2017 Mathcounts State Prep: Volume of a Regular Tetrahedron and Its Relationship with the Cube it's Embedded

How to find volume of a tetrahedron (right pyramid) with side length one.

The above link gives you a visual interpretation of the relationship of a regular tetrahedron, its
relationship with the cube that it is embedded and the other kind of tetrahedron (right angle pyramid).

The side of the cube is \(\dfrac {S} {\sqrt {2}}\) so the volume of the regular embedded tetrahedron is
\(\dfrac {1} {3}\times \left( \dfrac {S} {\sqrt {2}}\right) ^{3}\)=\(\dfrac {1} {3}*\dfrac {s^{3}} {2\sqrt {2}}\)= \(\dfrac {\sqrt {2}S^{3}} {12}\).

You can also fine the height of the tetrahedron and then \(\dfrac {1} {3}\)*base*space height to get the volume.
Using Pythagorean theory, the hypotenuse S and one leg which is \(\dfrac {2} {3}\) of the height of the equilateral triangle base, you'll get the space height.

2016 AMC-8 Prep

Interesting articles on math for this week:

How to Fall in Love With Math by Manil Suri from the New York Times

The Simpsons' secret formula : it's written by math geeks by Simon Singh from the Guardian

For this week's self studies (part I work):
Review:

2009 #25 : Review using the Harvey method. :D

2008 #25   Don't use the method on the link. Use the much faster method we talked about at our lesson.

2008 #24  Make a chart. Slow down on similar question such as this one. 
This type of problem is very easy to make mistake on under or overcounting. 
Skip first and definitely slow down and double, triple check. 

2007 #25 Read the solution if you don't get the method we talked about at our lesson.
It takes time to develop insights so you need to be patient.
If you understand the method, this question will be easy, right ?
Stay with this question longer.

2007 #24 
Aayush's method is faster.
(To get the sum of three digits that is a multiple of 3, you either get rid of 1 or 4 [do you see it jumps by 3, why?] ) , so the answer is 1/2.

2006 #25
I've seen other problems (AMC-10s) using the oddest prime, which is "2", the only prime number that is even.
Thus, make sure you understand this question.

2006#24  Taking out the factor question.
Also, learn "1001 = 7 x 11 x 13"
"23 x 29 = 667"

2005 #25 Venn diagram is your friend.

2005 #24 Working backwards is the way to go.

For part II work :
This week, work on the last 5 problems from AMC-8 year 2010, 2011, 2012, 1999 and 1998.
Here is the link from AoPs.

Sam' original question:
David has a bag of 8 different-colored six-sided dice. Their colors are red, blue, yellow, green, purple, orange, black, and white. What is the probability that David takes out a red die, rolls a 6, then takes out a purple die, and rolls another 6 without replacement?

Solution:
The probability of rolling a 6 on a red die is 1/8 * 1/6 = 1/48. The probability of rolling a purple die and rolling a 6 after that, without replacement, is 1/7 * 1/6 = 1/42. Therefore, to get both events, 1/48 * 1/42 = 1/2016.

Evan's compiled question:
\(\sqrt {18+8\sqrt {2}}=a+b\sqrt {c}\)
a, b and c are positive integers. Find a + b + c.
Solution:
Square both sides and you have \(18+8\sqrt{2}\) = \( a^2 + 2ab\sqrt {2} + b^2c\)
You can see ab = 4 = 4 x 1 and c = 2
a = 4, b = 1 and c = 2 so the sum is 7.

Sounak's problem:
A rhombus with sides 4 is drawn. It has an angle of 60 degrees. What is length of the longer diagonal?

Solution:
Well first you have to draw the rhombus's height .The resulting triangle will be 30,60, 90 triangle.
We know the hypotenuse is 4 so now we know the rest of the sides are \(2\sqrt {3}\) and 2.
Now if we draw the diagonal we see that it makes another right angle triangle.
We know the legs of this triangle are the same as the previous lengths so then we know the diagonal is \(4\sqrt {3}\).

First week's review reminders

For 16-17 Mathcounts handbook :

Warm-up 1 : review #3, 7, 9.
Think of ways to get these questions in seconds.

Warm-up 2 : review 17,18,19,20 -- slow down on #18, similar types are very easy to
get wrong the first try

direct and inverse relationship 

From AoPS videos :

7.1 Direct Proportion

7.2 Inverse Proportion

7.3 Joint Proportion

Instructions for reviewing 

2014 School Rounds link here 

Sprint : 

#18 (trial and error are faster), #20 (loose pennies out first)
#22, #26, #27, #28, #29, #30 -- Read the solutions and make sure to fully understand
what major concept(s) is (are) tested -- these are not hard problems.

Target : 

#1 : Calendar questions could be tricky, so slow down a bit.
       Check what stage it starts first.

#2 : WOW, this one is a killer, so tedious -- hard to get it right the first time. 

# 3 and 4 are in seconds questions.

# 5: One line, two numbers and you are done.
       With a calculator, you don't even need to write anything down, right?  :)

#6 : more interesting question

#7 and 8 : standard questions, not hard, just need to be careful and thorough.

Team : 

# 2, 3, 5, 6 (very tedious, I suggest skipping first)
8, 9 (more tedious than 8 and 10, so don't work on problems in order), and
#10 (very easy if you've noticed what is actually tested) It's in seconds question. 

Agai, scan those questions and only try those that are your weak spots or marked red.

Please check the solution files I'd sent you to learn the better methods.

Whenever you have extra time, use the other links (individually based) to keep learning.

Keep me posted. Have fun at problem solving.

Don't just do math.  :) 

Math related video : Making Stuff Faster, 

which includes "The Travelling Salesman Problem" and a competition between an astrophysicist and a paleontologist on how to move passengers boarding the planes faster 

Good luck, from Mrs. Lin 

2016 AMC-8 prep

Want to join our group for the up-coming AMC 8 test in Nov. ?

The problems are more complex, including many steps, occasionally not going in difficulty order, or/and there are troll/ trap questions, so it's GREAT to deter students to just memorize the formulas,

I do see some brilliant/ inquisitive students who are not good test takers, if you belong to that group, there are other ways to shine.

E-mail me on that. It's really much, much better to just sit back and enjoy the problems.
Check this awesome note from AoPS forum. 

You know, the most amazing thing about various competitions is the energy, the pleasure, the spontaneity, the camaraderie and the kindred spirits.

Thanks a lot to those diligent, inquisitive boys and girls for their impromptu, collaborated efforts.

You are one of its kind. :) 

2015 unofficial AMC 8 problems and detailed solutions from online whiz kids.

This year's AMC 8 official statistics is rolling in. 

Yeah, my boys' and girls' names are there. :) 

Move on to the most fun Mathcounts competition and of course, AMC 10 and AIME tests. 

My online whiz kids NEVER stop learning because ___ is who they are and it doesn't have to be all math and science related.

E-mail me at thelinscorner@gmail.com if you want to join us who LOVE problem solving (and many other areas equally challenging and engaging.)

Unofficial , official problems, answer key and detailed solutions to 2014 AMC8 test + official statistics.

11 Qualities Google Looks For In Job Candidates 

In Chinese : 《專題報導》亞裔尖子生 職場難出頭？

This year's (2013) AMC-8 results can be viewed here.

2013 AMC-8 problems in pdf format (easier to print out and work on it as a real test)

Try this if your school doesn't offer AMC-8 test.
40 minutes without a calculator.

If you want to use the test to prepare for Mathcounts, cover the multiple choice
options to make the questions harder unless you have to see the choices to answer
the question(s) asked.

2013 AMC-8 problems

2013 AMC-8 answer key

2013 AMC-8 problems with detailed (multiple) solutions 

Trickier problems : #18, 21(mainly the wording) and maybe 25 (slightly tricky)

2014 AMC-8 Result Statistics can be viewed here.

2014 AMC-8 problems and solutions from AoPS wiki. 

Comments :

#4 : We've been practicing similar problems to #4 so it should be a breeze if you see right away that the prime number "2" is involved. You'll get a virtual bump if you forgot about that again.

 #10 : Almost every test has this type of problem, inclusive, exclusive, between, calendar, space, terms, stages... It's very easy to twist the questions in the hope of confusing students, so slow down on this type of question or for the trickier ones, skip it first. You can always go back to it if you have time left after you get the much easier-to-score points.  (such as #12, 13, 18 -- if you were not trolled and others)

 #11 : Similar questions appear at AMC-10, Mathcounts.  For harder cases, complementary counting is easier.
This one, block walk is easier.

Counting Path on a Grid 

This one includes forbidden path finder

 #12: 1/ 3!

How about if there are 4 celebrities ? What is the probability that all the baby photos match with the celebrities ? only 1 baby photo matches,  only2 baby photos match, 3 baby photos match, or none matches ?

 #13: number theory

For sum of odd and /or even, it's equally likely --
odd + odd = even ; even + even = even
odd + even = odd ; even + odd = odd

For product of odd and/or even, it's not equally likely --
odd * even = even ; even * even = even
odd * odd = odd

For product probability questions, complementary counting with total minus the probability of getting odd product (all odds multiply)is much faster.

SAT/ACT has similar type of problems.

#15 : Central angle and inscribed angles --> Don't forget radius is of the same length.
Learn the basics from Regents prep 

#17: rate, time and distance could be tricky

Make sure to have the same units (hour, minutes or seconds) and it's a better idea writing down
R*T = D so you align the given infor. better.

Also, sometimes you can use direct/inverse relationship to solve seemingly harder problems in seconds.

 Check out the notes from my blog and see for yourself.

#18 : Trolled question. Oh dear !! 

1 4 6 4 1     , but it doesn't specify gender number(s) so 
(4 + 4)/2^4 is the most likely. 

#19 : more interesting painted cube problems --> one cube is completely hidden inside.

Painted cube animation from Fairy Math Tutors

Painted cube review   Use Lego or other plops to help you visualize how it's done.


#20 : Use 3.14 for pi and if you understand what shape is asked, it's not too bad.

#21: You can cross out right away multiples of three or sum of multiples of 3 by first glance.
For example 1345AA, you can cross out right away 3 and 45 (because 4 + 5 = 9, a multiple of 3).
You don't need to keep adding those numbers up. It's easier this way.

#22 : To set up two-digit numbers, you do 10x + y.
To set up three-digit numbers, you do 100x + 10y + z 

For those switching digits questions, sometimes faster way is to use random two or three digit numbers, not in this case, though.

#23 : This one is more like a comprehension question. Since it relates to birthday of the month, there are not many two digit primes you need to weight, so 11, 13 and 17. (19 + 17 would exceed any maximum days of the month). From there, read carefully and you should get the answer.

#24 : a more original question --

To maximize the median, which in this case is the average of the 50th and 51st term, you minimized the first 49 terms, so make them all 1s.
Don't forget the 51st term has to be equal or larger than the 50th term.

#25: The figure shown is just a partial highway image. 40 feet is the diameter and the driver's speed is 5 miles per hour, so units are not the same --> trap

I've found most students, when it comes to circular problems, tend to make careless mistakes because there are just too many variables.
Areas, circles, semi-circles, arch, wedge areas, and those Harvey like "think outside the box" fun problems.

Thus, it's a good idea to slow down for those circular questions. Easier said than remembered.

Happy Holiday !!

Mathcounts Strategy: Shoestring (or Shoelace) method of finding the area of any polygon

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

Shoelace formula from Wikipedia

More on Shoelace

Problems: Solutions below 

#1:  Find the area of a quadrilateral polygon given the four end points (3, 5), (11, 4), (7,0) and (9,8) in a Cartesian plane.

#2 2007 Chapter Target Round: A quadrilateral in the plane has vertices at (1,3),  (1,1), (2, 1) and (2006, 2007). What is the area of the quadrilateral?

#3: Find the area of a polygon with coordinates (1, 1), (3, -1),  ( 4, 4), and  (0.3)

#4: What is the number of square units in the area of the pentagon whose vertices are 

(1, 1 ), ( 3, -1),  (6, 2), (5, 6), and (2, 5)?

#5: Find the area of a polygon with coordinates ( -6, 0), (0, 5), (3, -2), and (4, 7)

#6: Find the area of a polygon with coordinates (20, 0), (0, 12), (3, 0), (4, -4)

#7: Find the area of a polygon with coordinates (-8, 4), (2, 12), (3, -5), (4, -4)

#8: Find the area of a triangle with coordinate (-8, -4), (-3, 10), (5, 6)

Solution I: Draw a rectangle and use the area of the rectangle minus the four triangles to get the area of the quadrilateral polygon. 

Solution II: Using shoestring method. First, plug in the four points. Second, choose one starting point and list the other points in order (either clockwise or counterclockwise)  and at the end, repeat the starting point. The answer is 33 square units.

Use this link to practice finding the area of any irregular polygon. Keep in mind that a lot of the times you don't need to use shoestring method. Be flexible!! Scroll to the middle section.





#2 Answer: 3008 square units

#3: Answer: 10.5 square units 

#4: Answer: 22 square units

#5: Answer: 45.5 square units

#6: Answer: 136 square units

#7: Answer: 98 square units

#8: Answer: 66 square units