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