Here is another UKMT Senior Challenge problem from 2017, which has a straight-forward solution:
“The diagram shows a circle of radius 1 touching three sides of a 2 x 4 rectangle. A diagonal of the rectangle intersects the circle at P and Q, as shown.
What is the length of the chord PQ?
__A_√5____B_4/√5____C_√5 – 2/√5____D_5√5/6____E_2”
See the Circle in Slot Problem
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.
This is a problem from a while back (2015) at Futility Closet.
“Which part of this square has the greater area, the black part or the gray part?”
See Modern Art
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.