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OMG Cphill! There are other methods besides coordinate geometry that are more efficient and easier to follow.

Either look at the diagram or read the question again. 

The moderators really are braindead. Take two seconds to look at the diagram and explain how the hell 1/2 makes sense. 

This is a rather intriguing (and ironic) development, maximum. 

...I suppose this works but you could've just used Power of a Point...

Total number  of ways to get from (0,0) to (6,7)  less the number of ways to get from (0,0) to ( 2,3)

C(13,6)  - C(5,2)  =   1716 -10  =  1706

This is AoPS homework.  Do not post AoPS homework.

Let BC  = x

Triangles AEC  and BDC  are similar

BC / BD  = AC / AE

x / 1 = (4 + x) / 3

3x = 4 + x

2x = 4

x = 2 =  BC

{ DC = sqrt (BC^2 - BD^2)  = sqrt ( 2^2 - 1^2)  = sqrt (3)

tan(angle DCB)  =  - ( BD / DC)  = -1/sqrt (3)

Equation of tangent line       : y = (-1/sqrt 3) ( x -6)   }

Since AN=3 and BN=4, we have AB=AN+BN=3+4=7. Let I be the incenter of triangle ABC, so IN is a radius of the incircle. Since AN=3, we have IN=3.

The incenter of a triangle is also the intersection of the angle bisectors. Therefore, ∠AIB=∠ACB/2. Also, since ∠INB=90∘, we have ∠ANB=90∘−∠AIB=90∘−∠ACB/2.

By the Law of Cosines on triangle ABC,

[\cos C = \frac{AC^2 + BC^2 - AB^2}{2 \cdot AC \cdot BC} = \frac{81 + BC^2 - 49}{18BC} = \frac{BC^2 + 32}{18BC}.]

By the Law of Cosines on triangle ANB,

[\cos (90^\circ - C/2) = \sin \frac{C}{2} = \frac{AN}{AB} = \frac{3}{7}.]

Therefore,

[\cos C = 1 - 2 \sin^2 \frac{C}{2} = 1 - 2 \cdot \left(\frac{3}{7}\right)^2 = \frac{47}{49}.]

We have [\frac{BC^2 + 32}{18BC} = \frac{47}{49}.]

Multiplying both sides by 882BC, we get 49BC2+1536=882BC. Then 49BC2−882BC+1536=0, which factors as (7BC−48)(7BC−32)=0. Therefore, BC=48/7 or BC=32/7.

Since AN=3<7/2, the incenter I is closer to A than to B. Therefore, I is inside the smaller of the two triangles formed by AC and AB.

This means that BC>AC=9. Therefore, BC = 48/7.

We can solve this problem by considering the number of moves right and up that Josh makes.

Josh needs to move 6 steps to the right and 7 steps up, for a total of 13 moves.

Each move can be either right or up, so there are 2 choices for each of the 13 moves.

Therefore, the total number of ways to reach his home is 2^13 = 8192.

However, this includes paths that go through the fire at (2,3). To exclude these paths, we need to consider how many paths go through the fire and subtract them from the total.

A path goes through the fire if it includes a move right at (1,2) and a move up at (2,2).

There are 11 ways to choose which of the remaining 11 moves will be the right move at (1,2).

For each of these 11 choices, there are 6 ways to choose which of the remaining 6 moves will be the up move at (2,2).

Therefore, there are 11 * 6 = 66 paths that go through the fire.

Finally, we subtract the number of paths that go through the fire from the total number of paths to find the number of paths that avoid the fire:

8192 total paths - 66 paths through the fire = 8126 paths that avoid the fire.

Therefore, there are 8126 ways for Josh to reach his home without going through the fire.

Here :

https://web2.0calc.com/questions/counting-problem_20562

We can find the  x  coordinate of A thusly

12^2   - x^2  = 17^2 - (18 - x)^2

144 - x^2  = 289 - (x^2 - 36x + 324)

144 = -35 + 36x

179 = 36x

x = 179 / 36

And we can find the y coordinate of A as follows

sqrt [ 12^2 - (179/36)^2 ] =  sqrt  [154583 ] / 36

So  A =  ( 179/36, sqrt [ 154583] / 36 )

Let the midpoint of  BC  = (9,0)  = D

AG will be 2/3 of the distance AD

So

AG  =  (2/3)sqrt  [ ( 179/36 - 9)^2  +  (154583/36^2) ]  = sqrt [ 542] / 3   ≈ 7.76

We can construct a trapezoid  with bases

CB = sqrt (32) =  4sqrt (2)

And  AD = sqrt (18) = 3 sqrt (2)

Let C = (-4sqrt (2) / 2 , 0) =  (-2sqrt (2),0)  = (-sqrt (8) , 0)

And by symmetry,  B =(sqrt (8) , 0)

Note that DB =   sqrt ( 3^2 + 4^2)  = 5

Construct circles  with radius = 5  at C , B

The equation for the circle at B =   (x -sqrt 8)^2 + y^2   = 25 

D will have the coordinates   ( 3sqrt (2)/2 , y)  =  (3/sqrt (2)  ,y)

And by symmetry A = ( -3/sqrt (2) , y)

We can  find the y coordinate for D   by solving this  for  y

(3/sqrt2 - sqrt 8)^2  + y^2  = 25     ⇒   y =  7/sqrt 2

So D = (3/sqrt (2) , 7/sqrt (2) )

The center of the circle through ABCD    =  E = (0, y)

We can find y by equating  the distance  from E to B  and  from E to D

So

(sqrt (8) - 0)^2 + y^2 =  ( 3/sqrt 2 - 0)^2 + (y - 7/sqrt 2)^2  

Solving this for  y  produces  3/sqrt (2)

So  E = (0, 3/sqrt (2))

The radius of the circle  can  be  found  as EB =   sqrt [  (sqrt (8) - 0)^2  + (3/sqrt (2) - 0)^2 ] =

sqrt [ 8  + 9/2]  = sqrt [ 25 / 2 ]  =   5 / sqrt (2)

So....the diameter  =  2 * 5 / sqrt (2) =  5 (2 / sqrt (2)  = 5sqrt (2)   { as Holtran found !!! }

If you arrange the small triangles with the medium size ones, you can create a right triangle. If you draw a line connecting C and G, you can divide the quadrilateral into 4 triangles. You can now see that the yellow triangle and the combined red triangle are equal, so the answer is 1/2.

For example, let us set the intersection of AE and BH to X. if you move XAH to the bottom of XHG, you can create triangle XAG. Lets set the intersection of lines CG and AE to Y. XYG will equal to XHG. so yea i think. 1/2.

                                  B

          A  45                D              45   C

AB = CB

Angle B  = 90°

Angle ABD = 45

So angle ADB = angle CDB  = 90

So  triangles ABD and CBD are congruent  by HL

AD = CD  =  6

BD = 6

Area  =  (1/2)AD * BD =   (1/2) (6)(6)  = 18

As this question is not solvable by plugging in numbers to a hand-held calculator (or the nasty one on this website), the answer bots unfortunately will not be able to answer your question. However, I will take the liberty to say that the answer is \(5\sqrt2\). Go figure.

A

     15

            X

      15

B                    C

AB =  sqrt (15^2 + 15^2)  = 15sqrt (2)

Triangles AXB and ABC  are similar

BX / AX =  BC / AB

15 / 15  = BC / 15sqrt (2)

1  = BC / 15sqrt (2)

15sqrt (2)  =  BC

We   find AD as follows

BD^2 = AD^2  + AB^2 - 2(AD * AB) cos (60°)

6^2 = AD^2 + 5^2 - 2 (AD * 5) (1/2)

36 -25 = AD^2 -5AD

11 = AD^2 - 5AD

AD^2 - 5AD  =  11

AD^2 - 5AD  + 25/4  = 11 + 25/4

(AD - 5/2)^2  =  [ 69 / 4]

AD - 5/2  =  sqrt (69) / 2

AD  = [ 5 + sqrt (69) ] / 2

The height =  (5/2)sqrt (3) =  5sqrt (3)/2 

Area  =  AD * height  =      [5sqrt (3)] / 2  *   [ 5 + sqrt (69)] / 2  =   25sqrt(3) / 4  + 5sqrt [ 207] / 4  =

25sqrt (3) / 4 + 15sqrt (23) / 4 =

(5/4) [ 5sqrt (3)  + 3 sqrt (23) ] ≈  28.8

sqrt [ 4x + 1 ]  / sqrt [ -4x - 1]

4x + 1 ≥ 0               -4x - 1 >  0

4x ≥ - 1                   -4x > 1            (divide by -4  and  change the  direction of the inequality sign )

x ≥ -1/4                      x <  - 1/4

There are no intersecting intervals....thus.....no values of  x that  make this well-defined

y = log ₂ (2x)

y = log ₄ (16 + x)

log ₂ (2x)  =  log ₄ (16 + x)         {change of base)

log (2x) / log 2  =  log (16 + x) / log 4

log (2x) / log 2  =  log ( 16 + x) / log 2^2

log (2x) / log 2  = log (16 + x) / [2 * log 2]

log (2x)  = log (16 + x) / 2

2 log (2x) = log (16 + x)

log (2x)^2  = log (16 + x)

4x^2  = 16 + x

4x^2 - x - 16  =  0

x must be positive.....so......

x =   [ 1 + sqrt ( 1 + 256 ) ] / 8  =   [ 1 + sqrt (257) ]  / 8

\(\log_2 (\log_3 x) = 2 \log_2 (\log_{27} x) \)

We can write

log₂ (log ₃x) = log₂ (log₂₇x)^2

The log bases are the same, so

log₃x = (log ₂₇ x)^2

log₃ x =  (log ₂₇ x) (log ₂₇ x)               {apply change-of-base theroem}

log x / log 3 =  ( log x / log 27) (log x / log 27)      {assume x not equal to 1} {divide out log x}

1/ log 3  =  (log x/ log 27) ( 1 /log 27)

log 27 / log 3  =  log x / log 27

(log 27)^2 / log 3 = log x

(log 3^3)^2   / log 3  = log x

(3 log 3)^2 / (log 3)  =  log x

9 (log 3)^2 / (log 3)  = log x

9 log 3  = log x

log 3^9  = log x

x = 3^9 =  19683