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Answered here    :  https://web2.0calc.com/questions/geometry-math-question-help

Two circles, each with radii 900, are tangent to each other and they are both tangent to a line. Find the radius of the circle which is tangent to both circles and also the line.

The closest I could get is:

Square side:  a = 3

Rectangle sides:   a = .8   and  b = 3

Therefore, the area of a rectangle is  2.4 units²

^{If the angles are 30° , 60° and 90° , it should work with any size of a square.}

The probability = 181 / 216 

I never would have guessed that, thanks Chris.

There are 52 ways of choosing the first card, and 3 ways of choosing its pair.

There are 48 ways of choosing the next number, and 3 ways of choosing its pair.

There are 44 ways of choosing the single card, which gives us 52*3*48*3*44 = 988416 ways.

There are 4*3*2*6*1 = 144 possible combinations.

Here are the possible outcomes :

We have 216 oucomes

I'll let you  count all the ones  whose products  are 10  or  greater.....

Your count  / 216  = the probability

Note that  in  both shaded triangles.....the  angles  that  are touching the base of the  figure  must   = 60°

So  the shaded area ( a triangle)  on the left  =  (1/2) (6)(3)sin(60°)

And   the shaded area on the right (a riangle)  = (1/2) * (6) (4) sin (60°)

So....the total area of the two shaded regions  =

(1/2)(6)sin(60°)  [ 3 + 4 ]  =

3(√3/2)  [ 7]  =

 21√3

_____     units^2    ≈   18.19 units^2

     2

Let the side of the octagon  = S

And  one of the yellow  regions  is a trapezoid

And the  height of one of these trapezoids =  S / √2

And the base  of one of these trapezoids =   S  + 2S/√2 =   (1 +√2)S

So....the area of one of the trapezoids =  (1/2) (S/√2) ( S + (1 + √2) S)  =

(√2/4) ( 2 + √2)S^2  =

(1/4) (2 + 2√2)S^2  =

(1/2) (  1 + √2)S^2

So  the two trapezoids  have an  area  of  ( 1 + √2)S^2 =  the yellow area

And the area of the blue rectangle =   S  * Base of the trapezoid =

S *  (1 + √2)S  =  (1 + √2)S^2

So.....the  yellow and blue areas are equal  !!!!

I count 10 too

See the following image :

Let I  be the center of the square = (10,10)

Let one of the quarter circles  have the equation x^2 + y^2  = 400

And let another of the quarter circles have the equation of (x - 20)^2 + y^2  =400

The x coordinate of the  intersection of these circles  = 10

So  the x coordinate of E   = 10

So...using the equation of the circle we can find the y  coordinate as

10^2 + y^2  =400

100 + y^2 = 400

y^2 = 300

y = 10√3

So  IE = the distance between I and E =   10√3 - 10   = 10 (√3 -1)

And by symmetry....this equals IF

So triangle  EIF is right and its area =   (1/2)(IE)(IF) = (1/2) [ 10 (√3 -1)]^2  =

50 (3 - 2√3 + 1)  = [200 - 100√3]units^2      (1)

And the area between sector AEF  and triangle AEF  = 

(1/2)*(20^2)(pi/6)  -  (1/2)(20^2)(sin ( EAF) )  =

(1/2) (20^2)(pi/6)  - (1/2)(20^2)(1/2)  = 

200 [ pi/6 - 1/2 ]  units^2   =

(100/3)pi - 100   units^2        (2)

So....by symmetry.....the yellow area  =  4 [ (1) + (2)  ] =

4 [ ( 200 -100√3 + (100/3)pi - 100]  units^2  =

4 [ 100 - 100√3 + (100/3)pi ] units^2  =

400 [ 1 - √3 + pi /3 ] units^2 ≈ 126.06 units^2

Oh, OK....we can use similar tiangles to solve this

We have that triangle EPC is similar to triangle COD

So

CP / PE  = CO / OD

CP/ PE = (CP + PO) / OD

CP /   3 =  (CP + 8) /  5       cross-multiply

5CP  = 3 (CP + 8)

5CP  = 3CP + 24

2CP  =24    divide both sides by 2

CP = 12

Thanks for responding Madyl,

Do you mean you have worked out all three for yourself?

If you have then that is excellent !

Do you have the address for the mathsisfun solution? 

I just think if may be laid out more simply than guests copied version.

The minimum value is 1, which you get for z = i.

Yes, I actually got it but forgot to post.

2019^8 + 1 = 276115350402095809603387042 = 2×97×1423275002072658812388593 (3 distinct prime factors)

97 is least odd prime divisor (factor)

Ah...thanks, Melody....I forgot about that.....!!!!

ok that is fine.

If what you say is correct then he/she was plaguizing.

You should have just accused him outright rather than asking him if he was guilty.

That does explain you attitude, thanks for that explanation.

He/she should not be rewared for plaguizing.

a) There is a circle propery that just says that the 2 tangential distances from an external point to a circle are equal.

Using this   AM=AP   and   BM=PM

so    AM=MB 

so if M is on AB it follows that M is the midpoint of AB

I haven't done the other 2. They would take more effort on my part.  Are you happy with my part A?

He simply copied and pasted this answer from the website Mathisfun. I know this because I've tried to find the answer to this question before and came across this exact post on the mathisfun forum. Technically, the guest is plagraizing so I should really thank the real writer of the guest's post I do not understand: bob bundy, an admin on Mathisfun. Basically, the guest's post was not original and he did not spend time on his/her post.