Tag Archives: MathsMonday

Pythagorean Parabola Puzzle

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Since the changes in Twitter (now X), I have not been able to see the posts, not being a subscriber.  But I noticed poking around that some twitter accounts were still viewable.  However, like some demented aging octogenarian they had lost track of time, that is, instead of being sorted with the most recent post first, they showed a random scattering of posts from different times.  So a current post could be right next to one several years ago.  That is what I discovered with the now defunct MathsMonday site.  I found a post from 10 May 2021 that I had not seen before, namely,

“The points A and B are on the curve y = x2 such that AOB is a right angle.  What points A and B will give the smallest possible area for the triangle AOB?”

See the Pythagorean Parabola Puzzle.

(Update 9/1/2023) Elegant Alternative Solution by Oscar Rojas
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Triangle Projection Problem

This is a Maths Item of the Month (MIOM) problem that seems opaque at first.  (“The Maths Item of the Month is a monthly problem aimed at teachers and students of GCSE and A level Mathematics.”)

“Two fixed circles, C1 and C2, intersect at A and B.  P is on C1.  PA and PB produced meet C2 at A’ and B’ respectively.  How does the length of the chord A’B’ change as P moves?”

Just start noticing relationships and the answer falls out nicely.

(MIOM problems often appear on MathsMonday and are also produced by Mathematics Education Innovation (MEI).)

See the Triangle Projection Problem

Another Christmas Tree Puzzle

This is a belated Christmas puzzle from December 2019 MathsMonday.

“A Christmas tree is made by stacking successively smaller cones. The largest cone has a base of radius 1 unit and a height of 2 units. Each smaller cone has a radius 3/4 of the previous cone and a height 3/4 of the previous cone. Its base overlaps the previous cone, sitting at a height 3/4 of the way up the previous cone.

What are the dimensions of the smallest cone, by volume, that will contain the whole tree for any number of cones?”

Recall that the volume of a cone is π r2 h/3.

See Another Christmas Tree Puzzle

Tethered Triangle Puzzle

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

Area vs. Perimeter Puzzle

This surprising, but simple, puzzle is from the 12 April MathsMonday offering by MEI, an independent curriculum development body for mathematics education in the UK.

“In the diagram various regular polygons, P, have been drawn whose sides are tangents to a circle, C.  Show that for any regular polygon drawn in this way:”

(Given that the polygons approximate the circle in the limit, it would not be surprising that this relationship would hold—in the limit.  It is surprising that it should be true for every regular polygon that circumscribes the circle.)

See the Area vs. Perimeter Puzzle

Diabolical Triangle 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