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2015 Mathcounts Chapter / State Prep : Octagon

January 29, 2015, 4:58 am
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Question #1: How to find the area of a regular octagon.
 method I:
The area of a regular octagon can be obtained by using 45-45-90 degree angle ratio to get the sides
and from there get the area.
method II: 
You can also use the area of the square minus the 4 triangles at the corner.
To get the length of the unknown
side length, again, use 45-45-90 degree angle ratio.

If the side is "a" unit, the area is \(2(1+\sqrt {2})*a^{2}\).



Question #2:  What is the area ratio of the regular octagon ABCDEFGH 
to the embedded square ACEG?

\(\Delta ATC\) is a right triangle and \(\overline {AC}\) is the hypotenuse.

Use the Pythagorean theorem:\(\left( \overline {AC}\right) ^{2}\) [same as the area of the square]
= \(\left( \sqrt {2}\right) ^{2}+\left( 2+\sqrt {2}\right) ^{2}\)
=\(8+4\sqrt {2}\) 
The aera ratio of the regular hexagon ABCDEFGH to the area of the embedded square ACEG
=\(\dfrac {8+8\sqrt {2}} {8+4\sqrt {2}}\) = \(\sqrt {2}\) 





Questions(answer key below): 
#1: What is the area of a regular octagon if each side is "2"
#2: What is the area of the rectangle HCDG? 
#3: What is the area of the trapezoid ABCH?
#4: What is the area of the triangle ADG?










Answer key:
#1:  \(8+8\sqrt {2}\)
#2: Area of the rectangle HCDG = \(4+4\sqrt {2}\) [1/2 of the area of the octagon]
#3: Area of the trapezoid ABCH is \(2+2\sqrt {2}\) [1/4 of the area of the octagon]
Notice the area of the HCDG = the sum of the other two congruent trapezoids
ABCH and FEDG, thus break the octagon to 1 : 1 ratio.
#4: The area is \(4+3\sqrt {2}\)
 




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This Week's Work : Week 1 -- for Inquisitive Young Mathletes

February 3, 2015, 1:13 pm
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Rate, Time, and Distance Question

February 6, 2015, 4:21 pm
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Question: Harder SAT question: Esther drove to work in the morning at an average speed of 45 miles per hour. She returned home in the evening along the same route and averaged 30 miles per hour. If Esther spent a total of one hour commuting to and from work, how many miles did Esther drive to work in the morning?

Solutions I:
Let t be the time it took Esther to drive to work and (1-t ) be the time it took him to return home.
Since rate times time = distance, we can set the equation as 45 t = 30 (1- t), 75t = 30 so
t = \(\frac{\Large{2}}{\Large{5}}\) Plug in t to the previous equation : 45 * \(\frac{\Large{2}}{\Large{5}}\) = 18 miles ,which is the answer.

Solution II: 
Let D be the distance from Esther's home to work. 
\(\frac{\Large{D}}{\Large{45}}\) +  \(\frac{\Large{D}}{\Large{30}}\) = 1 (hour)
Times 90 both sides to get rid of the denominator and you have 2D + 3D = 90 
D =18 miles
Solution III:
The rate ratio between driving to work and returning home is 45 : 30 or 3 :  2.
Since rate and time are inversely related (rt = d), the time ratio between the two is 2 : 3. 
45*\(\frac{\Large{2}}{\Large{5}}\) * 1 (hour) =18 miles

Solution IV: (from Varun in FL)
The average speed can be found by harmonic mean: \(\dfrac {2\times 45\times 30}{45+35}\) = 36 mph. She drives for a total of 1 hour, so she drives 36 miles. Since she drives the same distance both ways, the distance to work is \(\dfrac {36} {2}\)=18 miles.

Question 1991 Mathcounts National #28 : A man is running through a train tunnel. When he is \(\frac{\Large{2}}{\Large{5}}\) of the way through, he nears a train that is approaching the tunnel from behind him at a speed of 60 mph. Whether he runs ahead or runs back, he will reach an end of the tunnel at the same time the train reaches that end. At what rate, in miles per hour, is he running? (Assume he runs at a constant rate.)





Solution : When the man went back \(\frac{\Large{2}}{\Large{5}}\) of the length of the tunnel, the train is at the beginning of the tunnel. 

Same thing happened if the men went ahead \(\frac{\Large{2}}{\Large{5}}\) of the length of the tunnel. The time it took the train to run through the length of the tunnel, the man could only run \(\frac{\Large{1}}{\Large{5}}\) of the same distance. 

So the speed of the man is \(\frac{\Large{1}}{\Large{5}}\) of the train's speed, which is \(\frac{\Large{1}}{\Large{5}}\) * 60 or 12 mph.  








Question #3: Sally drives to her aunt's house, which is 84 miles away, at 40 miles per hour and comes back home at 60 miles per hour. What is her average speed? 
SolutionI: 
To find average speed, you use total distance over total time it takes Sally to drive. 
It takes Sally \(\frac{\Large{84}}{\Large{40}}\) + \(\frac{\Large{84}}{\Large{60}}\), or total 3.5 hours 
to drive 84 * 2 = 168 miles. 
\(\frac{\Large{168}}{\Large{3.5}}\) = 48 mph

Solution II: 
Similar to dimensional change question, as long as the segments are constant, the distance Sally drives is extraneous. Thus, it's much easier to use \(\frac{\Large{2}}{\frac{\Large{1}}{\Large{40}} - \frac{\Large{1}}{\Large{60}}}\)or \(\frac{\Large{2*40*60}}{\Large{40+60}}\) =48 mph


Question #4: If Sally drives to her work at 40 miles per hour, she will be 15 minutes late. If she drives to her work at 60 miles per hour, she will be 15 minutes early, what speed will she arrive at her work on time?  How far away is her office ? 
Solution I: 
Let t be the time (hpm) Sally drives to her work on time. According to the given, we can set up 
the following equation: 
40 (t + \(\frac{\Large{1}}{\Large{4}}\)) = 60 ( t- \(\frac{\Large{1}}{\Large{4}}\) )
40 t + 10 = 60 t - 15 ;      25 = 20t ;       t = 1.25
Plug in to get the distance as 40 (1.25 + 0.25) = 60 miles

To get the average, use total distance over total time she spent \(\rightarrow\) \(\frac{D}{T}\) = \(\frac{\Large{120}}{\Large{1.5 + 1}}\) = 48 mph

Solution II :
Again, have you noticed that if  both time (late or early) to arrive to the destination is the same, the easiest
method is to still use the harmonic mean. \(\rightarrow\) \(\frac{\Large{2*a * b}}{\Large{a + b}}\)
a and b are the rates. \(\frac{\Large{2*40*60}}{\Large{40+60}}\) = 48 mph
 
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Sequences and Series : level I

February 11, 2015, 8:59 am
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2014 Mathcounts State Prep : How to Avoid Making Careless Mistakes

February 19, 2015, 9:38 am
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Why Can Some Kids Handle Pressure While Others Fall Apart? from the New York Times

Relax! You'll Be More Productive from the New York Times

First, please read Mathcounts "Forms of Answers" carefully for a few times and remind yourself not to write down the unallowable answers.

Stop Making Stupid Mistakes by Richard Rusczyk from Art of Problem Solving

How to Avoid Careless Mathematical Errors from Reddit

Common errors from all my students: 

Number 1 issue with most of the boys/and a few girls who have strong intuition in math is their horrible handwriting since their minds work much faster than their hands can handle. (See!! I'm making excuses for them.)

I call some of my students "second try ___ [Put your name here.] or third try ____ because I don't need to teach them how to solve a problem but the normal pattern is that it has to take them twice or three times to get it finally right. Thus, they need to slow down and organize their thoughts.

Practice your numbers:  4 (it's not 21), 7, 9 (don't dislocate the circle on top), 2 (don't make it looks like a 7), etc...  

35 cents is very different from 0.35 cents.

Unless stated, you write improper fraction as your answer. Make sure to simplify the answer.

Many mistakes happen when it involves fractions, negative numbers, parenthesis, and questions that
have radicals on the denominator and you need to simplify it, so make sure to double check your math.

Some problems takes many steps to reach the final solutions. Some involve lots of data, extreme long strings of information so if those are your weakness, skip them first and go back to them later.

Every point is the same so don't stay with a tedious question too long and panic later not able to finish the last few harder questions which might be much easier to solve arithmetic wise if you know how.

Circle questions trouble many students. Make sure you are aware of what's being tested.
Is it circumference or area, is it diameter or radius, is it linear to area (squared) or vice versa ?
Are the units the same (most frequently appeared errors)?

Don't forget to divide by 2 for the triangle, times 6 if you use area of an equilateral triangle to get the hexagon. For geometry questions or unit conversions, don't do busy work, write down the equivalent equations and cancel like crazy.

Same goes to probability questions, cancel, cancel and cancel those numbers. You don't need to practice mental math multiplication for most of the Mathcounts problems. Think Smart!!

The answer for probability questions can only be 0 to 1, inclusive.

For lots of algebra questions, manipulations are the way to go or number sense. Use the digit clue to help you narrow down the "trial and error" method. 

Make sure you have the terms and space type questions right. Calendar questions, how many numbers (inclusive, exclusive, between, ...), Stage doesn't necessarily start with 1 or 0.
Make sure you don't over-count or under-count.

Mos students got counting questions wrong when it involves limit.

Case in point : How many non-congruent triangles are there if the perimeter is 15? 
Answer: There are 7 of them.  Try it !!

To be continue...









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

February 19, 2015, 9:39 am
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Finally, 2014 Mathcounts state are clear for discussions.&nbsp You can download this year's Mathcounts state problems here at Mathcounts.org.

Links, notes, Hints or/and solutions to this year's 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 )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: 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 using the Pythagorean Theorem is \sqrt{10^2-5^2}=5\sqrt{3}. 
Using the diagram below, let r be the radius of the top cone and let h be the height of the top cone. 
Let s=\sqrt{r^2+h^2} be the slant height of the top cone.

Image

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 yields10r=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}\impli...We simply substitute the value of h=r\sqrt{3} from above to yieldr^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 is50\pi-119.874\approx 157.08-119.874=37.206.

All we have left is to add the two bases. The total area of the bases is 25\pi+\pi\cdot r^2\approx 138.477. So our final answer is37.207+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]
\(\dfrac {1} {2}\left( 2\times 10\pi + 2\times 10\times \dfrac {\sqrt [3] {2}} {\sqrt [3] {3}}\pi \right)\)* \(\left( 10-10\times \dfrac {\sqrt [3] {2}} {\sqrt [3]{3}}\right)\)





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2013 Mathcounts State Harder Problems

October 28, 2014, 8:53 am
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 You can download this year's Mathcounts state competition questions here.

Trickier 2013 Mathcounts State Sprint Round questions :
Sprint #14:  
From Varun: 
Assume the term "everything" refers to all terms in the given set.
1 is a divisor of everything, so it must be first.
Everything is a divisor of 12, so it must be last.
The remaining numbers left are 2, 3, 4, and 6.
2 and 3 must come before 6, and 2 must come before 4.
Therefore, we can list out the possibilities for the middle four digits:
2,3,4,6
2,3,6,4
3,2,4,6
3,2,6,4
2,4,3,6
There are 5 ways--therefore 5 is the answer.

From Vinjai:
Here's how I did #14:
First, notice that 1 must be the first element of the set and 12 must be the last one.
So that leaves only 2,3,4,6 to arrange.
We can quickly list them out.
The restrictions are that 2 must be before 4, 3 must be before 6, and 2 must be before 6:
2,3,4,6          3,2,4,6          4,2,3,6            6,2,3,4
2,3,6,4          3,2,6,4          4,2,6,3            6,2,4,3
2,4,3,6          3,4,2,6          4,3,2,6            6,3,2,4
2,4,6,3          3,4,6,2          4,3,6,2            6,3,4,2
2,6,3,4          3,6,2,4          4,6,2,3            6,4,2,3
2,6,4,3          3,6,4,2          4,6,3,2            6,4,3,2

Only the bold ones work. So, the answer is 5.

#17: Common dimensional change problem
\(\overline {ZY}:\overline {WV}=5:8\) -- line ratio
The volume ratio of the smaller cone to the larger cone is thus \(5^{3}: 8^{3}\).
The volume of the frustum is the volume of the larger cone minus the volume of the smaller cone
= \(\dfrac {8^{3}-5^{3}} {8^{3}}\times \dfrac {1} {3}\times 8^{2}\times 32\times \pi\) = 516\(\pi\)

More problems to practice from Mathcounts Mini 

#24:  The answer is\(\dfrac {1} {21}\).

#28: Hats off to students who can get this in time !! Wow!!
From Vinjai:

For #28, there might be a nicer way but here's how I did it when I took the sprint round:

# 4's     # 3's     # 2's       # 1's     # ways
   1         2          0            0          3
   1         1          1            1          24
   1         1          0            3          20
   1         0          3            0          4
   1         0          2            2          30
   1         0          1            4          30
   0         2          2            0          6
   0         2          1            2          30
   0         2          0            4          15
   0         1          3            1          20
   0         1          2            3          60
   0         0          3            4          35

TOTAL: 277

#29: 
\(\Delta ADE\) is similar to \(\Delta ABC\)
Let the two sides of the rectangle be x and y (see image on the left)

\(\dfrac {x} {21}=\dfrac {8-y} {8}\)
x =\(\dfrac {21\left( 8-y\right) } {8}\)

xy =  \(\dfrac {21\left( 8-y\right) } {8}\)  * y = \(\dfrac {-21y\left( y-8\right) } {8}\) =
\(\dfrac {-2l\left( y-4\right) ^{2}+21\times 16} {8}\)

From the previous equation you know when y = 4, the area \(\dfrac {21\times 16} {8}\)is the largest. The answer is 42. 





Here is a proof to demonstrate that the largest area of a rectangle inscribed in a triangle is
half of the area of that triangle.

#30:
Solution I :
If (x, y) are the coordinates of the center of rotational points, it will be equal distance from A and A' as well as from B and B'.
Use distance formula, consolidate/simplify and solve the two equations, you'll get the answer (4, 1).

Solution II:

How to find the center of rotation from Youtube.

From AoPS using the same question

To sum up:
First, connect the corresponding points, in this case A to A' and B to B'.
Second, find the equation of the perpendicular bisector of line \(\overline {AA'}\), which is
y =  - x + 5
and \(\overline {BB'}\), which is y = 5x - 19
The interception of the two lines is the center of rotation.
The answer is (4, 1).

2013 Mathcounts Target :
#3:
RT = D, unit conversions and different rates are tested here:

Make Joy's rate (speed) uphill be x m/s, his downhill speed be 2x m/s.

It takes Greg 3000 seconds (time) to reach the starting point and that is also what it takes Joy to
ride up to the hill and down to the same point.

\(\dfrac {7000} {x}+\dfrac {10000} {2x}=3000\) \(\rightarrow\) x = 4 m/s 

#8:


Using "finding the height to the hypotenuse".( click to review)

 you get \(\overline {CD}=\dfrac {7\times 24} {25}\).

Using similar triangles ACB and ADC, you get  \(\overline {AD}=\dfrac {576} {25}\).
[\(\dfrac {24} {x}=\dfrac {25} {24}\)]

Using angle bisector (click to review),

you have \(\overline {AC}:\overline {AD}=\overline {CE}:\overline {ED}= 24: \dfrac {576} {25}\) = 600 : 576 = 25 : 24

\(\rightarrow\)\(\overline {ED}= \overline {CD}\times \dfrac {24} {24+25}\) = \(\dfrac {7\times 24} {25}\times \dfrac {24} {24+25}\) = \(\dfrac {576} {175}\)

 
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2015 Mathcounts Competition Preparation Strategies

March 11, 2015, 5:26 am
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It's the Mathcounts state, AIME/USAJMO/USAMO and science fair/bowl/olympiad/robotics/spelling bee/... season. Yeah !! 

Don't forget other equally interesting activities/contests, which engage your creativity  and imagination. Some also require team work. Go for those and have fun !! 

Don't just do math.  



Welcome Back !! Hope during the summer, you are still working on some challenging word problems
(my honor students do; they never stop learning ) + trying some new activities (don't just do math).

E-mail me at thelinscorner@gmail.com if you want to join us. Currently I'm running different levels of problem solving group lessons, and it's lots of fun learning along with students from different states.

My most advanced group students are just AMAZING !! to say the least. Ha ha, we are using
AMC-10, 12 questions as countdown round practices and some can solve the first few AIME problems in less than a minute. Oh dear !!

Before going full throttle mode for competition math, please spend some time reading this
well- thought-out article from BOGTRO at AoPS "Learn How to Learn".

It will save you tons of time and numerous, unnecessary hours without a clear goal, better method in mind.

Less is more. My best students make steady, very satisfactory progress in much less time than those
counterparts who spent double, triple, or even more multiple times of prep with little to show.

It's all about "deliberate practices", "tenacity", and most of all, "the pleasure of finding things out on your own".

Take care and have fun problem solving.

I have been coaching students for many years. By now, I know to achieve stellar performance you need : Grit (from TED talk), not only that but self-awareness (so you can fairly evaluate your own progress) and a nurturing-caring environment. (Parents need to be engaged as well.)
               
Thanks a lot !!  Mrs. Lin

"Work Smart !!" , "Deliberate practices that target your weakness ", " Relax and get fully rested.", "Pace your time well", "Every point is the same so let go of some questions first; you can always go back to them if time permits."

"It's tremendous efforts preparing for a major event on top of mounting homework and if you are the ones who want to try that, not your parents and you work diligently towards your goal, good for you !!"

"Have fun, Mathcounts changes lives, because at middle school level at least, it's one of those rare occasions that the challenges are hard, especially at the state and national level."

Now, here are the links to get you started: 

Of course use my blog.  Whenever I have time I analyze students' errors and try to find better ways (the most elegant solutions or the Harvey method I hope) to tackle a problem. Use the search button to help you target your weakness area.

Last year's Mathcounts competition problems and answer key

This year's handbook questions.
Near the end of the handbook, there is a page called  problem index (page 82 and 83 for 2013-2014 handbook).
For state/national prep, find your weakness and work on the problems backwards, from the hardest to the easiest. 

Here are some other links/sites that are the best.

Mathcounts Mini : At the very least, finish watching and understanding most of the questions from 2010 till now and work on the follow-up sheets, since detailed solutions are provided along with some more challenging problems.

For those who are aiming for the state/national competition, you can skip the warm-up and go directly to "The Problems" used on the video as well as work on the harder problems afterward.

Art of Problem Solving 

The best place to ask for help on challenging math problems. Some of the best students/coaches/teachers are there to help you better your problem solving skills.

Register for Alcumus and start using the great tool to practice deliberately.
Change the setting based on the levels of your proficiency of different topics.                                                                     Do Not Rush !!

Awesome site!!
       
For concepts reviewing, try the following three links.

Mathcounts Bible
 
Mathcounts Toolbox
 
Coach Monks's Mathcounts Playbook
 
You really need to understand how each concept works for the review sheets to be useful.

To my exasperation, I have kids who mix up the formulas without gaining a true understanding and appreciation of how an elegant, seemingly simple formula can answer myriads of questions.

You don't need a lot of formulas, handbook questions, or test questions to excel.

You simply need to know how the concepts work and apply that knowledge to different problems/situations.

Hope this is helpful!!

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2015 Mathcounts State Prep: Simon's Favorite Factoring Trick

March 16, 2015, 11:21 am
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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. 
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Some Articles to Read and Ponder after Mathcounts Chapter Competition

March 24, 2015, 5:04 am
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For those students who really love problem solving, who care and are inquisitive but are disappointed at their chapter's/state's performance, I want to let you know that it's an honor meeting you online and learning from/along with you.

I know no matter how sincere I write here, it won't help much, so I'll just shut up.

However, I'm also disappointed at some students who said they need to take a break for the foreseeable future.

Well, if you love ___ (fill in the blank), you won't stop if you don't get to the chapter/state/nationals.
So there...

Here are some articles that after your taking a break from problem solving (I hope it's not too long), hope to see you come back and read them. Best of luck !!  Keep me posted !!

Pros and Cons of Math Competitions

Dealing with Hard Problems

What is Problem Solving?

Life After Mathcounts

Great Mathematicians on Math Competitions and "Genius"

Math Contests Kind of Suck from Mathbabe

TEDxCaltech -Jordan Theriot- The Pleasure of Finding Things Out

Why Physics? Skateboarding Physicist and Educator Dr. Yung Tae Kim

Richard Feyman --The Pleasure of Finding Things Out

Hope it helps !! 


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Mass Points Geometry

April 16, 2015, 9:20 am
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Some of the harder/hardest questions at Mathcounts can be tackled at ease using mass point geometry
so spend some time understanding it.

Mass Point Geometry : This one is easier to understand with questions that have detailed solutions.

Mass Point Geometry : another link for basics

This year's 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

Mass Points by Kevin An

Videos on Mass Point :

Basics 

Mass Point Geometry Part I 

Mass Point Geometry : Split Masses Part II 

Mass Point Geometry : Part III 


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

April 20, 2015, 6:22 am
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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.
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Harder Mathcounts State Question

May 3, 2015, 12:48 pm
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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
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Games: logical puzzle: the wolf, the sheep and the cabbages

May 14, 2015, 2:27 pm
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Show Your Work, Or, How My Math Abilities Started to Decline

May 16, 2015, 3:28 pm
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Show your work, or, how my math abilities started to decline

I think it's problematic the way schools teach Algebra. Those meaningless show-your-work approaches, without knowing what Algebra is truly about. The overuse of calculators and the piecemeal way of teaching without the unification of the math concepts are detrimental to our children's ability to think critically and logically.

Of course eventually, it would be beneficial to students if they show their work with the much more challenging word problems (harder Mathcounts state team round, counting and probability questions, etc...), but it's totally different from what some schools ask of our capable students.

How do you improve problem solving skills with tons of worksheets by going through 50 to 100 problems all look very much the same? It's called busy work. 

Quote from Einstein. "Insanity: doing the same thing over and over again and expecting different results."

Quotes from Richard Feynman,the famous late Nobel-laureate physicist. Feynman relates his cousin's unhappy experience with algebra:

My cousin at that time—who was three years older—was in high school and was having considerable difficulty with his algebra. I was allowed to sit in the corner while the tutor tried to teach my cousin algebra. I said to my cousin then, "What are you trying to do?" I hear him talking about x, you know."Well, you know, 2x + 7 is equal to 15," he said, "and I'm trying to figure out what x is," and I say, "You mean 4." He says, "Yeah, but you did it by arithmetic. You have to do it by algebra."And that's why my cousin was never able to do algebra, because he didn't understand how he was supposed to do it. I learned algebra, fortunately, by—not going to school—by knowing the whole idea was to find out what x was and it didn't make any difference how you did it. There's no such a thing as, you know, do it by arithmetic, you do it by algebra. It was a false thing that they had invented in school, so that the children who have to study algebra can all pass it. They had invented a set of rules, which if you followed them without thinking, could produce the answer. Subtract 7 from both sides. If you have a multiplier, divide both sides by the multiplier. And so on. A series of steps by which you could get the answer if you didn't understand what you were trying to do.
So I was lucky. I always learnt things by myself.
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Base Numbers: Level 1

May 18, 2015, 7:47 am
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Base numbers: 

Convert other bases to base 10. 
Example #1:  
 1  | 0 |02 
 22|21 |20                  0*20 + 0*21 + 1*22 = 4

Thus, 1002  = 4 in base 10

The first few base 2 (or binary system) numbers that are equivalent to their base 10 (or decimal) counterpart.
 
base 2                       base 10

0                               0

1                               1

10                             2

11                             3

100                           4

101                           5

110                           6

111                          7

1000                         8

1001                         9

1010                       10    

Example #2:
 
1|  2|  45    
52|51|50                    4*50+ 2*51 + 1*52  = 4 + 10 + 25 = 39 

Thus, 1245 = 39 in base 10   

Some important points: 
 
* All the digits in the number has to be smaller than the base. (Why?)
* Hexadecimal is base 16. In that system, you can use numbers from "0" to "9", the same as base 10.
  However, numbers 10 to 15, you use letters: A for 10, B for 11, C for 12, D for 13, E for 14 and last, F      for 15. 

Short cut to some base conversions:

 #1: Convert 11011112to base 8.

Since 8 = 23, you can convert base 2 to base 8 by regrouping every 3 digits.
Thus the original number will become 1578
because if you regroup the original number starting from the right, you have 12-1012-1112and it's 157 in base 2's term.

#2: Try convert 111100112to base 4, which is 22 so you regroup by 2.

11 (3 in base 2)-11(3 in base 2)-00 (0 in base 2)-11(3 in base 2) so the answer is 33034. 

Convert a number from base 10 to any other base

Here is a link to help you practice more of bases conversion.
 
Practice Problems: (answers below)

Convert other bases to base 10.
#: 101112                                           
#2:  3304
#3:  1005 
#4:  2103
#5:  246
#6: Convert101112  to base 4 
#7: Convert 1233214  to base 16 (or hexadecimal) [student Andrew's problem]
#8: 110100112  is what to base 16 ?
#9:  Convert1648 to binary (or base 2) [Daniel's question]
#10: What is 100005  minus 15 ? [Remaining :Willie's questions]
#11: What is0.18to base 10?
#12: What is 5378  divided by 108 ?
#13: 3716  is what in base 4? what in base 10?
 
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Testimonials for my services. So far, most through words of mouth or chance meets online and it's great.

May 19, 2015, 6:50 am
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Dear Mrs Lin,

I just did not get the opportunity earlier to thank you for all your help. _____ was able to make to the National MathCounts largely because of your excellent guidance and coaching. I do not know how to thank you. He had a great once-in-a-lifetime experience there and he really loved the competition as well as meeting other people.
If you can please provide me your mailing address, he wants to send you a gift as a token of thanks for your guidance and tutoring. 

Sincerely,


Dear Mrs. Lin,

I have been meaning to tell you, but just didn't get the time. He just 'LOVES' your sessions. He said he is learning so much and gets to do lots of problems and he likes all the tips/shortcuts you are teaching. He looks forward to your session - he was so upset when we couldn't get back on time from _____ because he didn't want to miss your session. Honestly we came back Tue night only because he cried so much that he didn't want to miss your class:) 

In school, his teacher focuses more on details like, put all the steps, write neatly, don't disturb the class by asking unnecessary questions, don't ask for more work, behave properly etc....so he is not too happy with math in school.

Thank you so much for making such a big difference in his life. He  enjoys doing the homework you are assigning him and has not complained at all. He said "Mrs Lin is so smart and I love her classes...wish she lived next to our house...so I can go to her and have live classes'[disclaimer : I'm not ; my students are much smarter than I and I learn along with them and it's exhilarating ]   Thank you so much!!

One thing he said was it would be helpful if he can have targeted practice worksheets on the tips covered during the class, after that class, so he can practice those shortcuts/tips more.

Hello Mrs. Lin,

This is ______ 's mom. We greatly appreciate your help in working with
______, keeping him motivated and providing him wth constant
encouragement. We are fortunate to have great mentors like you who
are so selfless in their services to our younger generation.

We have a request, can you please share your address. ______ wanted
to send you a card to express his thanks.

BTW, _______ has his chapter level competition tomorrow.

Warm Regards

Mrs. Lin,

Good Afternoon.
Me and my son, ___ and I, have used your blog pages and got a great insight into several of the techniques that you use to solve the problems. I am glad to inform you that _____ was placed 14th in the State Mathcounts for ____. Your blog information has helped us a lot in this preparation and we really want to appreciate all that you do in sharing the information. 

_____ is completing Geometry this year and he hopes to have a exciting next year for AMC 8 and Mathcounts. We hope to learn a lot more from your blog pages.

Regards,

Hi Mrs Lin,

I just finished The One World School House by Salman Khan and I have been thinking of you and the online community you created. Even though _____ and ____ have taken web based courses before, what makes your class different is your inspiration and enthusiasm - you really care about them as unique individuals and you sincerely expect the best from them, more than they (and I sometimes) think they can achieve. Thank you for encouraging them to become a more responsible and self driven learner.

There are many more but I'll take my time to update/upload these infor. sheets. 

To be continued ... 

I'm quite busy these days with resuming our Math Circle + many other projects (my students don't just except at math, but many other areas equally fun and challenging + most Asian students have much bigger problems with critical reading/not to mention writing, taking initiatives and being strong leaders, and those are my other projects. 

Work-Life balance is upmost important. Less is more. 

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Problem Solving Strategies : Complementary Counting

June 16, 2015, 2:40 pm
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Check out Mathcounts here, the best competition math program for middle school students.

Download this year's Mathcounts handbook here.

Video to watch on complementary counting from "Art of Problem Solving"

Part 1

Part 2

Question: How many two-digit numbers contain at least one 9?

 At beginning level, kids start to write down all the numbers that contain 9. However, this turns into impossible task if it's a three-digit or four-digit number. So let's try other ways to do it.

9 _ , if 9 is placed as the tens digit, unit digit can be chosen from 0 -9, altogether 10 numbers.
A lot of kids think there are only 9. Be careful. To find how many consecutive terms from number a to b, you do (b-a) +1

_ 9, there will be 9 choices (1-9) this time as the tens digit.(Why?) Zero can't be placed other than unit digit.
So total you have 10 + 9 - 1 (you've counted 99 twice) = 18

This way is better than the first one, but once the numbers become large, you will easily lose track of those double-counting, triple-counting numbers and over count your answers.

Here is a better way to tackle this type of problem: Think about the case of numbers that contain no 9s, and subtracting this value from the total number of two-digit numbers will give you the answer.

There are 99-10 +1 =90 total two digit numbers. There are 8 (the tens digit) x 9 (unit digit) = 72 numbers that contain no 9s. So 90-72 = 18 gives you the answer.

Try this question: 
How many three-digit numbers contain at least one 9?
There are 999 - 100 + 1 or 999 - 99 = 900 three-digit numbers. 
Or 9 x 10 (you can use the digit "0" now" x 10 = 900 three-digit numbers
900 - 8 x 9 x 9 = 252numbers 

This is called "complementary counting" and there are numerous problems that you can use this strategies to simplify the reasoning. 

Here is another harder problem from 2003 Mathcounts Chapter Sprint Round #29:
Each day, two out of the three teams in a class are randomly selected to participate in a MATHCOUNTS trial competition. What is the probability that Team A is selected on at least two of the next three days? Express your answer as a common fraction.

Solution:  
Use complementary counting. 
If each day two of the team will be chosen, there will be 3C2 = 3 ways to choose the team -- AB, BC, or AC, so 1/3 of the chance that team A won't be chosen and 2/3 of the chance that team A will be chosen. 

Case 1: Team A is not chosen on any of the three days. The probability is (1/3) 3= 1/27.
Case 2: Team A is chosen on one of the three days : The probability is (2/3) times (1/3)2 times 3C1 = 6/27 (A - -, - A - or - - A, which is 3C1 = 3 ways)

Total possibilities - none - at least 1 time = at least two times Team A will be chosen 
so the answer is 1 - 1/27 - 6/27 = 20/27

Other applicable problems: (answer key below)

#1: 2006 AMC10 A: How many four-digit positive integers have at least one digit that is a 2 or 3? 

#2: What is the probability that when tossing two dice, at least one dice will come up a "3"? 

#3: If {x,y} is a subset of S={1,2,3,....50}. What is the probability that xy is even?















Answer key: 

#1:  9000 - 7 x 8 x 8 x 8 = 5416

#2:  The probability of the dice not coming up with a "3" is 5/6.
       1 - (5/6)2 = 11/36

2. 1 - (25/50) (24/49) -- only odd times odd will give you odd product, the others will all render even product, so the answer is 37/49.



 
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Problem Solving Strategy: Complementary Counting

June 24, 2015, 5:18 am
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#: How many three digit numbers contain the digit "9" at least once?
Solution:
For the "at least" questions: a lot of the time, the easiest way to solve the problems is to use the total number of ways minus the number of ways which do not satisfy the criteria we're looking for.

There are 999 -100 + 1 = 900 three digit numbers.

There are 8 x 9 x 9 = 648 numbers that do not contain the digit "9" at all, so the answer is 900 - 648 =252

#2: How many three digit numbers have at least two digits that are the same? 
Solution:
Use complementary counting to find how many three digit numbers have no digit that are the same.
9 digit choices for the hundredth digit (no 0), 9 for the tenth digit (0 is allowed) and 8 digit left for the unit digit, so 9 * 9 * 8 = 648
900 (3 digit numbers) - 648 =252  

#3: How many of the natural numbers from 1 to 600, inclusive, contain the digit "5" at least once?
(The numbers 152 and 553 are two natural numbers that contain the digit 5 at least once, but 430 is not.)
Solution: 
From 1 to 600 inclusive, there are 600 - 1 + 1 = 600 numbers.

___ ___ ___ If we're looking for all the three-digit numbers which do not contain any 5s: In the hundreds place, you can put 0, 1, 2, 3, or 4 (we'll get rid of '000' later). In the tens place, you can put 9 digits, and there are another 9 possible digits for the units digit, so 5 x 9 x 9 = 405.

405 - 1 (to get rid of '000') + 1 (to compensate for not counting the number '600') = 405, which is the total number of numbers which don't use the digit '5'.

600 - 405 =195, which is the answer.

#4: If you toss three coins, what is the probability that at least one coin lands heads up? 
Solution:
There is a \(\dfrac {1} {2}\times \dfrac {1} {2}\times \dfrac {1} {2}=\dfrac {1} {8}\)chance to get all tails = no heads, so if we want at least one head, the answer is 1 - \(\dfrac {1} {8}\)= \(\dfrac {7} {8}\).

Other applicable problems (answers below):
#1: Amy tosses a nickel four times. What is the probability that she gets at least as many heads as tails ?
#2: What is the probability that the product of the top faces on 2 standard die is even when rolled?
#3: 3 numbers are selected at random, with replacement, from the set of integers from 1 to 600 inclusive. What is the probability that the product of the 3 numbers is even ? Express your answer as a common fraction in lowest terms.
#4: 9 fair coins are flipped. What is the probability that at least 4 are heads? 
#5: How many 3 digit numbers does not contain the digit 1 but have at least one digit that is 5?







Answer: #1-\(\dfrac {11} {16}\) ; #2-\(\dfrac {3} {4}\) ; #3- \(\dfrac {7} {8}\) ; #4-\(\dfrac {191} {256}\) #5:217
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Logical games, puzzles and links for summer learning or/and having fun together

July 1, 2015, 4:51 pm
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