Mark:
MAPS
Yes.
Simple constrained version: The orientation of the small map is the same as the large map.
Let (0,0) be the lower left corner of the large map.
Let S (0 < S < 1) be the scale factor of the smaller map.
Let (Tx,Ty) be the lower left corner of the small map.
Let (x,y) and (X,Y) be coordinates of the small and large maps repectively.
Then,
x = S*X + Tx
y = S*Y + Ty
We want x=X and y=Y, so we have
x = S*x + Tx
y = S*y + Ty
x = Tx / (1-S)
y = Ty / (1-S)
Note that S could be negative (flip the map upside down), and different scale factors could be used in the X and Y directions.
If the small map is rotated, then the equations must be modified to include rotation (trig functions), but there will still be two equations and two unknowns. It might get a bit messy making sure the small map is completely contained within the large map.
Also: The small map could be crumpled up and placed over the large map, and the situation would still apply.
That, if I may say; is a beautifully crafted answer.
Seems to me that this is a two-dimensional analog of the one-dimensional problem that asks:
If a person hikes up a mountain from 12:00 to 4:00, then hikes down the same path the next day from 12:00 to 4:00, prove that there must be a point on the trail that the person occupied at the same time each day.
(True)
A slightly different take:
The point is obviously unique, because the two maps have different scales (but if P and Q where two fixed points the distance between them would be the same on both maps).
Let the small map square be A'B'C'D' and the large be ABCD, where X and X' are corresponding points. We deal first with the special case where A'B' is parallel to AB. In this case let AA' and BB' meet at O. Then triangles OAB and OA'B' are similar, so O must represent the same point. So assume A'B' is not parallel to AB.
Let the lines A'B' and AB meet at W, the lines B'C' and BC meet at X, the lines C'D' and CD meet at Y, and the lines D'A' and DA meet at Z. We claim that the segments WY and XZ meet at a point O inside the smaller square. W cannot lie between A' and B' (or one of the vertices A', B' of the smaller square would lie outside the larger square). If it lies on the opposite side of A' to B', then Y must lie on the opposite side of C' to D'. Thus the segment WY must cut the side A'D' at some point Z' and the side B'C' at some point X'. The same conclusion holds if W lies on the opposite side of B' to A', because then Y must lie on the opposite side of D' to C'. Similarly, the segment XZ must cut the side A'B' at some point W' and the side C'D' at some point Y'. But now the segments X'Z' and W'Y' join pairs of points on opposite sides of the small square and so they must meet at some point O inside the small square.
Now the triangles WOW' and YOY' are similar (WW' and YY' are parallel). Hence OW/OY = OW'/OY'. So if we set up coordinate systems with AB as the x-axis and AD as the y-axis (for the large square) and A'B' as the x'-axis and A'D' as the y'-axis (for the small square) so that corresponding points have the same coordinates, then the y coordinate of O equals the y' coordinate of O. Similarly, XOX' and ZOZ' are similar, so OX/OZ = OX'/OZ', so the x-coordinate of O equals its x'-coordinate. In other words, O represents the same point on both maps.
STEPS
34 + 18e = s
26 + 30e = s
e = 2/3 steps per second (escalator speed)
s = 46 steps
Congratulations, that is the same number as Butterfly came up with, so it must be right.
It was getting near bedtime and I had just finished a glass of wine. Shari was reading the last sentence of War and Peace. I thought this was the perfect time to make my move. However, she pre-empted my suggestion with a; no we are not subscribing to the pay to view channels unless you can tell me:
What is the smallest number of integer-sided rectangles that tile a rectangle so that no two rectangles share a common length
Needless to say I didn't know what she was talking about. However, with the whole future of broadcasting hanging in the balance, can anyone assist?
