Friday, July 25, 2014

Shortest Way to Help Grandma

Good news: The solution time has been extended till Friday as by technical reasons the newsletter didn't go on time.

You and your grandma live on the same side of the river. You need to find the shortest way to walk from your house to the river (with an empty bucket), fill the bucket with water and bring the water to your grandma. The river is fierce and you cannot swim in it.  Disregard the fact that it is much harder to walk with a full bucket than with an empty one. Just look for the shortest path.


Note: the solution does not involve any equations, just one-two simple lines.... Feel free to print the image or redraw it. Submit your answers as images via email to me or in the comments on this blog post. Please explain your answer. The answers are accepted any time until midnight Eastern Time on Friday, on our Family Puzzle Marathon

Friday is here and below is the answer drawing:


The shortest way is marked with a red line.
To find it draw a reflection of the grandma's house across the river and connect with a straight line Your house and the Reflection of Grandma's house. It surely is the shortest distance between these two points. 
But Grandma lives on this side of the river! OK, just reflect the interval of the line from the river to Grandma's house.

9 comments:

Jerome said...

Introduction and Wild Comments.
========================
Nice to see this problem floating around. I thought the only way to prove this was with calculus. Not so. Before I start, this problem is very important. In my mind it is one of God's greatest miracles. It represents the shortest distance a reflected beam of light can travel from a source to a mirror and then to an observer. I call it a miracle because it almost seems like the light beam has intelligence. It seems to know how to perform this shortest distance problem. For more wonders like this one, I recommend The Dancing Wu Li Masters by Gary Zukov -- a classic of it's kind.

But I digress.

I could not scan the diagram so it would show up, so I'll just have to describe it. Do it using the following steps.

Diagram
======
1. Draw a straight horizontal line. Label the ends A and B. A is on the left, B is on the right.
2. About one inch to the right of A and about 1.5 inches down put a point. Label it M for my place.
3. About one inch to the left of B and 3 inches down put another point. label it G for Grandmama's place.
4. Measure the perpendicular distance from the River to M. Label the River point as R1. Measure the same distance from R1 to M' where M R1 and M' are on the same straight line. MR1 = M'R1.
5. From M' draw a line to G.
6. Where M'G crosses AB label this point R3.
7. The minimum distance traveled is M R3 G.

Proof (Without Calculus).
==================
*MR1 = M'R1 by construction.
*MR1R3 = 90o Also by construction MR3 was made to be a right angle.
*M'R1R3 = 90o This is also by construction, but if you want to get really technical it is because two equal supplementary angles are 90o each.
*R1 R3 = R1 R3. A line can be equal to itself.

*Triangle M' R1 R3 = Triangle M R1 R3 Side angle Side.
*M'R3 = MR3 Corresponding Sides of Congruent triangles are congruent.

What this proof means.
=================
*A distance is a minimum when all point are on the same straight line. The minimum distance of M' to R3 and R3 to G is the straight line they are on. Since all three are co-linear, the distance must be a minimum.

*Since M'R3 = MR3 the minimum distance from my place to Grandmama's while touching the river must be M R3 G.

Note
===
You could have done the same thing from Grandmama's house. R3 is on AB no matter which way you do it.

Baskar Amaladosan said...

Project the perpendicular distance between the nearer river edge & grandma's house to the other side. Join the end point and my house using a straight line. The point of intersection of this line with the river edge is the point where I have to collect water and then walk towards grandma's house. Mail also sent....!!!!!

Anonymous said...

Hi! The yellow route is the shortest route because it has less dots than the red route. I think. I'm not sure because I didn't print it out and measure it. Please let me know. Thanks. Love the math puzzles btw. -AnonymousH

Tom said...

I'd use that map.
Assume North is up on the map.
I'd draw a duplicate location for Grandma's house, straight "North" of the river, the same distance from the river.

Then plot a straight line from "My house" to the duplicate Grandma's, diagonally across the river.

I believe the shortest path will be to begin on that plotted line, to the river, and then to Grandma's. Could tweak this to be still better I suppose, as the line will be diagonal across the river; could split the difference on the near riverbank, to a point halfway east...

Thad said...

This problem is actually super-simple once you know the secret. Reflect grandma's house over the riverbank. Draw a line from your house to the reflection of grandma's house. The point where the line intersects the riverbank is the point we should walk to.

Maria said...

Now the secret is out - check out the excellent explanations above and the drawing I added to the puzzle. Puzzle point for Jerome, Baskar, Tom and Thad. This one was a beauty. Let's remember it and share.

Anonymous said...

I see from my inbox that there is a solution at hand to the Shortest Route to Grandma’s puzzle, and before I take a peek I want to write up some comments on my experience with the puzzle. This puzzle has been a most excellent math learning experience for me. In particular I am astonished at how science-like the process has been.

I spent most of the day last Sunday working on it, and have come back to it periodically during the week. In the past, during math courses I have taken, I’ve had the sense that any empirical method of coming up with an answer is only second-best. Knowing “the answer” or quickly figuring out the answer about “how to derive the answer” has always seemed to be the primary goal in math. This time I set a challenge for myself that I would not look up any formulas online, but rather use only what I could come up with or remember for myself.

In the process of charting a 1,001 paths with my water bucket I became reacquainted with a compass, measured rectangular objects in my home and their diagonals, tied string loops and traced ellipses around my and my family members’ fingers, figured sine and cosine values for a few angles and plotted them on a graph, remembered and used geometric formulas, made up charts of squares and square roots, employed corners of envelopes, rulers and poster tack, dreamed up the geometry class project I wish I’d done in middle school, launched about a half dozen good guesses and tested them each using known answers for given configurations, and rejected them all.

The excitement of discovery, the thrill of original creation and sense of mastery and connection with other mathematicians throughout the ages, have been my constant companions. You might laugh, but the puzzle and its features even took on metaphorical dimensions in my life, waking me up in the morning to whisper quiet advice in my ear on relationships, attitude, and paths toward promising solutions to important life challenges.

I think I could work on the puzzle fruitfully for another year without exhausting its possibilities! Truly the value has been in the journey for me. Thank you!!!

Margaret (of Margaret & Fiona, though Fiona is on a road trip with her grandparents)

Anonymous said...

P.S. Here is what I think I know about this scenario: The problem is complex. It may be seen as an infinite SET of possible configurations with its primary Variables comprising the shortest distance from each point (I called the two houses points M and G) to the straight line (river), the distance between M and G and, implicitly, the angle that line MG forms with the river.

There is no reason to believe that the answer set has disjunctions or big leaps in it. That is, if you change the location of either house by increments, the answer A ought to change incrementally as well.

There are at least three known answers for three specific configurations, and any solution must yield these answers: 1) when G is on the river then the answer point which I call A will be the same as G, similarly when M is on the river then A = M; 2) If M = G (i.e. when the two houses are the same, that is, I live with Grandma) then A is the point for which line AM is perpendicular to the river; 3) if M and G are equidistant from the river, making line MG parallel to the river, then A is the point for which AW is perpendicular to both river and MG, W being the midpoint of line segment MG.

If you take a knotted string and rotate it around two points you will see that the set of points described by the string’s maximum reach (each of which may be seen as a summation of distances between each of the two fixed points and the moving peripheral point, which sum stays constant in the case of a knotted string) is elliptical. Therefore, wherever you place a straight line (the river) in relation to these two points, the minimal composite distance as requested by the puzzle makers is likely to be found by drawing the smallest possible ellipse, centered on those points, with the river as a tangent. Answer point A will be the place where the tangent touches the ellipse.

The tangent-to-an-ellipse solution satisfies the three known configurations and their required answers.* It also yields a smoothly changing answer set, as surmised must be the case, and one that “behaves” or moves in the proper directions between known answer points.

This is my best guess yet; I just don’t know how to draw my solution using “one-two simple lines” as stated in the instructions. All my other guesses involving straight lines only I have proven false to my satisfaction. There might be a formula for ellipses out there that would help, but I haven’t tried looking it up yet!

* 1) If point M or G is on the river, the ellipse collapses to a simple line segment and the river may be at any angle to line MG intersecting at either point. 2) If river is parallel to line MG then the tangent touches the broad side of the ellipse, at point A of the perpendicular line AW running between midpoint W and river. 3) If M = G then the ellipse turns into a circle with M and G at the center and a radius that is perpendicular to the tangent/river; again A is where the tangent touches.

Mathodically yours,

Margaret (of Margaret & Fiona) :)

Anonymous said...

To be clear, that was a knotted loop of string I was talking about! :) You subtract the length of MG and the remaining length of string stays constant, so the smallest possible loop (and the ellipse it describes) will give you the shortest distance to a point and back.

Fascinated by the explanation of why the reflect-G-across-river-and-find-congruent-triangle solution should be true….Intuitively it seems quite plausible but I am not yet entirely convinced by the logic. You actually could take both houses and reflect them across, and draw both long diagonals which are also equivalent. But why would the shortest line crossing the river to a reflection necessarily also be the shortest line bounced off the river to the actual house?

Margaret

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