to larger cone = 2 to 3, the side ratio of the

two radius is $$\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}$$.

Using this ratio, we can get the radius of the smaller circle as 10 * $$\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}$$ and the radius of the top circle of the frustum as 5 * $$\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}$$.

Now we can solve this :

$$\dfrac {1 } {2 }$$$$\left[ 10^{2}\pi -\left( 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\right) ^{2}\pi \right]$$ + $$5^{2}\pi +\left( 5\times \dfrac {\sqrt [3] {2}} {\sqrt {3}}\right) ^{2}\pi$$ = about 176 (after you round up)ional change and ratio, proportion.

Cut the cone and observe the shape.

The circumference of the larger circle is 20pi (10 is the radius) and the base of

the cone circle circumference is 10pi (5 is the radius), which means that the cut-off cone shape is a half circle because it's $$\dfrac {10\pi } {20\pi }$$ or $$\dfrac {1 } {2 }$$ of the larger circle. (180 degrees)

To find the part that is the area of the frustum not including the top and bottom circles,

you use the area of the half circle minus the area of the smaller half circle.

Since the volume ratio of the smaller cone to larger cone = 2 to 3, the side ratio of the

two radius is $$\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}$$.

Using this ratio, we can get the radius of the smaller circle as 10 * $$\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}$$ and the radius of the top circle of the frustum as 5 * $$\dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}$$.

Now we can solve this :

$$\dfrac {1 } {2 }$$$$\left[ 10^{2}\pi -\left( 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\right) ^{2}\pi \right]$$ + $$5^{2}\pi +\left( 5\times \dfrac {\sqrt [3] {2}} {\sqrt {3}}\right) ^{2}\pi$$ = about 176 (after you round up)

Solution III : Another way to find the surface area of the Frustum is :
median of the two half circle [same as median of the two bases] * the height [difference of the two radius]
$$\dfrac {1} {2}\left( 2\times 10\pi + 2\times 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\pi \right)$$* $$\left( 10-10\times \dfrac {\sqrt [3] {2}} {\sqrt [3]{3}}\right)$$