Christmas conundrum

[Greyhound Dude]

If you assume oak should be replaced with "orange", the distance is 61.61 ft. which is approximately 20.53 ft. which rounds up to 21 yds.

[andrewjackson]

If you assume oak should be replaced with "orange", the distance is 61.61 ft. which is approximately 20.53 ft. which rounds up to 21 yds.

I'd be curious to see how you get one exact answer rather than four.

[BackInTex]

the distance is 61.61 ft. which is approximately 20.53 ft.

I'll never understand this dog math! I thought you divided by 7.

[Greyhound Dude]

For those of you who wonder how I got an "exact" answer, I am a computer drafting and design technician and have CAD software that allows me to lay out points based on distances from known objects.
Pretty simple.

So much for doggie math and multiplying by 7.

So there!!!!!!!!

[andrewjackson]

For those of you who wonder how I got an "exact" answer, I am a computer drafting and design technician and have CAD software that allows me to lay out points based on distances from known objects.
Pretty simple.

So much for doggie math and multiplying by 7.

So there!!!!!!!!

I'm not questioning the math or that there is a possible arrangement that will result in a RB distance of 21 yards. My issue is that it looks to me like there are four possible arrangements of the trees that results in four different distances for RB. Here is one example. It is possible to draw the arrangement of trees like this. Ignore the dots since they are there only to correctly place the letters.



.......................................O



......................................................R
..........................Y................................P

.............B



The OR distance is slightly shorter than OP (10 to 11) and YR is slightly shorter than YR (12 to 13) so R can wind up inside the triangle OYP but very close to P. Since OB is much longer than OY and PB is much longer than PY it winds up down in the lower left. However R is slightly closer to B than P so RB must be around 16 yds. Ignore the dots since they are there only to correctly place the letters. I've drawn this out and done the math and I get 16.4 yards for RB.

[andrewjackson]

Or you can put the Blue tree up to upper right and get the same distance RB.

...................................................................................B




.......................................O



......................................................R
..........................Y................................P





But once again RB must be similar to the length of PB. There is no way it can be 21 yards.

[Greyhound Dude]

Basically, it's all about intersecting circles at different radii. You need to be drawing circles, not straight lines (if I'm understanding your diagram properly). If you have two objects at a certain distance, think of one object being the center of the circle and the other object being along the circumference of said circle with whatever radius you want. With only (2) objects, there are infinite possibilities (placement of the second object along the circumference). When you start adding other objects based on those original two objects, you narrow down the possible intersection points of the circumferences of the circles.

[andrewjackson]

Basically, it's all about intersecting circles at different radii. You need to be drawing circles, not straight lines (if I'm understanding your diagram properly). If you have two objects at a certain distance, think of one object being the center of the circle and the other object being along the circumference of said circle with whatever radius you want. With only (2) objects, there are infinite possibilities (placement of the second object along the circumference). When you start adding other objects based on those original two objects, you narrow down the possible intersection points of the circumferences of the circles.

I understand all that and that's how I did it.

I just seems to me like you can get different distances based on different arrangements of the trees that still respect the originally specified distances.

In the example that I tried to illustrate above I don't see how RB can be that much greater than PB.

But maybe not. Maybe I'm making some kind of fundamental mistake that I'm just not seeing.