CPhill

5 yards  =  180  in

We have a right triangle with a hypotenuse drawn from the vantage point to the center of the circle =  (2 in + 180 in)  =  182 in   and a leg of 2 in which meets a tangent drawn to the circle from the vantage point

We can use the arcsin function to  find  (1/2) of the measure of the angle  we need......so we  have

arcsin (2/182)  = θ  ≈  .63°

So....twice the measure of this angle is what we need  ≈ 1.26°

-2x + y =   k            0.5x + y  = 14

These intersect when  x  = -8.4

So...this means that

0.5 (-8.4) + y  = 14    simplify

-4.2 + y  =  14      add 4.2 to both sides

y  =  14 + 4.2   = 18.2

So....the intersection point  is (-8.4, 18.2)

And we have that

-2(-8.4)  + 18.2  =  k

16.8 + 18.2  = k

35  = k

Here is the graph : https://www.desmos.com/calculator/xryjrho9rx

Thanks, heureka.....I wasn't familiar with the "hockey stick" identity.......very informative....!!!!

For this problem....I'm assuming that the mother and daughter normally arrive at the train station at the same time

I'm also assuming that "between the station and their house" means halfway between the station and the house

Call the distance from the house to the train station , D

So....the distance that the mother would normally drive is 2D

And let her rate be   R

And let the time that she would normally drive be = T   (in hours)

So  we have that

2D / R   = T      (1)

But..on this day.....the mother only drives  (1/2)D + (1/2)D   = D

And....if she normaly returns at 5PM, but now only returns at 4:48PM....then her normal driving time must be 12 minutes less than normal  = 1/5  hr

So.....we have that

D / R  =  T - 1/5    multiply through by 2

2D / R  = 2(T - 1/5)      (2)

This implies that   (1)  = (2)...so....

T = 2(T - 1/5)

T = 2T - 2/5

2T - T  = 2/5

T  = 2/5.......this means that her normal round-trip driving time  is 2/5 hr  =  24 min

So....she must normally leave her house at   4:36 PM   (since 24 minutes after this = 5PM)

And she would normally reach the station at 12 minutes after 4:36 PM =  4:48 PM

But....this means that she reaches the halfway point at  6 mnutes after 4:36 PM  = 4:42 PM

Now.....let's look at this from the daughter's point of view..... she would normally meet her mother at the station at  4:48 PM.....but.....since she gets there an hour early......she must reach the station at 3:48 PM  and she immediately begins walking

So.....she walks from  3:48 PM   and meets her mother at the halfway point at  4:42 PM

So....she walks for 54 min

THX to Melody and heureka.....!!!!

THX, tom  for that helpful hint !!!

To generate a "closed" form solution that the Guest found......, we can use something known as the " sum of differences"

We have the following  sequence of row sums

1      5       15     34      65       111    take the positive difference between terms until we get a "constant" row

    4      10     19     31        46

         6       9      12     15

              3      3       3

It took us three  rows of differences to get to a "constant" row

This means that we will have a cubic (3rd power) "generating" polynomial  of the form

an^3  + bn^2  + cn + d = S(n)         where n is the row  number and S(n) is the row sum

So...we have this system of  equations 

a(1)^3 + b(1)^2 + c(1) + d  = 1

a(2)^3 + b(2)^2 + c(2) + d = 5

a(3)^3 + b(3)^2 + c(3) + d  = 15

a(4)^3  + b(4)^2 + c(4) + d  = 34   which gives us

a  +  b  +  c  + d   =   1

8a  + 4b + 2c + d  = 5

27a + 9b + 3c + d  = 15

64a + 16b + 4c + d  = 34

I'll  spare you the gory details of solving this by hand  [ but...I can go through it...if you want to ]

The solutions  are   a = 1/2, b = 0, c = 1/2   d = 0

So....the polynomial that generates the sum of any nth row is given by

(1/2)n^3 + (1/2)n  =  S(n)

[ n^3 +  n ]

________    =   S(n)

      2

Just as the "Guest" found!!!

Look   at   the last number in each row  and notice this pattern

               1     =     sum of the 1st positive integer

                  3   =    sum of the first two positive integers

                     6 =    sum of the first three positive integers

                        10  =  sum of the first 4 positive integers

So.....the last entry in the nth row apears to be just the sum of the 1st  n positive integers

And the sum of the first n positive integers is just   :  n (n + 1)   /2

So....the last entry  in  the 2019th row is the sum of the first 2019 positive integers  = 

(2019) (2020)  / 2  =   2039190

And the nth row contains n integers...so....

We will have 2019  integers in this row and we need to find the first one....

Note that, after the first row....the first entry on any row is given by  :

Last entry  - row number + 1 

So...the first entry on the 2019th row must be :  2039190 - 2019 + 1  = 2037172

So...finally....the sum of the terms in the 2019th row is given by

[ first entry + last entry ] * number of terms / 2  =

[2037172 + 2039190 ] * [ 2019] / 2  = 

4,115,087,439

Working backwards....the last number displayed before pressing the  \(\textcolor{blue}{\bf\circledast}\) key would have been 324

Taking the square root of this we get 18

So...before pressing  the  \(\textcolor{blue}{\bf\circledast}\)    key   we would have had 81

Taking the square root of this  = 9  =  the original integer

2y = -x + 3        y = 5x + 1

Rearranging the frst equation we have   x  = 3 - 2y

Sub this into the second equation for x and we have that

y = 5(3-2y) +1

y  =  15 - 10y + 1

11y  = 16

y = 16/11

So

x  =  3 - 2(16/11)

x = 3 - 32/11

x = 33/11 - 32/11  =   1/11

So...the intersection point is  ( 1/11, 16/11)