This is an interesting problem from Posamentier and Lehmann’s Mathematical Curiosities.
“In the figure we have a semicircle with the point P randomly placed on the diameter. Points A and B are situated on the circle such that they form angles of 60° with the diameter as shown in the figure. This problem asks us to show that the length of AB is equal to the radius of the semicircle.”
See the Ubiquitous Radius Problem
This is another problem from MEI’s MathsMonday.
“Two equilateral triangles share a common vertex. Show that the lengths marked a and b are equal for any such arrangement.”
This seems quite amazing at first. One can picture the small triangle swinging back and forth with red bungee chords tethering its bottom vertices to the bottom vertices of the large triangle. It would seem remarkable that the lengths of the chords would remain equal to each other throughout.
See the Tethered Triangle Puzzle
Here is an intriguing problem from the 2021 Math Calendar.
“If the smaller circle of diameter 7 rotates without slipping within the larger circle, what is the length of the path of P?”
The problem did not state clearly how far the smaller circle should rotate. Its answer implied it should complete just one full (360°) rotation within the larger circle.
Recall that all the answers are integer days of the month.
See the Wandering Epicycle
(Update 1/3/2022) First, this problem is dealt with in more detail and more expansively on the Mathologer Youtube website by Burkard Polster in his 7 December 2018 post on the “Secrets of the Nothing Grinder” (Figure 1). A further, deeper discussion of epicycles is given in the Mathologer’s 6 July 2018 post on “Epicycles, complex Fourier and Homer Simpson’s orbit” (Figure 2). And finally, a panoply of related puzzles is given in the 30 December 2021 Mathologer post “The 3-4-7 miracle. Why is this one not super famous” (Figure 3).
This last post reveals the ambiguity of the idea of “one full (360°) rotation” I disingenuously added to the problem to try to get the answer given in Math Calendar version.
For a complete explanation see the Wandering Epicycle Addendum.
This is a nice geometric problem from the Scottish Mathematical Council (SMC) Senior Mathematical Challenge of 2008.
“Mahti has cut some regular pentagons out of card and is joining them together in a ring. How many pentagons will there be when the ring is complete?
She then decides to join the pentagons with squares which have the same edge length and wants to make a ring as before. Is it possible? If so, determine how many pentagons and squares make up the ring and if not, explain why.”
See the Polygon Rings
Here is another problem from the “Brainteasers” section of the Quantum magazine.
“Side AE of pentagon ABCDE equals its diagonal BD. All the other sides of this pentagon are equal to 1. What is the radius of the circle passing through points A, C, and E?”
See the Circumscribed House Problem
Here is yet another collection of beautiful, stimulating geometric problems from Catriona Agg (née Shearer).
This is a nice brain tickling problem from Presh Talwalkar.
“A circle contains two tangent semicircles whose diameters are parallel chords. If the circle has an area equal to 1, what is the combined area of the two semicircles?”
See the Two Curious Semicircles
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 is another simple problem from Five Hundred Mathematical Challenges:
“Problem 57. Let X be any point between B and C on the side BC of the convex quadrilateral ABCD (as in the Figure). A line is drawn through B parallel to AX and another line is drawn through C parallel to DX. These two lines intersect at P. Prove that the area of the triangle APD is equal to the area of the quadrilateral ABCD.”
See the Triangle Quadrangle Puzzle
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.