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. [$\mathrm{\Delta }BMT$ and $\mathrm{\Delta }BCN$ are similar triangles ]

$\overline{NT}=\overline{TB}={\displaystyle \frac{4a+5a}{2}}=4.5a$

$\overline{ST}=0.5a$

 $\overline{MT}:\overline{AB}$ = 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.]


$\overline{TB}=4.5a$  [from previous conclusion]

Using 5 to 2 line ratio [similar triangles $\mathrm{\Delta }ARB$ and $\mathrm{\Delta }MRT$ , you get $\overline{BR}={\displaystyle \frac{5}{7}}\times 4.5a={\displaystyle \frac{22.5a}{7}}$ and $\overline{RT}={\displaystyle \frac{2}{7}}\times 4.5a={\displaystyle \frac{9a}{7}}$

Thus, x : y : z = 4a : ${\displaystyle \frac{1}{2}}a+{\displaystyle \frac{9a}{7}}$ : ${\displaystyle \frac{22.5a}{7}}$ = 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 ${\displaystyle \frac{x}{y+z}}={\displaystyle \frac{5}{9}}$.
9x = 5y + 5z ---- equation I

Using similar triangles ASB and CSN and you have ${\displaystyle \frac{x+y}{z}}={\displaystyle \frac{5}{4}}$.
4x + 4y = 5z  ---- equation II

Plug in (4x + 4y) for 5z on equation I and you have 9x = 5y + (4x + 4y) ; 5x = 9y ; x = ${\displaystyle \frac{9}{5}}y$
Plug in x = ${\displaystyle \frac{9}{5}}y$ to equation II and you have z  =  ${\displaystyle \frac{56}{25}}y$

x : y : z = ${\displaystyle \frac{9}{5}}y$  : y  :  ${\displaystyle \frac{56}{25}}y$ =  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 

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

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

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

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

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

Solution I: 

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

* * * |    | * * * * *

  ^       ^       ^

  a       b       c

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

Solution II: 

But this can also be done using the grid technique:

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

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