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From the center M of the circles, draw rays to the corner E and to the center S of the side belonging to the corner. The result is a right-angled triangle M, E, S with sides s/2, r and R.

Then:

\(R^2=(\frac{s}{2})^2+r^2\\r=\sqrt{R^2-(\frac{s}{2})^2}\\ \frac{s}{2}=R\cdot sin(60^{\circ})\)

\(r=\sqrt{R^2-R^2\cdot sin^260^{\circ}}\\ r=R\cdot \sqrt{1-sin^260^{\circ}}\\ r=R\cdot cos\ 60^{\circ}\\ \color{blue}r=\frac{1}{2}R\)

or very simply:

In the right triangle SME, angle SME=60°.

So is:

\(r=R\cdot cos\ 60^{\circ}\\ \color{blue}r=\frac{1}{2}R\)

!

1 -3x < x+8 ≤ 7 -x

Split this  up

1 -3x < x +8                             x + 8 ≤ 7 - x 

-7 < 4x                                     2x  ≤ -1

x > -7/4                                      x  ≤ -1/2

( -7/4 , -1/2 ]

-1 < 2x < 12    divide through by 2

-1/2 < x < 6

And

-1 < 3x  -7 < 12       add 7

6 < 3x < 19      divide through by 3

2 < x  < 19/2 

2 < x < 6   

We have a lower degree polynomial over a higer degree polynomial....in this situation,the limit  always =  0 

x^2 + mx + 4 =  2x^2 + 17x + 8        rearrange as

x^2 + (17 - m)x  + 4  =  0

If we have two distinct roots, the discriminant > 0  ...... so....

(17 - m)^2  - 4(1)(4)  >  0

(17 - m)^2  >  16          take the positive root

17 - m   >    4

13 > m

m <  13

ab - 3a + 4b  =  131

b ( a + 4)  =  131  + 3a

b =  (131  + 3a)  / ( a + 4)

a         b

3       20

13     10

min  l a - b l  = l 13  -10 l   =  3

        A

        17

   x       x

B           D            C

17

                    E

"Line segment AD is extended to E, such that angle DBE = angle BAD = 17 degrees."  This has no bearing on the problem.  

If AD  = AB  then angle  ABD  =  angle ADB

So  angle   ABD  = (180 - 17)  / 2   =  81.5°  

A

    2

       H

           x

C     1      B

Let HB = x

AC^2 = AB^2 - BC^2  =   (2 +x)^2 - 1^2  =  x^2 + 4x + 3

CH^2  = BC^2 - BH^2 =   1 - x^2      (1)

CH^2 = AC^2 - AH^2  =  (x^2 + 4x + 3) - 2^2  = x^2 + 4x -1    (2)

Set (1)   =  (2)

1-x^2  = x^2 + 4x  -1

2x^2  + 4x -2  =  0

x^2 + 2x - 1  =  0

x^2 + 2x  = 1      complete the square on x

x^2 + 2x + 1  = 1 + 1

(x + 1)^2   = 2          take the positive root

x + 1  =sqrt (2)

x = sqrt (2)  - 1

CH^2   =  BC^2 - BH^2 =   1^2 - (sqrt 2 -1)^2   =  1 - (2 - 2sqrt 2 + 1)  =  2sqrt (2) - 2  

CH =   sqrt  (2sqrt (2)  - 2)  =  sqrt (sqrt (8) - 2)  ≈  .91

Use the following:

Label the tangent point between Semicircle AB and the red circle as D, the tangent point between arc BC and the red circle as E, and the tangent point between the red circle and semicircle AC as F.  Label the center of semicircle AB to C₁ the center of semicircle of BC to C₂ and the radius of the red circle to C₃.

Additionally, set the radius of the red circle is x.

We know BF is 2, and so BC₃ is 2-x. 

Similarly, C₁C₃ is 1+x.

Construct the right triangle, C₁C₃B.

Using the pythagorean theorem, we can set an equation:

\({1}^{2}+{(2-x)}^{2}={(1+x)}^{2}\). In the equation, the x's conveniently cancel out, and we get radius = 2/3.

Note that angle CEA = angle DCA + angle DAC and angle BDE = angle DCE + angle EDC, so angle DCA = angle CEA - angle DAC and angle DCE = angle EDC - angle CED.

Also, angle BDE = angle ABD + angle ADB = angle AEC + angle DAC, so angle DCE + angle DCA = (angle EDC - angle CED) + angle DCA.

Therefore, angle ACE = angle ACD + angle DCE = angle CEA, which implies that triangle ACE is isosceles.

Another way  using similar triangles

Draw altitudes AM   and FN to the base DE

Triangles DAM  and DFN are similar

AD = (3/5) AF

angle DFN = 30°....DN = 2.5

angle FDN  = 60°....FN  = 2.5 sqrt (3)  = (5/2)sqrt (3)    

So   AM = (3/5)FN =   (3/5)(5/2)sqrt 3  =  (3/2)sqrt 3 =  sqrt (27/4)

DN = 5/2  

So

DM = (3/5)(5/2)  =  3/2

So  ME = DE - DM =    5 - 3/2 =  7/2

So

AE  = sqrt [ ME^2 + AM^2 ]  = sqrt [ (7/2)^2 + 27/4 ]  =  sqrt [ 49 + 27 ] / 2  = sqrt [ 76] /2  = 

sqrt [ 19 * 4 ]  / 2   =  2 sqrt [ 19]   / 2  =   sqrt (19)

We can  solve this with the Law of Cosines

AF =2   FE  =5    angle DFE = 60

cos (DFE) =cos (60)  =1/2

So

AE^2  = AF^2 + FE^2  - 2 (AF *FE) cos (60)

AE^2  = 2^2 + 5^2 - 2(2 * 5) (1/2)

AE^2 = 29 - 10

AE^2  = 19

AE = sqrt (19) ≈ 4.36

CPhill if you are here do you mind checking my work? I think its wrong but I think I used the right steps..

I guess Heron's formula could work.... oh boy. Gimme a sec. 

MENTION: I set AE as 2x so its easier.

\(\sqrt{(4+x)(4-x)(1+x)(x-1)}+\sqrt{(3.5+x)(1.5+x)(x-1.5)(3.5-x)} = \sqrt{18.75}\)

\(x=\sqrt{\frac{2.25322E15}{1.0E15}},x=\sqrt{\frac{-2.25322E15}{1.0E15}}\)

\(ans=\frac{\sqrt{\frac{2.25322E15}{1.0E15}}}{2}\)

ahem....

this is 100% wrong..

approx. \(\frac{\sqrt{112661}}{100\sqrt{5}}\)

    Q

                    5     

3           X      

    P                 R

               4

Y = P

Angle QPR  = 90°

sin angle (QRP) = (3/5)

Angle RPX  = 45°

sin angle RPX = 1/sqrt 2

Because PX is an angle bisector

QP / QX  = PR / RX

Let RX = x     QX =  5 - x

So

3/ (5-x)  =  4 / x

3x = 4 (5-x)

3x = 20 - 4x

7x = 20

x = 20/7

Law of Sines

(YX)  /sin (QRP)  = RX / sin (RPX)

YX / (3/5)  = (20/7) / ( 1/sqrt 2)

YX  =  (3/5)(20/7) * sqrt 2  =   (60/35)sqrt 2  =  (12/7)sqrt (2) ≈ 2.42

(2^1 - 3) + (2^2 -3) +(2^3 - 3) + (2^4 -3) + (2^5 - 3)  =

-1 + 1   + 5  + 13 + 29  = 

47

5

∑    3^(n -1)  / 27

n=1

8th term  =  -4300 (-0.1)^7  =  0.00043

Thanks, proyaop...here's another way

(1) Extend  BA in  the direction of A   and  let it meet the extension of segment connecting the centers of the circle.....call this intersection point,  O

Call the center of the larger circle M  and the center of the smaller circle, N

Let x be the distance from  the left edge of the small circle to O 

Triangles   BMO and ANO are similar such that

BM / MO  = AN / NO

4 / (4 + 2 + x)  =  1 / ( 1 + x)

4 / ( 6 + x)  = 1 /(1 +x)

4(1 + x)  = 1 (6 + x)

4 + 4x  = 6 + x

3x = 2

x = 2/3

OA =  sqrt [ NO^2 - NA^2 ] =   sqrt [ (1 + 2/3)^2  - 1^2] =  sqrt [ 25/9 -1 ] = sqrt [ 16/9]  =  4/3

OB  = sqrt [ MO^2 - MB^2 ]  [  (6 + 2/3)^2 - 4^2  ]  = sqrt  [ 400/9 -16] = sqrt [ 256/9] = 16/3

AB =  OB = OA  = 16/3 - 4/3  =  12/3  = 4  

Call the first term, F

Call the common difference, d

F + 22d  = 4    (1)

F + 24d =  5    (2) 

Subtract (1) from (2)

2d = 1

d = 1/2

F + 22 (1/2) = 4

F + 11  = 4

F = 4 -11  =   -7

48th term =   -7 + 47(1/2)    -7 + 23.5   =  16.5

Using the Law of Cosines

PQ^2  = PA^2 + AQ^2  - 2(PA*AQ) cos (PAQ)

13  =  20  + 17  - 2 (sqrt (20) * sqrt (17))cos(PAQ)

[13 - 20 - 17 ] / [ -2 sqrt (340) ]  =  cos (PAQ)  = 6/sqrt (85)

sin PAQ  = sqrt [ 1  - (6/sqrt85)^2 ] / sqrt (85) = sqrt [ 1 -36/85 ] / sqrt (85)  = sqrt [49 / 85 ] / sqrt [85] =

7 / 85

Limit  =  the ratio of the coefficients of the n^4 terms in the num/den = 

 [2n^4] / [ 6n^4 ]  =  2/6 =  1/3