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.”
See Triangle Acute-Angle Problem
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 really was trying to stop including Catriona Shearer’s problems, since they are probably all well-known and popular by now. But this is another virtually one-step-solution problem that again seems impossible at first. Many of her problems entail more steps, but I am especially intrigued by the one-step problems.
“What’s the sum of the two marked angles?”
See Kissing Angles.
Since everyone by now who has any interest has gone directly to Catriona Shearer’s Twitter account for geometric puzzles, I was not going to include any more. But this one with its one-step solution is too fine to ignore and belongs with the “5 Problem” as one of the most elegant.
“Two squares sit on the hypotenuse of a right-angled triangle. What’s the angle?”
See the Two Block Incline Puzzle
(Update 4/26/2019) Continue reading
I came across this problem in Alfred Posamentier’s book, but I remember I had seen it a couple of places before and had never thought to solve it. At first, it seems like magic.
In any convex quadrilateral (non-intersecting sides) inscribe a second convex quadrilateral with its vertices on the midpoints of the sides of the first quadrilateral. Show that the inscribed quadrilateral must be a parallelogram.
See the Magic Parallelogram.
Here is a problem from the famous (infamous?) Putnam exam, presented by Presh Talwalkar. Needless to say, I did not solve it in 30 minutes—but at least I solved it (after making a blizzard of arithmetic and trigonometric errors).
“Today’s problem is from the 1978 test, problem B1 (the easiest of the second set of problems). A convex octagon inscribed in a circle has four consecutive sides of length 3 and four consecutive sides of length 2. Find the area of the octagon.”
My solution is horribly pedestrian and fraught with numerous chances for arithmetic mistakes to derail it, which happened in spades. As I suspected, there was an elegant, “easy” solution (as demonstrated by Talwalkar)—once you thought of it! Again, this is like a Coffin Problem. See the Putnam Octagon Problem.
I was astonished that this problem was suitable for 8th graders. First of all the formula for the volume of a cone is one of the least-remembered of formulas, and I certainly never remember it. So my only viable approach was calculus, which is probably not a suitable solution for an 8th grader.
Presh Talwalkar: “This was sent to me as a competition problem for 8th graders, so it would be a challenge problem for students aged 12 to 13. When a conical bottle rests on its flat base, the water in the bottle is 8 cm from its vertex. When the same conical bottle is turned upside down, the water level is 2 cm from its base. What is the height of the bottle? (Note “conical” refers to a right circular cone as is common usage.) I at first thought this problem was impossible. But it actually can be solved. Give it a try and then watch the video for a solution.”
See the Conical Bottle Problem.
This is an old problem I had seen before. Here is David Wells’s rendition:
“Johannes Müller, named Regiomontanus after the Latin translation of Körnigsberg, his city of birth, later made famous by Euler, proposed this problem in 1471. … it is usually put in this form …: From what distance will a statue on a plinth appear largest to the eye [of a mouse!]? If we approach too close, the statue appears foreshortened, but from a distance it is simply small.”
I have added height numbers in feet for concreteness (as well as the mouse qualification, since the angles are measured from ground level). So the problem is to find the distance x such that the angle is maximal. See the Regiomontanus 1471 Problem
Catriona Shearer has come up with another challenging but elegant geometric problem. In some ways, it is similar to the famous Russian Coffin Problems that have an obvious solution—once you see it—but initially seem impenetrable. I really marvel at Catriona Shearer’s ability to come up with these problems.
“What’s the area of the parallelogram?”
See the Parallelogram Problem
The following interesting behavior was found at the Futility Closet website:
“A pleasing fact from David Wells’ Archimedes Mathematics Education Newsletter: Draw two parallel lines. Fix a point A on one line and move a second point B along the other line. If an equilateral triangle is constructed with these two points as two of its vertices, then as the second point moves, the third vertex C of the triangle will trace out a straight line. Thanks to reader Matthew Scroggs for the tip and the GIF.”
This is rather amazing and cries out for a proof. It also raises the question of how anyone noticed this behavior in the first place. I proved the result with calculus, but I wonder if there is a slicker way that makes it more obvious. See the Straight and Narrow Problem.
(Update 3/25/2019) Continue reading