# Puzzle: Area = 1 (refined)

January 25, 2007 am31 8:10 am

Find the set of all points (x,y), such that the figure formed by (1,0), (0,0), (0,1) and that point, not necessarily in that order, has area = 1. The figure formed should be a closed, non-intersecting polygon.

Bonus: Find the additional set of points (x,y) that taken with (1,0), (0,0), (0,1) and that point, not necessarily in that order, form a self-intersecting polygon with area = 1.

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In order to save some typing, let O be the point (0, 0), let A be the point (1, 0), let B be the point (0, 1) and let X be our mobile point (x, y). Also, everything in this post is for non-crossing polygons; maybe I’ll try and do the crossing polygons in a later comment.

So, we can divide it into cases as follows (some were already solved by commenters in the previous thread):

If X and O are opposite vertices, and they lie on opposite sides of line AB, we have that the area of AOBX is the combined areas of triangles AOB and AXB. Thus, AXB must have area 1/2. It can be seen that the only such points are on the line x + y = 2, and that any such point works.

If X and O are opposite vertices and they line on the same side of line AB, the area of AOBX is the area of AXB minus the area of of AOB, so AXB must have area 3/2. It can be seen that all such points lie on the line x + y = -2, but not all such points work: in order to yield a non-crossing polygon, X must also be in the 4th quadrant.

Now, we have other possibilities: what if X and A are opposite vertices? (Luckily, we won’t have to separately consider X and B as opposite vertices, since the figure is symmetric.)

If X and A are on opposite sides of line OB, the area of ABXO is equal to the area of ABO plus the area of XBO, so the area of XBO is 1/2. It can be seen that any such point lies on the line x = -1, and that all such points work.

If X and A are on the same side of line OB, the area of ABXO is the area of XBO minus the area of ABO, which is 1/2. So the area of XBO is 3/2. Every such point lies on the line x = 3, but not all such points work: in order for the polygon to be non-crossing, X must lie between the lines OA and BA, so we have here that x = 3 and y is between -3 and 0.

As noted above, the case where X and B are opposite is just a reflection of this case.

So: in each case, a line and a line segment. I assume that the crossing cases will start at the endpoints of the line segments, but I haven’t tried to work it out yet. Jonathan, if I make a diagram of this, is there any way to upload it in a comment?

A small error in the previous post: in my third paragraph, I wrote “4th quadrant;” I should have written “3rd quadrant.” (The quadrant where both coordinates are negative.)

Now, for one case of self-crossing polygons:

If X is opposite O, in order to get a self-crossing polygon we must have that X lies in the 2nd or 4th quadrant, below the line x + y = 1. Let’s deal with the quadrant 2 case first (the quadrant 4 case is symmetric): Let the intersection point of AX and BO be P. Then we can solve for the coordinates of P; they are (0, y / (1 – x) ). The area of the quadrilateral is the area of XBP plus the area of AOP. Since we know the coordinates of P, we can calculate both of these areas easily (in the first case, by considering BP as a base): they are -x(1 – x – y) / (2 – 2x) and y / (2 – 2x). (The minus sign in the first one is because the height is |x| = -x, since x is negative.) Adding them, setting the sum equal to 1 and solving for y gives the curve y = -(x – 1)(x + 2)/(x + 1), and the part of this which falls in the needed region is for x less than or equal to -2. (This is half of one branch of a rotated hyperbola.) The 4th quadrant case is just the reflection of this curve across the line y = x, by symmetry.

JBL,

if you make a diagram, I will try to put it in the comments. If not, I will put it in a follow up post, linking back and forth. Tonight I will try to digest your work (I completed the non-overlap case, but the overlap looks more involved)

Jonathan

you can use the “shoelace formula” to find the area of a convex polygon

and rather trivially derive the results obtained above at far greater cost (though no doubt with greater benefit)

http://www.mathwords.com/a/area_convex_polygon.htm