Dimentional Change Questions III: Similar Shapes

There are numerous similar triangle questions on Mathcounts.

Here are the basics:

If two triangles are similar, their corresponding angles are congruent and their corresponding sides will have the same ratio or proportion.

Δ ABC and ΔDEF are similar. \(\frac{AB}{DE}\) = \(\frac{AC}{DF}\) = \(\frac{BC}{EF}\)= their height ratio = their perimeter ratio.







Once you know the linear ratio, you can just square the linear ratio to get the area ratio and cube the linear ratio to get the volume ratio. 

Practice Similarity of Triangles here.  Read the notes as well as work on the practice problems.  There is instant feedback online. 

Other practice sheets on Similar Triangles                                                        

Many students have trouble solving this problem when the two similar triangles are superimposed. 

Just make sure you are comparing smaller triangular base with larger triangular base and smaller triangular side with corresponding larger triangular side, etc... In this case:

\(\frac{BC}{DE}\)= \(\frac{AB}{AD}\) = \(\frac{AC}{AE}\)




Questions to ponder (Solutions below)

#1: Find the area ratio of Δ ABC to trapezoid BCDE to DEGF to FGIH. You can easily get those ratios using similar triangle properties. All the points are equally spaced and line \(\overline{BC}\)// \(\overline{DE}\) // \(\overline{FG}\) // \(\overline{HI}\). 

#2: Find the volume of the cone ABC to Frustum BCDE to DEGF to FGIH. Again, you can use the similar cone, dimensional change property to easily get those ratios.Same conditions as the previous question.

 

Answer key: 

#1:

 #2:

Similar Triangles: Team question : Beginning level

9. In the figure below, quadrilateral CDEG is a square with CD = 3, and quadrilateral BEFH is a rectangle. If EB = 5, how many units is BH? Express your answer as a mixed number

Triangle BED is a 3-4-5 right triangle and is similar to triangle GEF.

BE : ED = GE : EF = 5 : 3 = 3 : FE

EF = 9/5 = BH  The answer

Harder Mathcounts State/AMC Questions: Intermediate level if you can solve in less than 2 mins.

2012 Mathcounts State Sprint #30: In rectangle ABCD, shown here, point M is the midpoint of side BC, and point N lies on CD such that DN:NC = 1:4. Segment BN intersects AM and AC at points R and S, respectively. If NS:SR:RB = x:y:z, where x, y and z are positive integers, what is the minimum possible value of x + y + z? 

Solution I :

\(\overline {AB}:\overline {NC}=5:4\) [given]

Triangle ASB is similar to triangle CSN (AAA)

\(\overline {NS}:\overline {SB}= 4 : 5\)

Let \(\overline {NS}= 4a,  \overline {SB}= 5a.\)

Draw a parallel line to \(\overline {NC}\) from M and mark the interception to \(\overline {BN}\)as T.

 \(\overline {MT}: \overline {NC}\) = 1 to 2. [\(\Delta BMT\) and \(\Delta BCN\) are similar triangles ]

\(\overline {NT} = \overline {TB}= \dfrac {4a+5a} {2}=4.5a\)

\(\overline {ST} = 0.5a\)

 \(\overline {MT} :  \overline {AB}\) = 2 to 5
[Previously we know  \(\overline {MT}: \overline {NC}\) = 1 to 2 or 2 to 4 and  \(\overline {NC}:\overline {AB}= 4 : 5\) so the ratio of the two lines  \(\overline {MT} :  \overline {AB}\) is 2 to 5.]


\(\overline {TB} = 4.5 a\)  [from previous conclusion]

Using 5 to 2 line ratio [similar triangles \(\Delta ARB\) and \(\Delta MRT\) , you get \(\overline {BR} =\dfrac {5} {7}\times 4.5a =\dfrac {22.5a} {7}\) and \(\overline {RT} =\dfrac {2} {7}\times 4.5a =\dfrac {9a} {7}\)

Thus, x : y : z = 4a : \( \dfrac {1} {2}a + \dfrac {9a} {7}\) : \(\dfrac {22.5a} {7}\) = 56 : 25 : 45

x + y + z = 126

Solution II : 
From Mathcounts Mini: Similar Triangles and Proportional Reasoning

Solution III: 
Using similar triangles ARB and CRN , you have \(\dfrac {x} {y+z}=\dfrac {5} {9}\).
9x = 5y + 5z ---- equation I

Using similar triangles ASB and CSN and you have \( \dfrac {x+y} {z}=\dfrac {5} {4}\).
4x + 4y = 5z  ---- equation II

Plug in (4x + 4y) for 5z on equation I and you have 9x = 5y + (4x + 4y) ; 5x = 9y ; x = \(\dfrac {9} {5}y\)
Plug in x = \(\dfrac {9} {5}y\) to equation II and you have z  =  \( \dfrac {56} {25}y\)

x : y : z = \(\dfrac {9} {5}y\)  : y  :  \( \dfrac {56} {25}y\) =  45 y :  25y  :  56y

45 + 25 + 56 = 126

Solution IV : Yes, there is another way that I've found even faster, saved for my private students. :D 

Solution V : from Abhinav, one of my students solving another similar question : 

Two other similar questions from 2016 AMC A, B tests : 

2016 AMC 10 A, #19 : Solution from Abhinav 

2016 AMC 10 B #19 : Solution from Abhinav 

2015 Mathcounts State Prep: Mathcounts State Harder Questions

Check out Mathcounts here, the best competition math program for middle school students.
Download this year's Mathcounts handbook here.

Question: 2010 Mathcounts State Team Round #10: A square and isosceles triangle of equal height are side-by-side, as shown, with both bases on the x-axis. The lower right vertex of the square and the lower left vertex of the triangle are at (10, 0). The side of the square and the base of the triangle on the x-axis each equal 10 units. A segment is drawn from the top left vertex of the square to the farthest vertex of the triangle, as shown. What is the area of the shaded region?

 There are lots of similar triangles for this question, but I think this is the fastest way to find the area.
  \(\Delta \)AGB is similar to \(\Delta \)DGC and their line ratio is 15 to 10 or 3 : 2.
     \(\Delta \)CGF is similar to \(\Delta \)CBE. 
     \(\dfrac {CG} {CB}=\dfrac {GF} {BE}\)
     \(\dfrac {2} {5}=\dfrac {GF} {10}\)\(\rightarrow GF=4\) From there you get the area =
      \(\dfrac{10\times 4} {2}=20\)

Question: 2010 Mathcounts State #30:  Point D lies on side AC of equilateral triangle ABC such that the measure of angle DBC is 45 degrees. What is the ratio of the area of triangle ADB to the area of triangle CDB? Express your answer as a common fraction in simplest radical form.

Since each side is the same for equilateral triangle ABC, once you use the 30-60-90 degree angle ratio and 45-45-90 degree angle ratio, you'll get the side.
Since area ratio stays constant, you can plug in any numbers and it's much easier to use integer first so I use 2 for \(\overline {CD}\).
From there you get the side length for each side is \(\sqrt {3}+1\).
\(\overline {AC}-C\overline {D}=\sqrt {3}+1-2\) = \(\overline {AD} = \sqrt {3}-1\)
\(\Delta ABD\) and \(\Delta CBD\) share the same vertex, so their area ratio is just the side ratio, which is \(\dfrac {\sqrt {3}-1} {2}\).

Dimensional Change

There are lots of questions on dimensional change and this is a very common one.

Make sure you understand the relationship among linear, 2-D (area) and 3-D (volume) ratio.

There are many similar triangles featured in the image on the left.
Each of the two legs of the largest triangles is split into 4 equal side lengths.





                                                                                            


Question : What is the area ratio of the sum of the two white trapezoids to the largest triangle? 
\(\dfrac {\left( 3+7\right) } {16}=\dfrac {10} {16}=\dfrac{5}{8}\)  

Question: If the area of the largest triangles are 400 square units, what is the area of the blue-colored trapezoid?
\(\dfrac {5} {16}\times 400\) =125 square units 

Again, each of the two legs are split into three equal segments. 

The volume ration of the cone on the top to the middle frustum to the 
bottom frustum is 1 : 7 : 19. 
 
Make sure you understand why.










 

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.

2014, 2015 Mathcounts state are harder 

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 )³  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: Lune of Hippocrates : in seconds solved 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 using the Pythagorean  is . 
Usingthediagram below, let  be the radius of the top cone and let  be the height of the topcone. 
Let  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 proportionCross multiplying yields This is what we need.

Next, the volume of the original cone is simply . 
The volume of the top cone is .
From the given information, we know that We simply substitute the value of  from above to yield We will leave it as is for now so the decimals don't get messy.

We get  and .

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 
. 
The surface area of the top cone is . 
So our lateral surface area is 

All we have left is to add the two bases. The total area of thebases is . So our final answer is 

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]
\(\dfrac {1} {2}\left( 2\times 10\pi + 2\times 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\pi \right)\)* \(\left( 10-10\times \dfrac {\sqrt [3] {2}} {\sqrt [3]{3}}\right)\)

Original Problem from a Student Problem Solver on Similar Triangles

An original problem on similar triangles from Varun (from FL)

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 

5 ways to Kill your dreams from TED talk 

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.