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?”

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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?”

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.

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.

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.

Catriona Shearer retweeted the following problem from Antonio Rinaldi @rinaldi6109

“My little contribution to @Cshearer41 October 7, 2018

A point D is randomly chosen inside the equilateral triangle ABC. Determine the probability that the triangle ABD is acute-angled.”

Another challenging problem from Presh Talwalkar. I certainly could not have solved it on a timed test at the age of 16.

“**One Of The Hardest GCSE Test Questions – How To Solve The Cosine Problem**

Construct a hexagon from two congruent parallelograms as shown. Given BP = BQ = 10, solve for the cosine of PBQ in terms of x.

This comes from the 2017 GCSE exam, and it confused many people. I received many requests to solve this problem, and I thank Tom, Ben, and James for suggesting it to me.”

See the Parallelogram Cosine Problem

I came across the following problem from an Italian high school exam on the British Aperiodical website presented by Adam Atkinson:

“There have been various stories in the Italian press and discussion on a Physics teaching mailing list I’m accidentally on about a question in the maths exam for science high schools in Italy last week. The question asks students to confirm that a given formula is the shape of the surface needed for a comfortable ride on a bike with square wheels.

What do people think? Would this be a surprising question at A-level in the UK or in the final year of high school in the US or elsewhere?”

I had seen videos of riding a square-wheeled bicycle over a corrugated surface before, but I had never inquired about the nature of the surface. So I thought it would be a good time to see if I could prove the surface (cross-section) shown would do the job. See Square Wheels.