Wednesday, March 19, 2008

A, B are two points on a circle... Extending Student Thinking in Geometry

One of my stellar SAT students (who has been mentioned before) led me to develop the following extension of a challenging SAT-type question. Problems similar to this have appeared on previous SATs and math contests. All geometry students can benefit from both the spatial reasoning inherent in this problem as well as the algebraic analysis needed to demonstrate the results numerically.

Part of the issue of problems like this is that test constructors often attempt to develop questions similar to previous questions that had proved effective. An earlier version of the problem below read something like this:

A, B are 2 distinct points on a line. How many points on the line are twice as far from A as from B?

Occasionally, the ratio changes: three times as far from A as from B...

I've even seen other locus versions of this basic premise:

If l and m are parallel lines in a plane, how many lines are twice as far from l as from m?

In both cases, there are two solutions, one involving internal division and the other involving external division.

One can make this a bit more challenging by asking the more general version:

How many points are twice as far from one point as from the the other. (Similarly for the parallel lines problem).

By symmetry, there would now be 4 solutions. Nice questions that promote spatial sense and the ideas of internal and external ratios. We could stop there, but changing the line to a circle adds another dimension to the question:


The Challenge:
A and B are distinct points on a circle of diameter 2.
The length of arc AB is 1.
Note: For this question, all distances are arc lengths, not chord lengths.

(a) How many points on the circle are three times as far from A as from B?

(b) Now specify the locations of such points. Thus, if P is a point such that the length of arc PA is three times the length of arc PB, determine the position of point P in each case and the actual lengths of these arcs.


Comments:

Do you think most students or even the author of this question (I've altered the original problem slightly and extended it with part(b)), considered what the student with greater insight might have considered, namely, the issue of major vs. minor arcs? Do some students see the more profound subtleties of these questions -- the ambiguities that the author or others do not consider? Worth discussing the topological similarities and differences between a circle and a line?

7 comments:

Florian said...

I like these geometry challenges! :)

The challenge problem is equivalent to two line problems
or if you want to one line problem with three points instead
of only two points (the third point position = 2*PI*r - (B-A)).

The two line problems are with points A and B and
with the points B and C where C = 2*PI*r - (B-A).

(Note: r=1 because diameter=2)

Since each line problem has two solutions there must
be 4 solutions in the circle problem.

Assuming that A < B < C we get:

CASE 1a: P < A < B
3*(A-P) = B-P <=> 2P =3A-B <=> P = (3A-B)/2

CASE 1b: A < P < B
3*(P-A) = B-P <=> 4P = 3A+B <=> P = (3A+B)/4

(Note: case A < B < P is the same as case 1a)

CASE 2a: P < B < C
B-P = 3*(C-P) <=> 2P = 3C-B <=> P = (3C-B)/2

CASE 2b: B < P < C
P-B = 3*(C-P) <=> 4P = 3C+B <=> P = (3C+B)/4

(Note: case B < C < P is the same as case 2a)

We dont know the positions of A and B, but we
know that their distance on the line is 1.
(and that the distance of B and C is 2*PI - 1).

Lets choose A to be at position zero => B is
at position 1 and C is at Position 2PI-1.

Now we can calculate P for the cases above:

CASE 1a: P = (3A-B)/2 = -1/2
CASE 1b: P = (3A+B)/4 = +1/4
CASE 2a: P = (3C-B)/2 = (6Pi-6-1)/2 = 3Pi-7/2 = 5.92...
CASE 2b: P = (3C+B)/4 = (6Pi-6+1)/4 = 3Pi-5/2 = 6.92...

On a circle with r=1 CASE 1a means that P = 2*Pi - 1/2 = 5.78...

Florian said...

In the above post the values for P are in relation to A at position zero. To get the arcs one has to work one more step and find the values of |A-P| and |B-P| for the various values of P.

Totally_clueless said...

>>How many points are twice as far from one point as from the the other.
>>By symmetry, there would now be 4 solutions.

If this is in 2-D space, shouldn't there be an infinite number of solutions (on a circle, I think jd2718 had posed a similar question before).


Also on the cirlce, do you define the distance as the shortest arc length between the two points?

I get three solutions for the circle question. Wonder if that is correct?

TC

Florian said...

TC, I think it's not in 2D its only 1D because the points only move along the circle (i.e. you can specify the location of a point with only one value) and the distances are always measuread along the arc and not the coord between two points.

Dave Marain said...

This is getting a lot more complicated than it seemed at first! Once we allow arcs to be either major or minor there should be several positions for point P. My first inclination was to consider that there were 3 locations for p as TC suggested, but I di not consider other orientations. I now see at least 4 possibilities.

I like Florian's idea of placing A at a fixed position like 12:00 or 3:00 or giving it a coordinate of zero (similar to the way the unit circle is set up in trig). I suspect there will be more dialogue here before this problem is laid to rest...

Eric Jablow said...

TC,

In two dimensions, the answer will indeed be a circle. Normalize the problem by letting A be (1, 0) and B be (-1, 0). Then, let's work with the squares of the distances, not the distances themselves. This avoids square roots. let P be (x, y).

PB = 2PA

PB² = 4 PA²

(x+1)² + y² = 4[(x-1)² + y²]

x² + 2x + 1 + y² = 4x² - 8x + 4 + 4y².

3x² + 3y² - 10x + 3 = 0.

x² + y² - 10/3 x + 1 = 0

You can assure yourself that this has solutions; either compute the determinant, or reason that it has solutions on the x-axis, so the discriminant of the x-terms is positive, and completing the square will give you an equation of the form

(x - k)² + y² = l² for some k and l.

You can compute k and l if you want to.

Dave Marain said...

Thanks, eric--
That's a nice locus problem with a result that is surprising to some students. some expect that the 2:1 ratio will lead to some other conic. If oyu change the ratio to k:1, it would appear you would still get a circle, but I haven't worked out the details to see if there are restrictions on k.

Eric, what about the given circle problem in this post. Does the ambiguity of major vs. minor arcs enhance this question or detract from its quality in your opinion? Should the original author of this problem specify that distance should be interpreted as the 'shortest distance'?