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Return to Session #927 at the NCTM 2002 Annual Meeting

Point P Perambulates - posted April 8, 2002

I found the basis for this problem on the 1972 "Annual High School Contest of the Mathematical Association of America." While the actual question they asked is the bonus here, I thought a shorter version of the question would be a good geometry problem to use.

Equilateral triangle ABP with side AB is placed inside square AXYZ so that B is on side AX. The sides of the triangle are 2 inches long, while the sides of the square are 4 inches. The triangle is then "rolled" around the interior of the square - rotate it first clockwise around B so that P lands on X, then rotate around P so that A lands on XY, etc. Continue until the triangle returns to the lower left-hand corner. What's the total length of the path traveled by P as the triangle rotates? Give your answer in exact form (you don't need a calculator!).

Bonus: Instead of stopping when the triangle returns to the lower left-hand corner, continue until P, A, and B all return to their original positions. Now what is the length of the path that P travels?

Hint 1:
One good way to tackle a problem like this is to actually "model" what's happening - cut out a square and a triangle of the appropriate size, label them, and then do exactly what the problem says. Be sure to label P so that you can easily keep track of it. (You could even trace P as you rotate the triangle.)
Hint 2:
The path P takes may be broken up into separate pieces, some of which are "long" and some of which are "short." What is the shape of the pieces of the path traveled by P?
Hint 3:
To find the lengths of the pieces, try to figure out the angle of rotation each time P moves. For example, the first time we rotate the triangle, P moves to X. What is angle PBX?
Hint 4:
Once you know the angle of rotation, what portion of a whole circle does that represent? How does that help us find the length of each piece of the path?
Ans 1:
We found that when the triangle returned to the lower left-hand corner, P was in the corner, A was on AX (which now looks like PX), and B was the top of the triangle (floating in the square).
Ans 2:
We noticed that the path P takes as it moves was composed of arcs of circles. We also found that there were 5 different arcs formed as the triangle is rotated once inside the square, 3 "long" and 2 "short."
Ans 3:
We found that the total length of the path traveled by P is 14pi/3 inches.
Solution 1

The length of the path traveled by P is 14pi/3.

I drew a bunch of pictures to figure this out. Basically I drew the triangle in the square as it rotated, with a new picture for each new position of the triangle. Then I drew in the paths (in red) of how P moved. The picture shows that sometimes P moved a lot, sometimes it didn't move at all when the triangle rotated, and sometimes it moved a little.

I noticed that there were two different movements done by P. Both of them are arcs of a circle, since you are rotating a segment (a side of the triangle) around a point, which makes a circle. For the big ones, P rotates 120 degrees (I know this because the angle of the triangle is 60, which leaves 120). 120 degrees is one third of a circle (which is 360), so those paths are 1/3 of the circumference of the circle. There are three of them, so that gives us 1 circumference.

P also moves in small arcs. These are rotations of 30 degrees, which is 1/12 of a circle (360/30 = 12). There are two of those, so that is 2/12, or 1/6 of a circumference.

When we add them together, we have 7/6 of the circumference of the circle. This circle has a radius of 2 (since that is the length of the side of the triangle), so the circumference is 4pi (circumference = 2 * pi * r, and r is 2). 7/6 of 4pi is 14pi/3.

Solution 2

The length of the path is (14pi)/3 inches. For the bonus, the answer is 40pi/3 inches.

I made a picture that shows all of the different positions of P, and then I drew in paths (the dotted red lines) each time P moves.

P is moving along the arc of a circle each time, because BP (or AP, depending on which rotation it is) is like the radius of a circle when you rotate around B. Since each arc will be part of a circle, we can find the circumference of the circle, then figure out how long each arc is.
 circumference = 2 * pi * r = 2 * pi * 2 = 4pi
The first time P moves, it rotates 120 degrees around point B. I know that it's 120 because angle PBA is 60, so angle PBX must be 120 (since the whole thing is 180). To find out what part of the circle this is, we divide 360 by 120, which is 1/3. So the first time, it moves 1/3 of the circumference of the circle.
    1           4pi
   --- * 4pi = -----
    3            3
The second time it moves it also goes 120 degrees, so that is another 4pi/3.

The third time it moves it only goes 30 degrees. I know that it's 30 because angle XYZ is 90 degrees and the the angle of the triangle is 60, so that leaves 30. We divide 360 by 30 to find that this is 1/12 of the circumference of the circle. The length of this arc is 1/12 * 4pi, which is pi/3.

The fourth time it also moves 30 degrees, so that is another pi/3.

The fifth time it moves 120 degrees, so that's 4pi/3.

Now we have to add them all up to find the total length.

  4pi     4pi     pi     pi     4pi     14pi
 ----- + ----- + ---- + ---- + ----- = ------
   3       3      3      3       3       3
This is all in inches, so the final answer is 14pi/3 inches.

For the bonus, I found that it's 40pi/3.

To do this, I made two more pictures - you have to rotate the triangle around the inside of the square two more times to get the vertices of the triangle back in their original positions. I will list out the amounts that P moves each time you rotate the triangle - some of them are 0 because that is when you are rotating the triangle around P, so it doesn't move. There are eight "measurements" for P during each trip around the square because that's how many times you have to rotate the triangle to get it back where it started.

Rotation 1 (which goes with the picture in the first part):

  4pi       4pi     pi         pi     4pi
 ----- + 0 ----- + ---- + 0 + ---- + ----- + 0
   3         3      3          3       3

Rotation 2:

  4pi     pi         pi     4pi         4pi     pi
 ----- + ---- + 0 + ---- + ----- + 0 + ----- + ----
   3      3          3       3           3      3

Rotation 3:

      pi     4pi         4pi     pi         pi
 0 + ---- + ----- + 0 + ----- + ---- + 0 + ----
      3       3           3      3          3
Add all that up and you have 8 long arcs (4pi/3) and 8 short arcs (pi/ 3).
            4pi         pi     32pi     8pi     40pi
       8 * ----- + 8 * ---- = ------ + ----- = ------ inches
             3          3       3        3       3
It is interesting to look at the pattern of numbers. Each time P moves to or from a corner of the square, the arc is long. Each time it moves to or from a midpoint of the edge of the square, the arc is short.
What's the length of the path traveled by the center of the triangle? What if we rotated the triangle inside a different polygon with edgelength 4, like an equilateral triangle or a pentagon?

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