Math Team

Skills forDecember Rounds

Round1

Round2

Round3

Round4

Round5

Round6

Round 1 – Trig: Right Angle ProblemsLaw of Sines and Cosines

For right triangles:

Pythagorean Theorem

For oblique triangles:(ASA, AAS, or SSA)

For oblique triangles:(SAS or SSS)

Round 2 – Arithmetic/Number Theory

Percent of Change = new value - old value

old value

Positive percent of change percent increaseNegative percent of change percent decrease

Number of FactorsTo find the number of positive integer factors of a number:• Find the number’s prime factorization (using exponents to describe repeating factors)•Add 1 to each of the exponents•Find the product of the numbers generated by adding 1 to each exponent

Example: 90 has 12 positive integer factors

1

2

2 1

90

2 3 5

2 3 5

1 2 1( 1)( 1)( 1)

2 3 2

12

Base 10 NumbersWe use base 10 numbers all of the time. We have memorized the places of base 10 numbers since we were young children.

For Example, the number has a 2 in the one’s place, a 9 in the ten’s place and a 6 in the hundred’s place. (The subscript of 10 simply means that it is a base 10 number. When there is no subscript, we always assume that the number is base 10)

10692

To understand other bases, we need to know where the names of the place values come from.

6 9 2010110210Base 10 place values:

Evaluate to get the names: 110100

So the number 69210 is really:

Six 100’s = 600+ Nine 10’s = 90+ Two 1’s = 2

= 692

Base 2 NumbersBase 2 numbers’ place values work the same as base 10:

When we are in base 10, remember that we can only use the numbers from 0 to 9. In base 2, we can only use numbers from 0 to 1. So to evaluate 101102 :

021222

Base 2 place values:

Evaluating this base 2 number: 13242

124816In base 2, each placevalue is worth…

0 1 1 0

16

0212223242

Using these place values:8 4 2 1

Yields: One 16 + Zero 8’s + One 4 + One 2 + Zero 1’s16 + 0 + 4 + 2 + 0Which is: =22

101102 = 2210Therefore:

Base 3 Numbers

If you can figure out the base 10 equivalent to 12013 then you’ve got it.

Solution: 27 + 18 + 0 + 3 = 48

Round 3 – Coordinate Geometry of lines and circles

Equation of a Circle: {with center (h,k) and radius r}

2 2 2( ) ( )x h y k r

2 2

2 2

0ax bx cy dy e

ax bx cy dy e

If given this form

Divide both sides by ac, then complete each square to change back to general form.

Ex) Circle with center (2, -1) and radius 4

2 2 2

2 2 2

2 2

( ) ( )

2 ( 1) 4

( 2) ( 1) 16

x h y k r

x y

x y

Equations of lines:

1 1( )

y x

Ax By C

m

my xy

b

x

Slope-Intercept

Standard Form

Point-Slope Form

slope y-intercept

Given Point (x1 , y1)

ASlope

B

Typically A > 0and A, B, C are Integers

Center of a circle is the midpoint of a diameter

1 2 1 2

1 2 1 2

,2 2

2 2m m

Midpoint Formula:

x x y y M=

x x y yor x and y

Slope of a line is constant.

If given slope, use to find additional pointsRise

Run

Round 4 – Log and Exponential Functions

logb x y yb x

Logarithmic Form vs. Exponential Form

base baseexponent

exponent

Remember: Log is exponent

logb x y

Properties of Logs

Properties of Exponents

log log log

log log log

log log

b b b

b b b

ab b

xy x y

xx y

y

a x x

x y x y

xx y

y

yx xy

b b b

bb

b

b b

log 6 log 2 3

log 2 log 3b b

b b

log 8 log 2

8log log 4

2

b b

b b

2

2log 3

log 3 log 9

b

b b

3 4 3 4 7x x x x

33 7 4

7 4

2 1 12 2

2 2 16

3 2 32 62 2 2 64

Ex)

Ex)

Ex)

Ex)

Ex)

Ex)

Inverse Properties:

log xb b x logb xb x

Ex)3

43

log 81 ?

log 3 4

Ex) 7log 57 5

Special Values:

10log logb b

ln logeb b

log 1 0b 0 1b

Special logs:

Round 5 – Alg 1: Ratio, Proportion or Variation

3

4

x

y

If the ratio of x to y is 3 : 4, then

Ratios can be reduced the same as fractions:

25 is to 100 as 1 is to 4. 2525 1

1 200 45

Cross multiply to solve proportions

Direct Variation“y varies directly as x”

Inverse Variation“y varies inversely as x”

y kx

ky

x

Direct variation is a line that intersects the origin (0, 0) and has slope (or constant of variation), k.

To solve variation problems, use the initial values of x and y to find the constant of variation, k. Then substitute k back into the equation.

Joint Variation“z varies jointly with x and y”

z kxy

Round 6 – Plane Geometry: Polygons (no areas)

Vocabulary: Midpoint, segment bisector, segment trisector, angle bisector, perpendicular, altitude, etc.

Sum of interior angles of a polygon with n sides: ( 2) 180n

Know your shapes and their properties: square, rhombus, rectangle, triangle, quadrilateral, etc.

For a regular polygon with n sides, each Interior angle of the polygon would be:

( 2) 180n

n

For a tangential polygon with an even number of sides, if you number the sides consecutively, the sum of the even sides is always equal to the sum of the odd sides.

Length of the sides of a Tangential (inscribed) polygon(a polygon in which each side is tangent to a circle)

a

b

cd

e

f In this example,

a + c + e = b + d + f