A Magic Trick with Geometry

I discovered this “magic trick” while painfully solving a related geometry puzzle.

First, some background. Given an arbitrary triangle ABC, draw squares on each side. Label the centers of the squares D, E, F. This forms another triangle DEF. Figure 1 indicates the basic construction (draw a perpendicular bisector half the length of the side ). ABC does not need to be a right angle triangle.


Figure 1: Constructing square centers

Now, it is fascinating to find that if squares are drawn on the second triangle DEF, but the squares are drawn inward rather than outward, the centers land on the mid points of the original triangle ABC. This is shown in Figure 2: JG is the inward perpendicular bisector of side EF, with JG = JE = JF. G is found to bisect side BC.


Figure 2: Outer squares -> Inner Squares -> midpoints

Now, let’s ask what would happen if we started with triangle ABC, but constructed the inner squares instead of outer squares, on it. Thus we have triangle DEF in figure 3 below, constructed on centers of inner squares.


Figure 3: Centers of Inner Squares on Triangle ABC

Now, it is natural to expect that the centers of outer squares drawn on DEF will fall on the midpoints of the original triangle ABC. That is, inner squares -> outer squares -> midpoints. Thus we have figure 4 below, and see that the square centers indeed fall on the midpoints of triangle ABC. outer-inner-squares-fig4

Figure 4: inner squares -> outer squares -> midpoints

So where is the magic trick, the sleight of hand? Well, examine Figure 4 again. Actually, we found the midpoints thus: inner squares -> inner squares -> midpoints.

And that was really amazing to discover, that the constructions break a natural symmetry:

outer squares -> inner squares -> midpoints

inner squares -> inner squares -> midpoints.

Note: the figures are drawn using a beautiful geometry construction software called C.A.R. (Compass and Ruler).

The puzzle

For the reader who wondered what the original puzzle was, it is a set of related puzzles:

(a) Construct outward squares on a triangle ABC of any shape. Let the centers of the squares be DEF. Now, given only DEF, reconstruct the original triangle ABC.

(b) Instead of squares, draw equilateral triangles on sides of ABC. Given the centers of the equilateral triangle, reconstruct the original triangle ABC.

(c) Generalizing further: On side BC with midpoint G, draw a perpendicular GD such that GD = m * GC, where m is a real number. This generates point D. Similarly draw points E and F using sides AC and AB respectively, using the same value of m.Given DEF, reconstruct the original triangle ABC. (setting m  = 1 gives puzzle (a), and so on.)

An alternative approach to draw points D E F is as follows: Draw two lines from the ends of each side at a constant angle to the side. The intersection of these two lines gives the points D E F. Then 45 degree angles gives case (a); 60 degree angles give case (b).


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Drive-in Movie Theater

A drive-in movie theater has a vertical screen whose top and bottom sides are at a height of h1 and h2 meters, respectively. Now if a viewer in a car is too close to the screen, it will look too foreshortened. On the other hand, if the viewer is too far away, the screen gets too small. Perhaps the best “sweet spot” is when the angle subtended by the screen at the viewer’s eyes is largest.

 drive-in-theater diagram

In the diagram, the viewer is at P and looking at the screen AB. Find the distance d such that the angle theta = angle APC is maximum.

This has a quickie solution; no need for calculus

(the finicky may measure the heights h1 and h2 from eye level instead of ground level)

Posted in Puzzle

A Quick Sum

Find the sum of the series involving factorials

1(1!) + 2(2!) + 3(3!) + … + n(n!)

[note: the notation r! means r(r-1)(r-2)…(1), just in case you didn’t know]


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Two Cylinders

The axes of symmetry of two right cylinders of equal diameters intersect at right angles. What volume do the cylinders have in common?

note: there is a quickie solution, no calculus required.

two cylinders

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Tangled Strings

Three tangled and interwoven strings are tied to three pegs on board 1.  It is required to tie three other strings to the three free ends (without moving the existing strings) and attach the free ends of the new strings (which may be tangled), to the the three pegs on board 2 in such a way that the resultant entanglement can be combed out to give three loosely parallel (untangled) strings.

How can this be done? Can it always be done, no matter how tangled and knotted the original strings were?


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The Gold Ring

A gold ring is made from a sphere of solid gold but with a cylindrical hole drilled through its center (That is, the cylindrical volume and the two spherical caps above and below the cylinder have been removed).


Now, the width of the ring (i.e. the height of the cylindrical hole) is given = b. What is the volume of the gold?

At first, it seems that there is not enough information. But with some calculation (with calculus, or by using volume formulas for sphere, spherical cap, and cylinder – see calculations) it is possible to show that the volume indeed only depends upon the value of b. I came across this puzzle many years ago. Although the calculation was not difficult, it was always a mystery as to why it should be so.

Then, I came across a fascinating explanation in the Mathematical Mechanic. Even more fascinating is how it is connected to the bicycle problem (riding-a-bicycle-contd). I won’t give away the solution here, but think of a rigid plane figure sweeping out a volume in 3D space. The swept out volume is not influenced by the sliding action of the plane figure.

Now all you have to do is to find the appropriate plane figure in the gold ring!

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Mathematical quickies #1

A gem of a book, “Mathematical Quickies” by Charles W Trigg, (1967), has several puzzles with very elegant “quickie” solutions. I had read it with pleasure many decades ago. I found some notes while cleaning out old papers recently and that triggered memories.
I plan to post some of those Quickies in this blog. Here is the first one.

A unit circle contains a set S of two million points placed at arbitrary fixed locations. Prove that there exists a straight line that partitions S in two equal halves. That is, there will be one million points on each side of the line (with no point from S falling on the line).

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