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
This is a somewhat elegant problem from the 1987 Discover magazine’s Brain Bogglers by Michael Stueben:
“Each dot in the figure at left represents a factory. On which of the city’s 63 intersections should a warehouse be built to make the total distance between it and all the factors as short as possible? (A much simpler solution than counting and totaling the distances is available.)”
Note that the distance is the taxicab distance I discussed in my article South Dakota Travel Problem rather than the distance along straight lines between the warehouse and factories.
See the Factory Location Problem
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
Here is another Brain Bogglers problem from 1987.
“Exactly four minutes after starting to run—when the take-up reel was rotating one and a half times as fast as the projecting reel—the film broke. (The hub diameter of the smaller take-up reel is 8 cm and the hub diameter of the projecting reel is 12 cm.) How many minutes of film remain to be shown?”
This feels like another problem where there is insufficient information to solve it, and that makes it fun and challenging. In fact, I was stumped for a while until I noticed something that was the key to completing the solution.
See the Movie Projector 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
Presh Talwalkar had another interesting problem.
“A triangle is drawn inside a square with sides 4, 3, and 5, as shown. What is the length of the square’s side?”
The problem looks simple at first, but it takes some care to avoid some hideous quartic equations.
See Tipsy 3-4-5 Triangle
This problem from the 1987 Discover magazine’s Brain Bogglers by Michael Stueben apparently traces back to 1770, though the exact reference is not given.
“Here’s an arithmetic problem taken from a textbook published in Germany in 1770. Three people are gambling. In the first game, Player A loses to each of the others as much money as each of them had when the game started. In the next game, B loses to each of the others as much money as each had when that game began. In the third game, A and B each win from C as much money as each had at the start of that game. The players now find that each has the same sum, 24 guineas. How much money did each have when play began?”
See the 1770 Card Game 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 Doppelgänger 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.