This is another imaginative puzzle from MEI’s MathsMonday. I didn’t know what topple blocks were at first, but a previous MathsMonday puzzle defined them as shown in the first frame. Apparently they derive from the game of Jenga created by Leslie Scott and launched in 1983.
See the Topple Blocks Puzzle
This simple-appearing problem is from the 17 August 2020 MathsMonday offering by MEI, an independent curriculum development body for mathematics education in the UK.
“The diagram shows an equilateral triangle in a rectangle. The two shapes share a corner and the other corners of the triangle lie on the edges of the rectangle. Prove that the area of the green triangle is equal to the sum of the areas of the blue and red triangles. What is the most elegant proof of this fact?”
Since the MEI twitter page seemed to be aimed at the high school level and the parting challenge seemed to indicate that there was one of those simple, revealing solutions to the problem, I spent several days trying to find one. I went down a number of rabbit holes and kept arriving at circular reasoning results that assumed what I wanted to prove. Visio revealed a number of fascinating relationships, but they all assumed the result and did not provide a proof. I finally found an approach that I thought was at least semi-elegant.
See the Diabolical Triangle Puzzle
(Update 1/30/2021) New MEI Solution
Here is another good problem from Five Hundred Mathematical Challenges:
“Problem 100. A hexagon inscribed in a circle has three consecutive sides of length a and three consecutive sides of length b. Determine the radius of the circle.”
This problem made me think of the Putnam Octagon Problem. Again my approach might be considered a bit pedestrian. 500 Math Challenges had a slightly slicker solution.
See the Lop-sided Hexagon Problem
This is another problem from the indefatigable Presh Talwalkar.
_ _____Hard Geometry Problem
“In triangle ABC above, angle A is bisected into two 60° angles. If AD = 100, and AB = 2(AC), what is the length of BC?”
See Hard Geometric Problem
(Update 7/18/2020, 7/20/2020) Alternative Solution Continue reading
Having fallen under the spell of Catriona Shearer’s geometric puzzles again, I thought I would present the latest group assembled by Ben Orlin, which he dubs “Felt Tip Geometry”, along with a bonus of two more recent ones that caught my fancy as being fine examples of Shearer’s laconic style. Orlin added his own names to the four he assembled and I added names to my two, again ordered from easier to harder.
See Geometric Puzzle Munificence.
(Update 4/16/2020) Ben Orlin has another set of Catriona Shearer puzzles 11 Geometry Puzzles That Drive Mathematicians to Madness which I will leave you to see and enjoy. But I wanted to emphasize some observations he included that I think are spot on. Continue reading
This is another fairly simple puzzle from Futility Closet.
“If an equilateral triangle is inscribed in a circle, then the distance from any point on the circle to the triangle’s farthest vertex is equal to the sum of its distances to the two nearer vertices (q = p + r).
(A corollary of Ptolemy’s theorem.)”
See A Tidy Theorem
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
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.
It is always fascinating to look at problems from the past. This one, given by Thomas Whiting himself, is over 200 years old from Whiting’s 1798 Mathematical, Geometrical, and Philosophical Delights:
“Question 2, by T. W. from Davison’s Repository.
There are two houses, one at the top of a lofty mountain, and the other at the bottom; they are both in the latitude of 45°, and the inhabitants of the summit of the mountain, are carried by the earth’s diurnal rotation, one mile an hour more than those at the foot.
Required the height of the mountain, supposing the earth a sphere, whose radius is 3982 miles.”
See the Mountain Houses Problem
This is a problem from the UKMT Senior Challenge for 2001. (It has been slightly edited to reflect the colors I added to the diagram.)
“The [arbitrary] blue triangle is drawn, and a square is drawn on each of its edges. The three green triangles are then formed by drawing their lines which join vertices of the squares and a square is now drawn on each of these three lines. The total area of the original three squares is A1, and the total area of the three new squares is A2. Given that A2 = k A1, then
_____A_ k = 1_____B_ k = 3/2_____C_ k = 2_____D_ k = 3_____E_ more information is needed.”
I solved this problem using a Polya principle to simplify the situation, but UKMT’s solution was direct (and more complicated). See the Six Squares Problem.