Solving Geometry Problems in Everyday Life
James M. Parks
SUNY-Potsdam
parksjm@potsdam.edu
Abstract: Many examples of max/min geometry problems from everyday life can be solved using transformational geometry. The ideal format is to illustrate these on Geometers SketchPad, which we will do. As many examples (with solutions) will be given as time allows.
1. Reflections
A. Heron’s Principle of Light Reflection.
An important result that can be used in conjunction with many applications, the Principle of Light Reflection says that for a given ray of light the angle of incidence i° to a surface/mirror equals the angle of reflection r° (measured from the normal to the surface, see figure 1).
Physicists tell us that the light ray will travel along a path which minimizes distance. Thus
the distance along the path from point A to P and then from P to B is to be a minimum.
Fig. 1.
But this is the same idea as reflection about a line. If we reflect ray PB about the mirror line to get ray PB’, we know from Euclidean geometry that r’° = r°, and thus the shortest path from A to B via point P is equal to the straight line distance from A to B’.
This means that we must choose point P to be the intersection of segment AB’ with the mirror, and that then i° = r’° = r°.
As we shall see below, this result can be used in applications with many different representatives for the light ray, such as the shortest path traveled by an object between two points. The following example is just one such case.
B. The Burning Tent Problem.
Suppose you are camping, and you set up your tent near a river, you start a small camp fire in front of it, and then you go off down the river to see if anyone has caught any fish yet.
Upon returning, you notice that the campfire has set your tent on fire!
Luckily you have a bucket in your hand, and you instinctively race for the river to get some water (figure 2).
Fig. 2.
Quick! What is the shortest path to get some water and put out your tent?
If we use reflections to reflect the path RT about the edge of the river to obtain a new path RT’, and a polygonal path XRT’ (figure 3).
This gives us an easy way to find the shortest path, since the shortest
path from X to T’ is a straight line?
This will in turn correspond to the path which satisfies the above Principle of Light Reflection.
Fig. 3.
Therefore, you should always head for the point R’ on the river, which is the intersection of the line XT’ and the edge of the river! (try this on GSP to discover how changes in the position of points X or T changes the position of point R’).
Now, suppose you don’t have a bucket? What would be your shortest path if you have to first run to a fence on your left to get a bucket? (see figure 4).
Fig. 4.
C. Designing a “cheap” diamond pendant (the inscribed triangle of smallest perimeter).
Suppose that you are designing a new piece of jewelry which is in the shape of a triangle pendant and you want to make an inscribed triangle with diamonds around it’s perimeter? Then finding the inscribed triangle of smallest perimeter is definitely of some interest ($$)?
Fig. 5.
So the problem is: given a triangle ABC is it possible to inscribe a triangle DEF such that the vertices D, E, and F are on different sides of ABC, and DEF has the smallest possible perimeter (see figure 5 above)?
The Principle of Light Reflection can be used here also, for we can “unfold” DEF by reflecting about an edge of ABC (this example is best carried out by following along on GSP).
The solution given here follows Yaglom, Geometric Transformations II, MAA Press.
Specifically, assume that we have an acute triangle ABC, and reflect ABC along edge BC to get a congruent A’BC with inscribed D’EF’ (figure 6).
So what have we accomplished?
The path around DEF is now equal to the paths DE, EF’ & F’D’.
If we now reflect A’BC about the other side, A’C, to get A’B’’C with inscribed D’’E’’F’. We will have completely “unfolded” DEF along paths DE, EF’, F’D’’ (the bold lines in figure 7).
Fig. 6.
As in the Burning Tent Problem, we wish to find the shortest path between two points, in this case the path which is equivalent to the shortest path from D to D’’ in our unfolded triangle (the dashed straight line in figure 6).
Therefore we simply move point E on BC to coincide with the intersection of BC and DD”, and we move point F’ on A’C to coincide with the intersection of A’C with DD’’ (and this can always be done?, see figure 7).
Notice how *DEF automatically adjusts to these changes?
Fig. 7.
But have we found the triangle of minimum perimeter? The results we have so far depend on the starting position of vertex D??
So now the problem is to find the position of vertex D which gives the desired minimum perimeter for *DEF. Another observation we can now make is that when we move vertex D, we notice that the length of DD’’ changes.
Also, vertex C is a “pivotal” point with respect to the position of DD’’. In fact if we connect CD and CD’ we have an isosceles triangle, DCD’’, with base DD’’ (why?, see figure 8).
The desired position for vertex D therefore corresponds to that point which makes the legs of DCD’’ a minimum, which is at the foot of the point where the altitude (the perpendicular from C to AB) of ABC at vertex C occurs!
Since the choice of side BC to reflect about in the beginning was arbitrary, this means that the solution is achieved by choosing the vertices for DEF at the feet of the respective altitudes of the respective vertices of ABC.
Recall we assumed that ABC was an acute triangle in the beginning? Now what if Fig. 8.
one of the angles is now 90° or larger?
If you follow through the construction given above with this type of triangle you will discover that one of the vertices of DEF will coincide with one of the vertices of ABC (try this one out on GSP to see how these choices make the solution obvious!).
D. Which set of paths/communication lines is the fastest/shortest? (the Erdös-Mordell Problem)
An example of comparing lengths of “sets” of paths is illustrated in the following problem (conjectured by P. Erdös in 1932, and first proved by Mordell in 1935 using trigonometry).
Suppose you have a triangle of “outposts” A, B, & C with connecting communication lines/paths a, b, c (see figure 9 below). A “central” command post P is to be built in the central region of the outposts. Connecting lines are needed between this central post P and the outposts A, B, C.
There are also some constraints on these connecting lines.
If these lines are built perpendicular to the lines between the outposts, a, b, c, then a second (return) line is required, and when a message is sent out, a return message is sent from these connections D, E, F to insure that the connections are working. If the connecting lines are built directly to the outposts, A, B, C, then only a single line is required, and no return message is needed (see figure 9).
There are at least two questions here. First, which set of connecting lines will be the fastest(shortest)? and second, does it matter where the center P is built?
Fig. 9.
Check this one on GSP to get a feel for what the result may be.
Choose a position for the “center” P. If we let PA, PB, PC denote the distances to the vertices A, B, C, and Pa, Pb, Pc denote the perpendicular distance to the sides a, b, c, respectively, then, since speed and distance are directly proportional in this case (why?) we need to compare the distance
PA + PB + PC to the distance 2(Pa + Pb + Pc).
The following proof is essentially due to V. Komornik , see American Mathematics Monthly, vol. 104, 1997, pp. 57-60.
So, suppose point P is on side a, then reflect ACP about side b to get a "kite" with area bPb (Pb is the perpendicular distance from P to c). Similarly, the area of the "kite" formed by reflected ABP about side c is cPc. Thus twice the area of ABC is bPb + cPc (see figure 10).
Now PA is at least as long as the altitude of ABC from vertex A, so we have the inequality aPA ≥ bPb + cPc.
If we move P to some other location in ABC this inequality still holds, since we can extend AP to the intersection with side a at Q and the inequality holds by similarity for AQ (figure 11 below).
Next reflect P about the angle bisector AH of angle CAB to get P''' with PA = P'''A, Pb = P'''c, and Pc = P'''b.
Then we have the inequality aPA ≥ bPc + cPb (figure 12 below). Similarly, for P inside ABC, we have bPB ≥ cPa + aPc and cPC ≥ aPb + bPa.
Fig. 10.
Fig. 11. Fig. 12.
Then, by substitution, we get PA+PB+PC ≥ (a/b+b/a)Pc+(a/c+c/a)Pb+(b/c+c/b)Pa ≥ 2(Pc+Pb+Pa).
Note that equality holds when ABC is equilateral and P is the center point?
Thus the shortest set of paths is {Pa, Pb, Pc} with length 2(Pa+Pb+Pc).
2. Translations
A. The new road problem (where to build the bridge?).
Two cities on opposite sides of a river (but not directly opposite each other) wish to build a bridge over the river with connecting roads to the cities.
Where should they put the bridge in order to make the distance between the cities as short as possible, assuming that the river is approximately of constant width in this location (figure 13)?
Fig. 13.
Try different paths (on GSP), and it quickly becomes clear that, since the bridge is (relatively) constant in the calculations, it can be ignored? In other words, it’s the location of the bridge which dictates the length of the roads.
In fact, the direction of the bridge over the river can be arbitrary?
Fig. 14.
One way to get around this “bridge” part of the problem is to take out the bridge!, that is, translate the bottom half of the figure up by the “bridge vector” ED, using the translationTED , so that the roads come together at the edge of the new “river/bridge” point D/E (see figure 14).
Now we are back to the straight line/shortest distance type of problem (recall the Burning Tent Problem?).
Fig. 15.
The answer therefore is clear, move D/E to the dashed line - river edge intersection and reinstall the bridge (translate the bottom half down by the translation TDE ).
Alternately, you can build a revised but equivalent “mathematical” route with the bridge at the beginning, which is equivalent to the original path traveled (see figure 15).
Fig. 16.
Then the solution is similar to that in figure 14.
An interesting variation would be to have a river branch out at the location where the bridge is to be built (see figure 16 below).
Fig. 17.
Then a similar technique to the above result will work.
If you remove the bridges and get down to just the roads (see figure 17) then all you have to do is move the points D/E and F/G to the dashed straight line, and then reinstall the bridges, as above.
B. The Shortest Shoelaces Problem.
Did you ever experiment with different ways to lace up your shoes? Maybe you’ve broken your shoelaces before and have had to use a shorter piece to make do?
Most people use the conventional zigzag pattern, but sometimes you see variations on this in shoe stores and other places.
See J. Halton, “The Shoelace Problem” in The Mathematical Intelligencer, vol 17, Fall 1995, pp 36-41, reprinted in Scientific American, July 1996 issue.
So the question is “What would be the shortest laces you could possibly use to tie your shoes, if you have to use all of the eyelets?”.
Of course we need to make some ground rules here?
So let’s agree that for a “legal lacing” we have to use all of the eyelets, each one exactly once, and the lacing has to start at the top left eyelet, zigzag in any right-left-right pattern, and end at the top right eyelet.
To make things even easier, we will also assume there are only 4 pairs of eyelets, and let’s assume that the bow is the same for all lacings, so that we can ignore the bow part of the lacing.
Fig. 18.
In fact all we need to concentrate on are the eyelets, so we will consider only the part of the problem pictured in figure 18.
Try making up some examples, like the zigzag pattern above, and a few others on your own (see figure 19 below).
How can we tell if we have the best (shortest) possible lacing?
We could mark each lacing’s beginning and end positions, then take them back out and measure them, or try to measure them in the eyelets? If we could “unfold” the lacings, using the Principle of Angle Reflection, we would have a way of comparing different lacings with each other? This might make the job of finding the minimum lacing easier?
Fig. 19.
Using the GSP Sketch “eyelets-4” and the GSP Script “4-eyelets” we can easily do exactly the “unfolding” operation mentioned above. That is, we will successively reflect the eyelets about the right-hand column and draw in the next lacing one step at a time. The comparison of the first three examples in figure 19 above is shown in figure 20 below.
Fig. 20.
The straight line path of the unfolded first example (zigzag pattern) in figure 20 catches your eye immediately. It is definitely a candidate for the minimum lacing!?!? To see why we have left out the last example in figure 19 here, try unfolding it for yourself on GSP and see what the pattern looks like compared to these.
If we reflect the “right half” of these examples in figure 20 above about the bottom row of eyelets we can then use the Principle of Reflection, as in the Burning Tent example above, to get a clearer picture of the shortest path from the top left eyelet to the top right eyelet (see figure 21).
It is now clear that, at least for these three examples, the first one is the shortest path from the
top left eyelet to the bottom right eyelet (the reflected top right eyelet), and since this path is equivalent to the original “first zigzag” path of lacing the eyelets, it is the shortest lacing.
Fig. 21.
But how do we know if this is the overall minimum for all possible (legal) lacings?
Suppose we have another lacing which is vying for the honor of “shortest” lacing? Then it will have to vary from the “zigzag” pattern at some point, which means that it will have to either go straight across, which will correspond to a horizontal line segment in the unfolded graph (figure 20), or it will have to go “up”, which will correspond to a line segment going up in the unfolded graph (figure 21).
If there is a horizontal segment, then we can remove the only horizontal line segment (in the middle) from the “zigzag” pattern and the horizontal segment from the new “pretender” lacing. If we connect up the ends, the resulting path will still go from the upper left eyelet to the lower right eyelet, but now the “new” zigzag path is a straight line, and it is clearly the shortest possible path!
If the new lacing has a segment which goes up (and then it must have a corresponding segment after this point which goes down?) then I leave it to you to argue that the new path for this case is longer than the “zigzag” path.
3. Rotations
A. Where to build the Mall? (Finding the Fermat/Torricelli point).
This is a well-known problem.
Three cities wish to build a mall (or airport, or school, or - ??) in a central location between them (see figure 22).
Of course this means building new access roads. In order to save money the “Mall Committee” decides that the roads should be as short and as straight as possible, and they decide
that this means that the total length of the three roads should be as short as possible (maybe there is a mathematician on the committee?).
Fig. 22.
Therefore, they wish to find the spot for the Mall which will meet these requirements.
So the problem is to find the point M for the Mall which makes the sum of the paths TM, AM, and SM a minimum.
To start with, let’s assume that the triangle *SAM is acute.
It would be nice if we could use the “straight line” solution, as in some of the applications above? But the neat thing is, if we use a rotation, then we can use the “straight line” solution!
So here’s the key, consider *SAM contained in *ATS (figure 23). If we rotate this triangle 60° counter-clockwise about point A, then *M’AM is an equilateral triangle, so AM = M’M, and M’S’ = MS by the properties of rotations (notice that the position of point S’ is unaffected by the position of point M?).
Fig. 23.
Thus the total length of the roads, TM+AM+SM = TM+M’M+M’S’, and obviously the shortest path from T to S’ is a straight line (the dashed line), so we have the solution? All we need to do is move point M down to the dashed line S’T so that M’ is also on S’T (and this can always be done?).
It will be obvious, once this point M is found, that the angle <SMA is 120°?, therefore each of the central angles is 120°? (the point M is sometimes called a Steiner point).
Now, what happens if we don’t have an acute triangle?
In particular, if one of the angles of *ABC is 120° or larger the above construction will lead to another result.
What is it? (check this one out on GSP!).
B. Finding the largest rectangular lot in a triangular piece of land (the inscribed parallelogram/rectangle of largest area)
In his book, Stories about Maxima and Minima, AMS-MAA Press, 1990, V. M. Tikhomirov claims that this is the only max/min problem in Euclid’s Elements.
The problem is to find the parallelogram (rectangle) ADEF of largest area which can be inscribed in a given triangle *ABC (figure 24).
An application is to find the largest rectangular lot which can be made from a triangular shaped piece of land.
Since a parallelogram and a rectangle of the same base and height have the same area, we shall see that the solution of one problem also solves the other problem (figure 25).
Fig. 25.
If you make the construction in figure 24 on GSP it quickly becomes apparent that a good candidate for a solution is the parallelogram with vertex A, and points D, E & F near (at?) the midpoints of sides AC, CB and AB, respectively. It should also be clear that the parallelogram determined by any vertex of *ABC and the midpoints of the sides has the same area as this parallelogram (why?).
We will give an argument using parallelograms which is different from that given by Euclid.
Make a copy of *ABC which is rotated 180° and translated so that it shares side AB, but in the opposite direction (figure 26).
This makes ACBC’ a parallelogram(?).
Connect F & D’, and E’ & D’ to make another parallelogram(?) AE’D’F in *AC’B. Also notice that EFD’B is a parallelogram(?).
Let L be the length of AC’, the base of ACBC’, and let W be the height of parallelogram ACBC’.
Let M be the corresponding length of the base CE = DG, and X be the height of parallelogram DCEG (G the point on EF which makes DCEG a parallelogram), and notice that DCEG is equal to two copies of *CDE?
We will also assume that X ? W/2 and M ? L/2, since if not we can change the choices of X and M (switch sides) to make these inequalities hold.
Let N be the length of the base FD’ = EB, and let Y be the height of parallelogram EBD’F.
Then Y = W-X (E’G’D’C’ is a copy of DCEG?), and N = Y - M (figure 27).
Now, the area of the “white” part of AC’BC is XM + YN, and if D, E, & F (and thus also D’ & F’) are at the midpoints of the respective sides, then the area is 2(W/2)(L/2) = WL/2, since then X = Y = W/2, and N = M = L/2 (it’s helpful to follow along on GSP).
Thus, if we can show that the difference between these areas is always nonnegative, (XM + YN) - WL/2 * 0, then the maximum area is determined by the midpoints of the sides (and we’re done).
So, if we substitute the definitions of Y and N into the first area and simplify, we get the following result: (XM + YN) - WL/2 = XM + (W - X)(L - M) - WL/2 = WL/2 - XL - WM + 2XM = (W - 2X)(L/2 - M).
But these last two factors are always nonnegative?
Hence the maximum area is indeed determined by the midpoints of the sides of *ABC, and we’re done!