Wednesday, July 6, 2016

Mathcounts Prep : Algebra Manipulation

Note that: 

\(\left( x+y\right) ^{2}-2xy= x^{2}+y^{2}\)
\(\left( x-y\right) ^{2}+2xy= x^{2}+y^{2}\)
\(\left(x-y\right) ^{3}+3xy\left( x-y\right) =x^{3}-y^{3}\)
\(\left( x+y\right) ^{3}-3xy\left( x+y\right) =x^3 + y^{3}\)
\(\left( x+y+z\right) ^{2}-2\left( xy+yz+xz\right) =x^{2}+y^{2}+z^{2}\)

Applicable questions:

Question 1: If x + y = a and xy = b, what is the sum of the reciprocals of x and y?

Solution: 
\(\dfrac {1} {x }+\dfrac {1} {y}=\dfrac {x +y} {xy}\)= \(\dfrac {a} {b}\)

Question 2: If \(x^{2}+y^{2}=153\) and x + y = 15, what is xy?

Solution: 
\(\left( x+y\right) ^{2}-2xy= x^{2}+y^{2}\)
\(15^{2}-2xy=153\)\(\rightarrow xy=36\)

Question 3: If \(\left( x+y\right) ^{2}=1024\) , \(x^{2}+y^{2}\) = 530 and x > y , what is x - y? 

Solution: 
 \(\left( x+y\right) ^{2}-2xy=x^{2}+y^{2}\)
1024 - 2xy = 530\(\rightarrow 2xy=494\)
\(\left( x-y\right) ^{2}+2xy=x^{2}+y^{2}\)
\(\left( x-y\right) ^{2}=36\)
x - y = 6

Question 4: x + y = 3 and  \(x^{2}+y^{2}=89\), what is \(x^{3}+y^{3}\)? 

Solution: 
\(\left( x +y\right) ^{2}-2xy=x^{2}+y^{2}\)
9 - 2xy = 89 \(\rightarrow -2xy=80\) so xy = -40
\(\left( x+y\right) ^{3}-3xy\left( x+y\right) =27 - 3(-40)* 3 = 27 + 3*40*3 = x ^{3}+y^{3}\)
\(x ^{3}+y^{3}\)= 387

Question #5: If \(x+\dfrac {1} {x}=5\), what is \(x^{3}+\dfrac {1} {x ^{3}}\)?

Solution:
\(\left( x+\dfrac {1} {x}\right) ^{3}=x^{3}+3x^{2}.\dfrac {1} {x}+3x.\dfrac {1} {x^{2}}+\dfrac {1} {x^{3}}\)
\(5^{3}=x^{3}+3\left( x+\dfrac {1} {x}\right) +\dfrac {1} {x^{3}}\)
125 - 3*5 = \(x^{3}+\dfrac {1} {x ^{3}}\)
The answer is 110.

Question #6 : 2011 Mathcounts state sprint #24 : x + y + z = 7 and \(x^{2}+y^{2}+z^{2}=19\), what is the arithmetic mean of the three product xy + yz + xz? 

Solution: 
\(\left( x+y+z\right) ^{2}-2\left( xy+yz+xz\right) =x^{2}+y^{2}+z^{2}\)
\(7^{2}-2\left( xy+yz+xz\right) =19\)
xy + yz + xz = 15  so their mean is \(\dfrac {15} {3}=5\)



More practice problems (answer key below):

#1:If x + y = 5 and xy = 3, find the value of \(\dfrac {1} {x^{2}}+\dfrac {1} {y^{2}}\). 

#2: If x + y = 3 and \(x^{2}+y^{2}=6\), what is the value of \(x^{3}+y^{3}\)? 

#3: The sum of two numbers is 2. The product of the same two numbers is 5. 
 Find the sum of the reciprocals of these two numbers, and express it in simplest form. 

#4:If \(x-\dfrac {6} {x}\) = 11, find the value of \(x^{3}-\dfrac {216} {x^{3}}\)? 

#5: If \(x+\dfrac {3} {x} = 9\), find the value of \(x^{3}+\dfrac {27} {x^{3}}\)?

#6:If \(x+\dfrac {1} {x} = 8\), what is \(x^{3}+\dfrac {1} {x ^{3}}\)? 





Answers:
#1 :\(\dfrac {19} {9}\)
#2: 13.5
#3: \(\dfrac {2} {5}\)
#4: 1529 [ \(11^{3}\)+ 3 x 6 x 11 =1529]
#5: 648   [\(9^{3}\)-3 x 3 x 9 = 648]
#6: 488   [ \(8^{3}\)– 3 x 8 = 488]

Friday, June 3, 2016

Harder Mathcounts State Question

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.

Solution V : from Abhinav, one of my students solving another similar question : 

2016 AMC 10 A, #19 




2016 AMC 10 B #19 : Solution from Abhinav