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