Friday, December 29, 2017

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."

The Myth of the Science and Engineering Shortage from The Atlantic Magazine 


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

Friday, December 8, 2017

Harder Mathcounts State/AMC Questions

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 





Sunday, April 23, 2017

Tricky Algebra Mathcounts National Questions: Counting Backwards

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

#24 1993 National Sprint: Bottle A contains more Diet Coke than Bottle B. Now do the following:
a. Pour from Bottle A into B as much Diet Coke as B already contains.
b. Pour from B into A as much Diet Coke as A now contains.
c. Pour from A into B as much Diet Coke as B now contains.
Both bottles now have 64 ounces. How many more ounces were in A than in B at the beginning?

#30: 1993 National Sprint: Auggie spent all of his money in 5 stores. In each store, he spent $1 more than one-half of what he had when he went in. How many dollars did Auggie have when he entered the first store?

#25: 1998 AMC-8  Three generous friends, each with some money, redistribute the money as follow: Amy gives enough money to Jan and Toy to double each amount has. Jan then gives enough to Amy and Toy to double their amounts. Finally, Toy gives enough to Amy and Jan to double their amounts. If Toy had 36 dollars at the beginning and 36 dollars at the end, what is the total amount that all three friends have?












Solution I: Use Algebra:
#24: Let A contains x ounces and B contains y ounces and x > y (given).
After first pouring, A has (x - y) ounces left and B has 2y ounces (double the original amount)
After second pouring, A has ( 2x - 2y)(double the amount) ounces and B has (3y - x) ounces left.
After third pouring, A has (3x - 5y) ounces left and B has (6y - 2x) (double the amount)
3x - 5y = 64   times 2 for each terms      6x - 10y = 128  ----equation 3
6y - 2x = 64   times 3 for each terms      18y - 6x = 192  ---- equation 4
equation 3 + equation 4 and you have 8y = 320 and y = 40 ; Plug in any equation and you get x = 88
88 - 40 = 48 ounces 

Solution II: Solving it backwards: 
At the end,both A and B have 64 ounces, which is after same amount of Diet Coke being pour from A to B.
Thus before action C, A has 64 + half of 64 = 96 ounces and B has 32 ounces. [Make sure you understand this]
With the same reasoning, before action B, B has 32 + half of 96 = 80 oz. and A has 48 ounces.
Again, use the same strategy, you have before action A, A has 48 + half of 80 = 88 and 40.
The difference is 88 - 40 = 48 oz.

Solution I: Use Algebra  
#30: Let Auggie had x dollars at the beginning. At the first store, he would spent 1 + (x/2) = (2+x)/2 and would have x - (2 +x)/2 = (x-2)/2 left
At the second store, he would spend 1 + (x-2)/4 and would have (x-2)/2 - 1 - (x-2)/4 or (x-6)/4 left
At the third store, he would spend 1 + (x-6)/8 and would have (x-14)/8 left
It looks like there's a pattern. At the fourth store, he would spend (x-30)/16
and at the 5th store he would spent (x-62)/32 = 0 so x - 62 = 0 and x = 62 dollars

Solution II: Work backwards
Since Auggie spent all his money at the 5th store. If there are x dollars left before he spent the money all at the 5th store. You can set up the equation such as this:  x = 1 + 1/2 of x (according to the given)
So at the 5th store, he had 2 dollars. 
Use the same strategy, if he had y dollars before he spent the money at the 4th store, he had 
y = 1 + 1/2 of y + 2 ; y = 6
Use the same method, Aggie had 16 before he spent at the 3rd store, 30 before the 2nd store and finally, 
62 dollars at the beginning.  

#25: The total sum of what Amy, Jan, and Toy have stay constant so use Toy's amount to solve this problem.

                   Amy          Jan         Toy
                    ?               ?             36  
First round Amy gave Jan and Toy double the amount of what each of them has, so 
                  Amy           Jan         Toy
                    ?                ?            72  
Second round Jan gave Amy and Toy double the amount of what each of them has, so
                  Amy           Jan         Toy
                   ?                 ?            144
Third round Toy gave Amy and Jan double the amount of what each of them has an at the end Toy has 36 dollars        Amy           Jan            36   
That means that  at the second round, Amy + Jan = 144 - 36 = 108 dollars.
So they total have 108 + 144 = 252 dollars.                                                                                                    
 

Monday, April 10, 2017

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.

Tuesday, February 14, 2017

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

Links, notes, Hints or/and solutions to 2014 Mathcounts state harder problems.
2014, 2015 Mathcounts state are harder 

Sprint round:

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

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

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

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

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

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

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































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

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

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

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

Add them up and the answer is 1806.

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

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

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

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

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

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

Target Round : 

#3: Lune of Hippocrates : in seconds solved question.
^__^

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

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

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

//cdn.artofproblemsolving.com/images/ad1f21b9f50ef27201faea84feca6f2e6e305786.png

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

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

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

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


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

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

Cut the cone and observe the shape.

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

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

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

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

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

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

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


Now we can solve this :

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

Cut the cone and observe the shape.

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

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

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

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

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

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

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


Now we can solve this :

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


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





2013 Mathcounts State Harder Problems

 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

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