I will post answers to all six next week. Have a go at them!
What is this well-known book that is suggested by the images?
Here is the first Mumble of the series, titles of well-known books. I will post the answers separately, later on.
Mumbles are a restricted form of rebus (see Wikipedia entry on Rebus), where a set of visual images evoke a sequence of syllables that evoke the answer, especially when mumbled.
As an example of a mumble, a graphic with eight puppies can be read as “pup 8” i.e. puppet.
The second part of a mumble is the key: A phrase that should not allow the reader to guess the answer, but once the answer is found, the key confirms that the answer is indeed correct. For the “puppet” example above, a possible key could be: “master = controller”
Mumbles do not allow “negative” elements, where some graphics are to be subtracted. For example, “arrow” – “rho” can be read as “are” — but this is not allowed in Mumbles.
I had created many Mumbles during 2009-2014 in my other blog, for example, see ‘Famous People Mumbles’ but had turned to other activities since then. Now, we are back, and here is a new series after that gap.
The next few posts present some new puzzles based on the theme of titles of well-known books.
Note: Google search has become very smart such that it is hard to provide a key that will not give away the answer to a simple search query. So full marks for solving only if no internet search was used!
There is an interesting set of related coloring puzzles (mathematical problems) which require mostly logical thinking rather than geometrical acumen to solve.
Answers are given in hidden form like this: [Hidden answer here]; select the hidden text to make it visible.
Each point on a straight line is colored either red or blue. Prove that there exist three points of the same color such that one point is the midpoint of the other two.
[Proof: Find two distinct points A and B of the same color, say red. Consider their midpoint C. If C is red, we are done. Else consider a point D such that A is the midpoint of DB. If D is red, we are done. Else consider a point E such that B is the midpoint of EA. If E is red, we are done. Else D, E and C are all blue. But C is the midpoint of DE, so we are done.]
Each point on a circle is colored either red or blue. Prove that there exist three equally spaced points of the same color on the circle.
[Proof 1: Essentially the previous proof works, if we “zoom in” so that the curve being closed does not matter, and definition of “midpoint” is modified to “along the curve”.
Proof 2: Consider a regular pentagon inscribed in the circle. Three of its vertices have to be of same color. In each case, the three vertices form an isosceles triangle, hence proved.]
Each point of the plane is colored either red or blue. Prove that for a given distance d, there exist two points of same color at distance d.
[Proof 1: Consider a point A whose color is, say, red. Now consider the points on the circle of radius d and center at A. If there is even one red point on the circle, we are done. Else all points on the circle are blue. But then, consider one blue point B on this circle, and consider a second circle with center at B and radius d. Let one intersection of the two circles be C. Now consider the pair B, C. Both are blue and at a separation d, hence proved.
Proof 2: Consider an equilateral triangle with sides d on the plane. At least two points must be of the same color, hence proved.]
Each point of the plane is colored either red or blue. Prove that there exists an equilateral triangle with all three vertices of the same color (“monochromatic triangle”).
[Proof: Start with three equidistant points A B C of same color (say red) on a line (see first problem). Construct two equilateral triangles ADB and BEC on the same side of the line. Next, construct a third equilateral triangle DFE on top of the two triangles. Notice that these three triangles combine to form a bigger equilateral triangle AFC. Now consider the point D: if it is red, we are done since ADB is monochromatic. Else consider point E: if it is red, we are done since BEC is monochromatic. Else, both D and E are blue. But now consider point F. If it is blue, DFE is monochromatic, otherwise it is red but then AFC is monochromatic!]
Each point of the plane is colored either red or blue. Prove that there exists a rectangle with all four vertices of the same color (“monochromatic rectangle”).
[Proof: Draw three horizontal lines and nine vertical lines crossing the horizontal lines. Consider the three points of intersection of one vertical line with the three horizontal lines: they can only be colored in eight different ways. But there are nine such vertical lines, so two of these vertical lines must have the same pattern of coloring. Now of these two vertical lines, there must be two intersection points of the same color at the same position on each vertical line. These four points form the monochromatic rectangle.]
Is it possible to color a plane in three colors such that every straight line in the plane contains only two colors, that is, is “bi-chromatic”? Note: all three colors must be present on the plane.
[Answer: Take a point A and color it red. Color every line passing through A either blue or green except for point A that remains red. Now any line not passing through A is bi-chromatic, as is any line passing through A.]
Consider a plane with n arbitrarily placed straight lines. The lines divide the plane into several closed/open polygonal regions. Is it possible to always color each region either red or blue such that no adjacent regions (that share a side) have the same color?
[Answer: We add lines one by one and use induction. With one line, there are two regions, call them semi-planes, color them with different colors. Now add a second line; this generates two semi-planes on either side of the second line. Flip the colors on one of the new semi-planes, leaving the other semi-plane unchanged. This preserves the coloring property: it is automatically preserved in regions that remain entirely within each semi-plane, and for regions that were divided by the new line, the two sub divisions were earlier the same color but due to color flipping on one side, now they are of different colors. By induction, the result is established for n lines.]
Cut-the-knot has lots of fascinating stuff. I found the above puzzles on this site.
Here is an interesting puzzle: given a line segment AB and a line parallel to it, find points on the segment that divide it into six equal parts. However, you are allowed to perform only two actions:
We start with two marked points, A and B. Also remember that after a line is drawn by action #2, one can mark points anywhere on that line. The figure below shows a segment AB and a line parallel to it (green). It also shows an arbitrary point C being chosen on the parallel line and then line CA is constructed (blue). Line CA actually extends to infinity in both directions but, to reduce clutter, it shown as a ray.
The puzzle looks difficult because no measurements, compass, or angle measurements are available. However a bit of exploration with an amazing application called Zirkel (C.A.R.), an open-source compass-and-ruler construction program, can leads one to a solution. By the way – “Zirkel und Lineal”in German == “Compass and Ruler” in English.
Of course, proving the construction is correct will take more effort where Zirkel probably will not be of help!
[Note: I know of two different solutions. One is brute force, the other is a bit more elegant.]
Puzzles can sometimes be solved elegantly with just the right insight. Here are some lateral thinking puzzles to hone your out-of-box thinking.
Answers are hidden between square braces; select the text to make it visible, like: [ this]