The Grid Technique in Solving Harder Mathcounts Counting Problems : from Vinjai

The following notes are from Vinjai, a student I met online. He graciously shares and offers the tips here on how to tackle those harder Mathcounts counting problems. 

The point of the grid is to create a bijection in a problem that makes it easier to solve. Since the grid just represents a combination, it can be adapted to work with any problem whose answer is a combination.

For example, take an instance of the classic 'stars and bars' problem (also known as 'balls and urns', 'sticks and stones', etc.):

Q: How many ways are there to pick an ordered triple (a, b, c) of nonnegative integers such that a+b+c = 8? (The answer is 10C2 or 45 ways.)

Solution I: 

This problem is traditionally solved by thinking of ordering 8 stars and 2 bars. An example is:

* * * |    | * * * * *

  ^       ^       ^

  a       b       c

This corresponds to a = 3, b = 0, c = 5.

Solution II: 

But this can also be done using the grid technique:

The red path corresponds to the same arrangement: a = 3, b = 0, c = 5. The increase corresponds to the value: a goes from 0 to 3 (that is an increase of 3), b goes from 3 to 3 (that is an increase of 0), and c goes from 3 to 8 (that is an increase of 5). So a = 3, b = 0, c = 5.

Likewise, using a clever 1-1 correspondence, you can map practically any problem with an answer of nCk to fit the grid method. The major advantage of this is that it is an easier way to think about the problem (just like the example I gave may be easier to follow than the original stars and bars approach, and the example I gave in class with the dice can also be thought of in a more numerical sense).

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

Some articles on problem solving and parenting

Why America's Smartest Students Fail Math 

Are our kids failing in math because they can't read ? 

Kids of Helicopter Parents Are Sputtering Out 

The Juilliard Effect : Ten Years Later 

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 

A Skill for the 21st Century: Problem Solving by Richard Rusczyk

Does our approach to teaching math fail even the smartest kids ? 

Quotes from that article  "According to research from the University of California, Los Angeles, as many as 60 percent of all college students who intend to study a STEM (science, technology, engineering, math) subject end up transferring out. In an era when politicians and educators are beside themselves with worry over American students’ lagging math and science scores compared to the whiz kids of Shanghai and Japan, this attrition trend so troubles experts it has spawned an entire field of research on “STEM drop-out,” citing reasons from gender and race to GPAs and peer relationships."

Why Science Majors Change Their Minds (It's Just So Darn Hard) from the New York Times 

A Skill for the 21st Century: Problem Solving by Richard Rusczyk, founder of  "Art of Problem Solving".

Top 10 Skills We Wish Were Taught at School, But Usually Aren't 
from Lifehacker

Mathcounts prep

 Hi, Thanks for visiting my blog.

E-mail me at thelinscorner@gmail.com if you want to learn with me.  :) :) :) 

Currently I'm running different levels of problem solving lessons, and it's lots of fun learning along with students from different states/countries. 

Sequences and Series -- Arithmetic and Geometric Sequences

Sequences are fun to learn and not really that difficult. 
There are many similarities between arithmetic and geometric sequences, so 
learn both together. 

Enjoy !!!!! 

From Mthcounts Mini: Sequences and Series

Easier concepts:

Sequences

Arithmetic sequence/determine the nth term

Arithmetic and geometric sequences

Mathcounts strategies : review some sums 

Note : Don't just memorize, but really understand the concepts.

Harder concepts:

Sum and Average of An Evenly Space

Relationship between arithmetic sequences, mean and median

Sequences, series and patterns

Some Common Sums

Som sums' formula:

Pathfinder

From Mathcounts Mini :

Counting/Paths Along a Grid

From Art of Problem Solving

Counting Paths on a Grid 

Math Principles : Paths on a Grid : Two Approaches 

Question #1: How many ways to move the dominoes on a 6 by 6 checker board if you can only move the dominoes to the right or to the bottom starting from the upper left and you can't move the dominoes diagonally? 

Solution :
You can move the dominoes 5 times to the right at most and 5 down to
the bottom at most, so the answer is \(\dfrac {\left( 5+5\right) !} {5! \times 5!}\) = 252 ways

Question # 2: How many ways can you  move from A to B if you can only move downward and to right? 

Solution : There are \(\dfrac {\left( 4+4\right) !} {4!\times 4!}\) * 2 * \(\dfrac {\left( 4+4\right) !} {4!\times 4!}\) = 9800 ways from A to B

Sum of All the Possible Arrangements of Some Numbers

Check out Mathcounts, the best middle school competition math program up to the national level.

Questions to ponder: (detailed solutions below) 

It's extremely important for you to spend some time pondering on these questions first without peeking on the solutions. 

#1: Camy made a list of every possible distinct four-digit positive integer that can be formed using each of the digits 1, 2 , 3 and 4 exactly once in each integer. What is the sum of the integers on Camy's list?

#2: Camy made a list of every possible distinct five-digit positive even integer that can be formed using each of the digits 1, 3, 4, 5 and 9 exactly once in each integer. What is the sum of the integers on Camy's list? (2004 Mathcounts Chapter Sprint #29)

#3: 2020 Mathcounts state sprint #24 













Solutions:
#1:  
Solution I: 
There are 4! = 24 ways to arrange the four digits. Since each digit appears evenly so each number will appear 24 / 4 = 6 times.
1 + 2 + 3 + 4 = 10 and 10 times 6 = 60 ; 60 (1000 +100 +10 + 1) = 60 x 1111 = 66660, which is the answer.

Solution II: 

The median of four numbers 1, 2, 3, 4 is (1 + 2 + 3 + 4) / 4 = 2.5 and there are 4! = 24 ways to arrange
the four numbers. 
2.5 (1000 + 100 + 10 + 1) x 24 = 66660 

#2:  
Solution I: 
Since this time Camy wants five-digit even integer, which means that the number "4" has to be at the unit digit and only 1, 3, 5, 9 can be moved freely.
Again there are 4! = 24 ways to arrange the four numbers. 1 + 3 + 5 + 9 = 18 and 18 x 6 = 108 (Each number that can be moved freely appears 6 times evenly.)108 x 11110 + 4 x 24 = 1199976

Solution II: 

Since this time Camy wants five-digit even integer, which means that the number "4" has to be at the unit digit and only 1, 3, 5, 9 can be moved freely. 
There will be 4! = 24 times the even number 4 will be used so 4 x 24 = 96
As for the remaining 4 numbers, their average (or mean) is \(\dfrac{1 + 3+ 5 + 9} {4} = 4.5\)                               
4.5 * ( 10000 + 1000 + 100 + 10) * 24 (arrangements)  + 96 = 4.5 * 11110 * 24 + 96 = 1199976

#3: The answer is 101. 

Other applicable problems: (answers below)

#1: What is the sum of all the four-digit positive integers that can be written with the digits 1, 2, 3, 4 if each digit must be used exactly once in each four-digit positive integer? (2003 Mathcounts Sprint #30)

#2: What is the average (mean) of all 5-digit numbers that can be formed by using each of the digits 1, 3, 5, 7, and 8 exactly once? (You can use a calculator for this question.) (2005 AMC-10 B)

#3: What is the sum of all the four-digit positive integers that can be written with the digits 2, 4, 6, 8 if each digit must be used exactly once in each four-digit positive integer?

#4: What is the sum of all the 5-digit positive odd integers that can be written with the digits 2, 4, 6, 8, and 3 if each digit must be used exactly once in each five-digit positive integer?  

#5:What is the sum of all the four-digit positive integers that can be written with the digits 2, 3, 4, 5 if each digit must be used exactly once in each four-digit positive integer?

  





Answer key: 
#1: 66660
#2: \(\dfrac{1 + 3 + 5 + 7 + 8}{5} = 4.8\)
4.8 * 11111 =\(\color{red}{53332.8}\) 
#3: 133320
#4: 1333272
#5: 93324

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 \(\dfrac {\sqrt {3}-1} {2}\).

16 17 Mathcounts handbook more interesting questions that have nicer solutions

Thanks to Achuth for trying out these questions and time them as an actual Mathcounts test.  :) 

First week : warm up 1, 4, 7.  (time for 40 mins. like sprint)

Second week : warm up 2, 5, 8.

third week : workout 3 --> all right. (pair 1 to 6, 2 to 7, each time for 6 mins. as 

target) 

fourth week: workout 4 --> #95, then self correct. 

At lesson: workout 5 and other harder problems. 

These are nice questions that have various solutions, so it’s better to slow down and try them as puzzles.

Less is more and slow is fast.

If you are new to problem solving, one nice strategy is to make the question much simpler and explore ideas that come to your mind. 

Answer key down below. 

#66: A school of 100 fish swims in the ocean and comes to a very wide horizontal pipe. The fish have three choices to get to the food on the other side: swim above the pipe, through the pipe or below the pipe. If we do not consider the fish individually, in how many ways can the entire school of fish be partitioned into three groups with each group choosing a different one of the three options and with at least one fish in each group? 

 #105 When fully matured, a grape contains 80% water. After the drying process, called dehydration, the resulting raisin is only 20% water. What fraction of the original water in the grape remains after dehydration? Express your answer as a common fraction. 

 #112: Cora has five balls—two red, two blue and one yellow—which are indistinguishable except for their color. She has two containers—one red and one green. If the balls are randomly distributed between the two containers, what is the probability that the two red balls will be alone in the red container? Express your answer as a common fraction? 

 #116: A 12-foot by 12-foot square bathroom needs to be tiled with 1-foot square tiles. Two of the tiles are the wrong color. If the tiles are placed randomly, what is the probability that the two wrong-colored tiles share an edge? Express your answer as a common fraction.

#66: 4851

#105:  1/16

#112:  1/32

#116: 1/39 

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.










 

Mass Points Geometry

Some of the harder/hardest questions at Mathcounts can be tackled at ease using mass point geometry
so spend some time understanding it.

Basics 

2014-15 Mathcounts handbook Mass Point Geometry Stretch
from page 39 to page 40

(Talking about motivation, yes, there are students already almost finish
this year's Mathcounts' handbook harder problems.)

From Wikipedia

From AoPS

Mass Point Geometry by Tom Rike

Another useful notes 

Videos on Mass Point :

Mass Points Geometry Part I 

Mass Points Geometry : Split Masses Part II 

Mass Points Geometry : Part III 

other videos from Youtube on Mass Points

It's much more important to fully understand how it works, the easier questions the weights align
very nicely.

The harder problems the weights are messier, not aligning nicely, so you need to find ways to may them integers (LCM) for easier solving.

Let me know if you have questions. I love to help (:D) if you've tried.

Face Diagonal and Space Diagonal of a Rectangular Prism

Face diagonal and space diagonal of a cube 

Ways to calculate face and space diagonal.

Each side of the cube is x units long.

Use  45-45-90 degree angle ratio
( 1 - 1 - √ 2  ) or Pythagorean theorem to get the face diagonal.

Using Pythagorean theorem twice and you'll get the space diagonal.







Face diagonal and Space diagonal of a rectangular prism.

Same way to figure out the face

diagonal of a rectangle as well as

space diagonal of a rectangular prism.

Use Pythagorean theorem or

30-60-90 degree angle ratio

(1 -√ 3  - 2) to figure out the face

diagonal and Pythagorean theorem

twice to figure out the space diagonal.

2020 Mathcounts State Prep: Simon's Favorite Factoring Trick

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

The most common cases of Simon's Favorite Factoring Trick are:

I:  \(xy+x+y+1=\left( x+1\right) \left( y+1\right)\)

II:  \(xy-x-y+1=\left( x-1\right) \left( y-1\right)\)

It's easy to learn. Here is the best tutorial online, by none other than Richard Rusczyk.
The method Rusczyk uses at the second half is very nifty. Thanks!!

Questions to ponder:(answer key below)
#1: Both x and y are positive integers and \(x>y\). Find all positive integer(s) that \(xy+x+y=13\) 
#2: Both x and y are positive integers and \(x>y\). Find all positive integer(s) that \(2xy+2x-3y=18\)
#3: Find the length and the width of a rectangle whose area is equal to its perimeter.
#4: Twice the area of a non-square rectangle equals triple it's perimeter, what is the area of the rectangle? 













Answer key:
#1:  x = 6 and y = 1
#2: ( x, y ) = (4, 2) 
#3: Don't forget square is a kind of rectangle (but not the other way around) so there are two answers: 
4 by 4 and 3 by 6 units. 
#4: One side is 4 units and the other 12 units so the answer is 4 x 12 or 48 square units. 
There is another one, 6 by 6 that would fit if the question doesn't specify non-square rectangle. 

2019 AMC 8 problems, solutions and some thoughts

2019 AMC 8 problems and solutions, for students, by students

A student's reflection on this year's test : 

Mrs. Lin,
I did the AMC 8 yesterday, and it was actually quite easier than last year. I was reviewing my answers, and I believe I only got the last two wrong. I used stars and bars for the last one, but did 21C2 instead of 20C2. I could’ve done number 24, because geometry is really my best subject in math. I realized that I should’ve used mass points later on. It’s fine, though, because it’s still a good score. I think that many people could solve this test pretty well because in many of the last questions you could plug in the multiple choice answers and get the right answer. Also, a lot of it was just plain algebra. Question 20 was actually just an equation, which anybody who knows basic algebra can solve. I thought that I would never say this, but I honestly wish that it was harder, because I was hoping for some interesting problems. Those are the problems that get people’s gears turning; this year the problems were quite basic. I think many people will get really good scores on this test, which, along with a good thing, is also not so good because it brings down the credibility of the test.

Thanks,

some links that you can review those very basic, but extremely useful strategies on this 
year's seemingly harder, but not really last two questions. 

mass points  learn together with triangles sharing the same vertex 

dimensional change / scaling 

balls and urns, stars and bars  (lots of variations or twists on this one, so 
you need to fully understand the concept so to use it well. Be patient !!!!!) 

2012 Harder Mathcounts State Target Questions

Check out Mathcounts here -- the best competition math program for middle schoolers up to the 
state and national level. 

# 6: A semicircle and a circle are placed inside a square with sides of length 4 cm, as shown. The circle is tangent to two adjacent sides of the square and to the semicircle. The diameter of the semicircle is a side of the square. In centimeters, what is the radius of the circle? Express your answer as a decimal to the nearest hundredth. [2012 Mathcounts State Target #6]


#6:  Solution:
Using Pythagorean theory: (2 + r)² = (4-r)² + ( 2- r)²
4 + 4r + r² = 16 - 8r + r² + 4 - 4r + r²
 r² - 16 r + 16 = 0
Using the quadratic formula You have 8 ± 4√ 3
Only 8 - 4√ 3 = 1.07 works

There is a Mathcounts Mini #34 on the same question. Check that out !!

The above question looks very similar to this year's AMC-10 B #22, so try that one.
(cover the answer choices so it's more like Mathcounts)

2014 AMC-10 B problem #22 

#8: In one roll of four standard, six-sided dice, what is the probability of rolling exactly three different numbers? Express your answer as a common fraction. [2012 Mathcounts State Target #8]

Solution I : Permutation method
If order matters, there are 6 * 5 * 4 * 1 ways to choose the number, 1 being the same number as one of the previous one.
Let's say if you choose 3 1 4 1.

Now for the placement of those 4 numbers on the 4 different dice. There are 4C2 ways to place where the two "1" will
be positioned so the answer is : \(\dfrac {6\times 5\times 4\times 1\times 4C2} {6^{4}}\) = \(\dfrac{5}{9}\)

Solution II:  Combination method

There are 6C3 = 20 ways to choose the three numbers.

There are 3 ways that the number can be repeated. [For example: If you choose 1, 2, and 3, the fourth number could be 1, 2 or 3.]

There are \(\dfrac {4!} {2!}\) =12ways to arrange the chosen 4 numbers.[same method when you arrange AABC]

So the answer is\(\dfrac{20* 3 *12}{6^4}\) = \(\dfrac{5}{9}\)

Notes to 2018 Mathcounts chapter more interesting problems

2018 Mathcounts Chapter Spring problems : solutions down below 
Thanks to a boy mathlete who tried these problems and e-mail me for feedback. 

Please try these problems first before reading the explanations. :D

#25 : Three employees split a bonus valued at some number of dollars. Arman first receives $10 more than one third of the total amount. Bernardo then receives $3 more than one half of what was left. Carson receives the remaining $25. What is the total dollar value of the bonus?

#27: For a particular list of four distinct integers the mean, median and range have the same value. If the least integer in the list is 10, what is the greatest value for an integer in the list?

#29: There are two values of x such that \( |\dfrac {x-2018} {x-2019}|=\dfrac {1} {6}\). . What is the absolute difference between these two values of x? Express your answer as a common fraction.

Target #8 : Four congruent circles of radius 2 cm intersect with their centers at intersection points as shown. What is the area of the shaded region? Express your answer in terms of π.

#25 : It's easier if you go backward and use inverse operations to solve this question. 
(25 + 3) *2 = 56 and 56 + 10 = 66, which is \( \dfrac {2} {3}\) of the original bonus value, or 
what is left after \( \dfrac {1} {3}\) was given out. 

\( \dfrac {2} {3}\) of bonus is 66 dollars, so the answer is 99. 

#27: Let the average be x and the three other numbers be a, b, c and \( a < b < c \).
The least number is 10 (given), so 
\( 10+a+b+c \) = \( 4x \)---> equation 1 
\( \dfrac {a+b} {2}\) = \( x \) (how to find the median), so \( a+b\) = \( 2x \) --- (2)
\(c-10 = x\) (given because it's the range), \(c= x + 10\) ---(3)
Substitute (2) and (3) to equation one and you have \(10 + 2x + x + 10 = 4x\), so \(x = 20 \)
\(C = 20 + 10 = 30\), the answer 

#28 : Let \(x - 2018 = y\) , then \(x - 2019 = y -1\)
We then have either 
\( |\dfrac {y} {y -1}|=\dfrac {1} {6}\)  \(\rightarrow\) \(y\) = \( \dfrac {-1} {5}\)

or \( |\dfrac {y} {y-1}|=\dfrac {-1} {6}\) \(\rightarrow\) \(y\) = \( \dfrac {1} {7}\)

Their positive difference is \( \dfrac {12} {35}\) , the answer. 


Target #8 : Thanks to a 5th grader girl mathlete's solution: 

If you move parts around, you'll see the answer is exactly a semi-circle with a radius 2, so the answer is \(2\pi\), the answer. 






or check out another solution from me: 

Geometric mean of a right triangle

Geometric mean of a right triangle :

Video tutorial 

Notes

Practice I 

Practice II : focus on the height (or altitude)

Practice III : focus on the legs 

The Largest Rectangle Inscribed in Any Triangle

From Mathcounts Mini : Maximum area of inscribed rectangles and triangles

\(\Delta EHI\sim\Delta EFG\) \(\rightarrow\) \(\dfrac {a} {c}=\dfrac {d-b} {d}\)\(\rightarrow\) \(a=\dfrac {c\left( d-b\right) } {d}=\dfrac {-c\left( b-d\right) } {d}\)

We are going to find out what the largest area of a rectangle is with the side length a and b.
It can be shown that by substituting the side length "a" with the previous equation + completing the square that the largest area is half of the area of the triangle the rectangle is embedded.

\(a\times b=\dfrac {-c\left( b-d\right) \times b} {d}=\dfrac {-c\left( b^{2}-bd\right)} {d}= \dfrac {-c\left( b-\dfrac {1} {2}d\right) ^{2}+\dfrac {1} {4}dc} {d}\).

From there, you know that when \(b= \dfrac {1} {2}d\), it will give you the largest area, which is \(\dfrac {1} {4}dc\).

\(a=\dfrac {-c\left( b-d\right) } {d}= \dfrac {-c\left( \dfrac {1} {2}d-d\right) } {d}=\dfrac {c\left( d-\dfrac {1} {2}d\right) } {d}=\dfrac {1} {2}c\).

Thus, the maximum rectangle area occurs when the midpoints of two of the sides of the triangle were joined to make a side of the rectangle and its area is thus 50% or half of the area of the triangle or 1/4 of the base times height.

Proof without words from Mr. Rusczyk 

Try using different types of triangles to experiment and see for yourself.
Paper folding is fun !!!!!
It's very cool :D

Dimensional Change questions I:

Questions written by Willie, a volunteer.  Answer key and detailed solutions below.

1a. There is a regular cylinder, which has a height equal to its radius. If the radius and height are both increased by 50%, by what % does the total volume of the cylinder increase?

1b. If the radius and height are both decreased by 10%, by what % does the total volume of the cylinder decrease?

1c. If the radius is increased by 20% and the height is decreased by 40%, what % of the volume of the original cylinder does the volume of the new cylinder represent?

1d. If the radius is increased by 40% and the height is decreased by 20%, what % of the volume of the original cylinder does the volume of the new cylinder represent?

1e. If the height is increased by 125%, what % does the radius need to be decreased by for the volume to remain the same?

2. If the side of a cube is increased by 50%, by what % does the total surface area of the cube increase?

3a. If the volume of a cube increases by 72.8%, by what % does the total surface area of the cube increase?

3b. By what % did the side length of the cube increase?

4. You have a collection of cylinders, all having a radius of 5. The first cylinder has a height of 2, the second has a height of 4, the third a height of 6, etc. The last cylinder has a height of 50. What is the sum of the volumes of all the cylinders (express your answer in terms of pi)?

Answer key: (Each question should not take you more than 30 seconds to solve if you really understand the concepts involved.)

 

 

1a.  The volume of a cylinder is πr²x h (height). The radius itself will be squared and the height stays at constant ratio. The volume will increased thus (1.5)³ - 1³ -- the original 100% of the volume = 2.375

=237.5%

1b.  Like the previous question: 1³ - 0.9³ [when it's discount/percentage decrease, you use the 100% or 1 - the discount/decrease percentage] = 0.271 =  27.1% decrease

1c.  1.2² [100% + 20% increase = 1.2] x 0.6 [100% -40% = 0.6] = 0.864  or  

86.4% of the original volume

1d.  1.4² [100% + 40% increase = 1.4] x 0.8 [100% -20% = 0.8] = 1.568 = 156.8% of the original volume

1e.  When the height of a cylinder is increased 125%, the total volume is is 225% of the original cylinder, or 9/4.

Since the radius is used two times (or squared), it has to decrease 4/9^1/2 = 2/3 for the new cylinder to have the same volume as the old one. [9/4 times 4/9 = 1 or the original volume.]

1 - (2/3) = 1/3 = 0.3 = 33.3%

2. Surface area is 2-D so 1.5² - 1 = 1.25 = 125% increase

3a. If a volume of a cube is increased by 72.8 percent, it's 172.8% or 1.728 of the original volume. Now you are going from 3-D (volume) to 2-D (surface area). 1.728^2/3 = 1.44 or 44% increase. [Don't forget to minus 1 (the original volume) since it is asking you the percentage increase.]

3b. From surface area, you can get the side increase by using 1.44^1/2 = 1.2, so 20% increase.

Or you can also use 1.728^1/3 = 1.2;  1.2 - 1 = 20%

4. The volume of a cylinder is πr²x h . (2 + 4 + 6 + ...50) x 5²π = (25 x 26) x 25π =16250π

2011 Mathcounts Chapter Sprint Round solutions

#22: The answer is 2674.
 See left for explanations.














#23: Let the two consecutive positive integers be x and x + 1.
( x + 1 ) / x = 1.02, x + 1 = 1.02x, 0.02x = 1, x = 1 divided by 0.02 = 1 times 100/2 = 50
The two numbers are 50 and 51 and their sum is 50 + 51 = 101.

#24: The two x-intercepts when y is "0" are 10 or -10; the two y-intercepts when x is "0" are 5 or -5.
Area of a rhombus is D1 x D2 / 2 so the answer is [10-(-10)] x [5 -(-5)] = 100 square units.

#25: The area ratio of the two similar triangle is 150/6 so the line ratio is √ 150/6  or 5:1.
the length of the hypotenuse of the smaller triangle is 5 inches, so the other two legs are 3 and 4. 
(a Pythagorean triple)
The sum of the lengths of the legs of the larger triangle is (3 + 4) * 5 = 35.

#26: To have same number of boys and girls, the committee needs to consist of 3 boys and 3 girls. 
(6C3 x 4C3)/ 10C6 = 80/210 = 8/21

#27: When the point (3, 4) is reflected over the x-axis to B, B would = (3, -4). 
When B is reflected over the line y = x to C, C would = (-4, 3).

The area of the triangle is [4 -(-4)] x [ 3 - (-4)]/ 2 = 28 square units. 


#28: Tonisha is 45 miles ahead Sheila when Sheila leaves Maryville at 8: 15 a.m. 
Each hour Sheila will be 15 miles closer to Tonisha. 45/15 = 3, which means that 3 hours 
later Sheila will pass Tonisha. 
8:15 + 3 hours = 11: 15 a.m.


#29: Using 30-60-90 degree angle ratio, you can make the radius be √ 3 and half of the side of the hexagon would be 1 so each side of the hexagon
is 2.

The area of the hexagon is (√ 3/4) times 2² times 6 = 6√ 3.
The area of the circle is 3Π.
The fraction is 3Π/6√ 3 = √ 3 / 6
a = 3 and b = 6, ab = 18

#30:  Area of triangle KDC is easy to find once you realize the height is just the right triangle with a hypotenuse 6 and a leg 4. (half of the length of CD),

Using Pythagorean theorem, you get the height to be 2√ 5 .

8 x 2√ 5 /2 = 8 √ 5 .

Original Problem from a Student Problem Solver on Similar Triangles

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