I have been subverted again by a recent post by Ben Orlin, “Geometry Puzzles for a Winter’s Day,” which is another collection of Catriona Shearer’s geometric puzzles, this time her favorites for the month of November 2019 (which Orlin seems to have named himself). I often visit Orlin’s blog, “Math with Bad Drawings”, so it is hard to kick my addiction to Shearer’s puzzles if he keeps presenting collections. Her production volume is amazing, especially as she is able to maintain the quality that makes her problems so special.
The Stained Glass puzzle generated some discussion about needed constraints to ensure a solution. Essentially, it was agreed to make explicit that the drawing had vertical and horizontal symmetry in the shapes, that is, flipping it horizontally or vertically kept the same shapes, though some of the colors might be swapped.
See Geometric Puzzle Madness
This is truly an amazing result from Five Hundred Mathematical Challenges.
“Problem 119. Two unequal regular hexagons ABCDEF and CGHJKL touch each other at C and are so situated that F, C, and J are collinear.
(i) the circumcircle of BCG bisects FJ (at O say);
(ii) ΔBOG is equilateral.”
I wonder how anyone ever discovered this.
See the Magic Hexagons
Here is another simple problem from Futility Closet.
“Draw an arbitrary triangle [ABC] and build an equilateral triangle on each of its sides, as shown. Now show that [straight lines] AP = BQ = CR.”
See Threewise Problem
I was really trying to avoid getting pulled into more addictive geometric challenges from Catriona Shearer (since they can consume your every waking moment), but a recent post by Ben Orlin, “The Tilted Twin (and other delights),” undermined my intent. As Orlin put it, “This is a countdown of her three favorite puzzles from October 2019” and they are vintage Shearer. You should check out Olin’s website since there are “Mild hints in the text; full spoilers in the comments.” He also has some interesting links to other people’s efforts. (Olin did leave out a crucial part of #1, however, which caused me to think the problem under-determined. Checking Catriona Shearer’s Twitter I found the correct statement, which I have used here.)
I have to admit, I personally found the difficulty of these puzzles a bit more challenging than before (unless I am getting rusty) and the difficulty in the order Olin listed. Again, the solutions (I found) are simple but mostly tricky to discover. I solved the problems before looking at Olin’s or others’ solutions.
See the Geometric Puzzle Mayhem.
This is another problem from the Math Challenges section of the 2000 Pi in the Sky Canadian math magazine for high school students.
“Problem 4. From a point P on the circumference of a circle, a distance PT of 10 meters is laid out along the tangent. The shortest distance from T to the circle is 5 meters. A straight line is drawn through T cutting the circle at X and Y. The length of TX is 15/2 meters.
(a) Determine the radius of the circle,
(b) Determine the length of XY.”
See the Circle Tangent Chord Problem
I found this problem from the Math Challenges section of the 2002 Pi in the Sky Canadian math magazine for high school students to be truly astonishing.
“Problem 4. Inside of the square ABCD, take any point P. Prove that the perpendiculars from A on BP, from B on CP, from C on DP, and from D on AP are concurrent (i.e. they meet at one point).”
How could such a complicated arrangement produce such an amazing result? I didn’t know where to begin to try to prove it. My wandering path to discovery produced one of my most satisfying “aha!” moments.
See the Mysterious Dopplegänger Problem
Update (12/27/2019) I goofed. I had plotted the original figure incorrectly. (No figure was given in the Pi in the Sky statement of the problem.) Fortunately, the original solution idea still worked.
Here is yet another surprising result from Colin Hughes at Maths Challenge.
It can be shown that a unique circle passes through three given points. In triangle ABC three points A’, B’, and C’ lie on the edges opposite A, B, and C respectively. Given that the circle AB’C’ intersects circle BA’C’ inside the triangle at point P, prove that circle CA’B’ will be concurrent with P.”
I have to admit it took me a while to arrive at the final version of my proof. My original approach had some complicated expressions using various angles, and then I realized I had not used one of my assumptions. Once I did, all the complications faded away and the result became clear.
See Circular Rendezvous Mystery.
This is from the UKMT Senior Challenge of 2004.
“L, M, and N are midpoints of a skeleton cube, as shown. What is the value of angle LMN?
See Cube Slice Angle Problem.
In my search for new problems I came across this one from Martin Gardner:
“A square formation of Army cadets, 50 feet on the side, is marching forward at a constant pace [see Figure]. The company mascot, a small terrier, starts at the center of the rear rank [position A in the illustration], trots forward in a straight line to the center of the front rank [position B], then trots back again in a straight line to the center of the rear. At the instant he returns to position A, the cadets have advanced exactly 50 feet. Assuming that the dog trots at a constant speed and loses no time in turning, how many feet does he travel?”
Gardner gives a follow-up problem that is virtually impossible:
“If you solve this problem, which calls for no more than a knowledge of elementary algebra, you may wish to tackle a much more difficult version proposed by the famous puzzlist Sam Loyd. Instead of moving forward and back through the marching cadets, the mascot trots with constant speed around the outside of the square, keeping as close as possible to the square at all times. (For the problem we assume that he trots along the perimeter of the square.) As before, the formation has marched 50 feet by the time the dog returns to point A. How long is the dog’s path?”
See the Marching Cadets and Dog Problem.
I was reading yet another book on the Scientific Revolution when I came across a discussion of the mathematical significance of the invention of perspective for painting in the 15th century Italian Renaissance. The main player in the saga was Leon Battista Alberti (1404 – 1472) and his tome De Pictura (On Painting) (1435-6), which contained the first mathematical presentation of perspective. Even though mathematics was advertised, it was not at the level of trigonometry I used in my post “The Perspective Map”, but rather entailed simple Euclidean plane geometry. So the discussion was largely historical rather than mathematical. Nevertheless, I became curious to learn how much Alberti was able to discover about perspective without a lot of math. This essay is the result.
See Alberti’s Perspective Construction
(Update 7/29/2019) I got a response! Continue reading