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Power of a point rule:    O * W = O * W       ( outside * whole = outside * whole )  

3rd problem:   ST * SU = SV * SW  

By partial fractions, (A,B) = (4,-1).

Please explain more.

Is the a 2d object or is it a representation of a 3D object.

What object is it?

Attn:  grmolives,

Here is the last time that you asked this question.

By division

or you could use the factor function on your calculator

factor(1676962425600) = (((((2^8*3^2)*5^2)*7)*11^2)*37)*929

You do not need to know how many factors, that is determined by the rest of the question.

You only need to know if 1 is included as a factor or if it is not.

2a+3b=c (1)

a+b+c=42 (2)

c-a=19  (3)

let's start by (3), we can say that a-c=-19 (The reverse) by just multiplying by -1

so let a-c=-19  (4)

Add (2) and (4) 

a+b+c=42

a-c=-19 

indeed,

2a+b=23 

Relating it to (1) looks very similar, however, some steps are required.

2a+3b=c

2a+b=23

Let's before continuing this make some statements about c.

from (3) 

we can say that 

c=19+a

back to (1)

2a+3b=c, we can say 2a+3b=19+a 

simplify this by subtracting an a

a+3b=19

Back to the system of equations,

a+3b=19

2a+b=23

Let's multiply the second equation by -3 so the b's cancel

a+3b=19

-6a-3b=-69

Again back to the system of equation

a+3b=19

-6a-3b=-69

let's add both of them

-5a= -50

divide by -5

a=10 

Well we know that from (3) that c-a=19

so c-10=19

c=29 

From (2) we know that a+b+c=42

so 10+b+29=42

39+b=42

b=3

Now we got the values of a,b,c

a=10

b=3

c=29

The question says find (c+1)/(ab) 

(29+1)/10*3=30/30=1

Answer is =1

Multiplying both sides by 3x(x + 1)(x + 2): 3(x + 2) - 3x < (x + 1)(x + 2).

Simplify: x^2 + 3x - 4 > 0

Factor: (x + 4)(x - 1) > 0

Solution: (-inf,-4) U (1,inf)

n=1;p=0;b= cos(1/2*n*(n+1));p=p+b;n++;if(n<=360, goto2,0);printp

OUTPUT = - 2

(1 + 3 + 5 + ... + (2 n - 1)) = sum_(k=1)^n(-1 + 2 k) = n^2

(2 + 4 + 6 + ... + 2 n) = sum_(k=1)^n 2 k = n (n + 1)

n = 2017

Last one

By  the intersecting secants theorem

SU * ST   =   SW * SV

18 * 3  =  SW * SV

54  =  SW * SV

Let SV  =  x      so    SW = x + 3   .....so.....

54 = ( x + 3) * x     simplify

54  = x^2 +3x      rearrange as

x^2 + 3x  - 54   =  0       factor

(x - 6) ( x - 9)  =  0

Set each factor to  0  and solve for x and we have that   x  = 6  or x  = 9

Only  x = 6  = SV

So..SW =  6 + 3   = 9

Chords AB and CD of a circle meet at Q inside the circle. If CD = 38, AQ = 6, and BQ = 12, then what is the minimum length of CQ?

By the intersecting chord theorem

AQ * BQ   =  CQ  * DQ

6 * 12  = CQ * DQ

72  = CQ * DQ

Let CQ  = x    Let  DQ  =  38 - x    ....so we have

72 =  x  ( 38 - x)      simplify

x^2 - 38x  + 72  = 0       factor

(x -36) ( x- 2)  =  0

Setting each factor to 0   and solving for x we have that  x  = 36    or  x  = 2

So.....the minimum length  of  CQ   = 2

First one

By the tangent-secant theorem, we have that

WZ^2   = WY * WX

8^2  =  WY *  4

64  =  WY * 4

WY  = 16

And using the Pythagorean Theorem

YZ  = sqrt  [ WY^2  + WZ^2]   = sqrt  [ 16^2  + 8^2]  =sqrt [ 320 ] =  8sqrt(5)

Call the radius  of the small circles, R

Call the radius of the large circle, 2R

(1/2)  of one of the  yellow areas = [ (1/4)pi R^2  -  (1/2)R^2  ]

So   the total of the yellow area  = 8 [ (1/4)pi R^2   - (1/2) R^2]  = R^2 [ 2pi - 4] 

And  one of the pink areas  =   pi R^2  - [ pi R^2  - 2 R^2]   =  2R^2

So....the total of the pink areas =  8R^2

And the  area  of the large circle is  pi (2R)^2 =  4 pi R^2

So....the  blue area  =

Area of large circle -  pink areas  - yellow area

   4 pi R^2 - 8R^2  - R^2 [ 2pi - 4]  =

4pi R^2  - 8R^2  - 2pi R^2  + 4R^2  =

2 pi R^2  - 4R^2

R^2  [ 2pi  - 4 ] 

So....the areas are equal   !!!

Of course, thanks Tiggsy.  

Much appreciated.

Problem 1:

PoP: WX*XY = WZ^2, 4*XY = 64, XY = 16

Pythagoras: YZ = sqrt(WY^2 + WZ^2) = sqrt(20^2 + 8^2) = 4*sqrt(29).

Problem 2:

PoP: CQ can be 5 or 33, so the answer is 5.

Yah, too lazy to even get rid of the LaTeX.

17 * x = 100     divide both sides by 17

x = 100/17  = 5.8823529411764706

sin   *   sin/cos = 1

sin^2 / cos =1 

(1-cos^2) =cos

cos^2 + cos -1 = 0        Let    x = cos

x^2 + x -1 = 0    use quadratic formula to find  x = .618   or   -1.618   (throw out ...cos cannot be this magnitude)

(a) Let's call (a,b,c) a smiley face if b is less than a and b is less than c, because when we plot the graph, we get a happy face!  And if b is greater than a and b is greater than c, that's a frowny face, because we get a frowny face when we turn a smiely face up-side-down.

There are other kinds of faces like smirks (like a is less than b and b is less than c) and neutral faces (like when a is equal to b and b is equal to c).  If the face is neutral, then a equals b and b equals c, so when we choose a, b, and c are also chosen, and there are n choices for a, so there are n neutral faces.

Now we count the number of smirks.  There are n ways to choose a, and there are n - 1 ways to choose b.  We also multiply by 3, because the value that we chose for a could have also been the value of b, or the value of c.  So there are 3n(n - 1) smirks.

Now we count the number of smiley faces.  There are n ways to choose a, then n - 1 ways to choose b, then n - 2 ways to choose c.  So there are n(n - 1)(n - 2) = 3C(n,3) smiley faces.  By symmetry, there are 3C(n,3) frowny faces.

Therefore, the total number of faces is n^3 = n + 3n(n - 1) + 6C(n,3).

(b) To choose three numbers, we can choose two groups one group with two numbers and the other group has one number.  The total of n + 2 numbers can be separated into two groups.

In the first way, one group has 2 numbers, and the other group has n numbers.  They form a total of n + 2 numbers.  There are C(2,2) = 1 ways to choose two numbers from the 2 group.  There are C(n,1) = n ways to choose one number from the one group.  This gives us a first term of 1*n.

In the second way, one group has three numbers, and the other group has n - 1 numbers.  They form a total of n - 1 numbers.  There are C(3,2) = 2 ways to choose two numbers from the 2 group.  There are C(n - 1,1) = n - 1 ways to choose one number from the one group.  This gives us a second term of 2*(n - 1).  The pattern will continue until we reach n.  So the two sides are equal.

You are CORRECT !   My mistake......I corrected my answer above....  Thanx!    ~ EP    

"her on the th day of her life"......     On WHAT day?????  .

These are all AoPS homework problems.  We should not help students cheat on their homework.

For the third problem, by power of a point, ST*TU = SV*VW, so SV = 15.

The length of a chord  YZ = 17.889