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 

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.                                                                                                    
 

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

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

2013 Mathcounts Nationals Sprint #29 and similar problem hints

Mathcounts problems : state/nationals level

2013 Mathcounts Natinals Sprint # 28

2013 Mathcounts National Sprint #28 : In right triangle ABC, shown here,  line AC = 5 units and line BC = 12 units. Points D and E lie on  line AB and line BC respectively, so that line CD is perpendicular to line AB and E is the midpoint of line BC. Segments AE and CD intercept at point F. What is the ratio of AF to FE ? Express your answer as a common fraction.

                                                     Solution I : Using similar triangles

                                                        Solution II : Use Mass Point Geometry