This seemingly magical result from Futility Closet defies proof at first. Go to the Wolfram demo by Jay Warendorff and then …
“Grab point B above and drag it to a new location. Surprisingly, M, the midpoint of RS, doesn’t move.
This works for any triangle — draw squares on two of its sides, note their common vertex, and draw a line that connects the vertices of the respective squares that lie opposite that point. Now changing the location of the common vertex does not change the location of the midpoint of the line.
It was discovered by Dutch mathematician Oene Bottema.”
As we shall see, Bottema’s Theorem has shown up in other guises as well.
See Bottema’s Theorem
This is a nice puzzle from Alex Bellos’s Monday Puzzle column in the Guardian.
“My cultural highlight of recent weeks has been the brilliant BBC documentary Gods of Snooker, about the time in the 1980s when the sport was a national obsession. Today’s puzzle describes a shot to malfunction the Romford Robot … and put the Whirlwind … in a spin.
A square snooker table has three corner pockets, as [shown]. A ball is placed at the remaining corner (bottom left). Show that there is no way you can hit the ball so that it returns to its starting position.
The arrows represent one possible shot and how it would rebound around the table.
The table is a mathematical one, which means friction, damping, spin and napping do not exist. In other words, when the ball is hit, it moves in a straight line. The ball changes direction when it bounces off a cushion, with the outgoing angle equal to the incoming angle. The ball and the pockets are infinitely small (i.e. are points), and the ball does not lose momentum, so that its path can include any number of cushion bounces.
Thanks to Dr Pierre Chardaire, associate professor of computing science at the University of East Anglia, who devised today’s puzzle.”
See the Snooker Puzzle
This surprising, but simple, puzzle is from the 12 April MathsMonday offering by MEI, an independent curriculum development body for mathematics education in the UK.
“In the diagram various regular polygons, P, have been drawn whose sides are tangents to a circle, C. Show that for any regular polygon drawn in this way:”
(Given that the polygons approximate the circle in the limit, it would not be surprising that this relationship would hold—in the limit. It is surprising that it should be true for every regular polygon that circumscribes the circle.)
See the Area vs. Perimeter Puzzle
This is a simple problem from Five Hundred Mathematical Challenges:
“Problem 24. Let P be the center of the square constructed on the hypotenuse AC of the right-angled triangle ABC. Prove that BP bisects angle ABC.”
See the Center of Square Problem
Here is yet another collection of beautiful geometric problems from Catriona Agg (née Shearer). For some reason I found these a bit more challenging than the previous ones. Some of them required more time to “see” the breakthrough.
See Geometric Puzzle Magnificence
This is another Brainteaser from the Quantum math magazine .
“How can a polygonal line BDEFG be drawn in a triangle ABC so that the five triangles obtained have the same area?”
I found this problem rather challenging, especially when I first tried to solve it analytically (using hyperbolas). Eventually I arrived at a procedure that would accomplish the result. The Quantum “solution,” however, was tantamount to just saying divide the triangle into triangles of equal area—without providing a method! That is, no solution at all.
See the Equitable Slice Problem
These two interesting problems were posed on MEI’s MathsMonday site on 3 February 2020 and 2 March 2020, respectively. MEI and readers posted various approaches, but I used a method suggested by another problem whose origin I no longer recall.
See the Nested Polygons Puzzle
Here is a problem from the Quantum magazine, only this time from the “Challenges” section (these are expected to be a bit more difficult than the Brainteasers).
“Three circles with the same radius r all pass through a point H. Prove that the circle passing through the points where the pairs of circles intersect (that is, points A, B, and C) also has the same radius r.”
Indeed, I found this quite challenging. It took me several weeks to work out my approach and details.
See Three Equal Circles
This is an interesting problem from the 1977 Canadian Math Society’s magazine, Crux Mathematicorum.
“206. [1977: 10] Proposed by Dan Pedoe, University of Minnesota.
A circle intersects the sides BC, CA and AB of a triangle ABC in the pairs of points X, X’, Y, Y’ and Z, Z’ respectively. If the perpendiculars at X, Y and Z to the respective sides BC, CA and AB are concurrent at a point P, prove that the respective perpendiculars at X’, Y’ and Z’ to the sides BC, CA and AB are concurrent at a point P’.”
See the Twin Intersection Puzzle
Puzzles and Problems: plane geometry, Dan Pedoe, Crux Mathematicorum
Here is yet another problem from Presh Talwalkar. This one is rather elegant in its simplicity of statement and answer.
“Solve For The Angle – Viral Puzzle
I thank Barry and also Akshay Dhivare from India for suggesting this problem! This puzzle is popular on social media. What is the measure of the angle denoted by a “?” in the following diagram? You have to solve it using elementary geometry (no trigonometry or other methods). It’s harder than it looks. I admit I did not solve it. Can you figure it out?”
See the Shy Angle Problem.