A convex polygon is one that has no reentrant corners, or to put it differently, has no interior angles greater than . Figure 1 below is convex; Figure 2 is not.
An acute angle is one that is less than a tight angle, that is, less than .
Prove that a convex polygon with sides can never have more than acute angles and that for any it is possible to construct a polygon with exactly acute angles.
For the proof, see the Comments.
At the start of this month I wrote about the following problem.
Inscribe an equilateral triangle in a circle. Draw a line through the midpoints and of two of its sides, to meet the circle at .
Show that is equal to , the number of the Golden Section.
This elegant result was proposed by George Odom as problem E3007 in the American Mathematical Monthly in 1983 and the best solution out of the 74 receive was published in 1986. Here is the miraculous solution given by Jan van de Craats. I have drawn my own diagram and provided slightly more explanation but not changed the underlying proof at all. Here is the diagram above but with some additional lines and labels.
Take the length of as and suppose . By symmetry . Because the small triangle is equilateral and is the midpoint of we have .
Because and are subtended on the circumference of the circle by the same arc the angles there are equal. Angles and are equal, because they are vertically opposite. Hence triangles and are similar.
The ratios of corresponding pairs of sides in these two triangle must therefore be equal, so we have
and rearranging gives or , the equation whose positive root is , the number of the Golden Section.
(Jan van de Craats' write-up was terser; he just gave the diagram above and underneath wrote
relying on the Intersecting Chords Theorem, but I thought it would be better to use similar triang;es and not assume knowledge of that theorem.)
Footnote: I quite liked my own proof but the one above is far, far nicer, which I suppose is the reason it got published. Mine was a worthy effort but really just an 'also ran', along with the 74 other ARs.
This is a solution to the problem I posted 16 May 2026.
In the diagram triangles ABC and CDE are equilateral, with points A, C and E lying on a straight line. The problem is to prove CP and CQ have the same length.
There are probably many proofs - for example using coordinate geometry or complex number - but here is a short one using Euclidean geometry.
In the second diagram the coloured triangles ACD and BCE are congruent ('two sides and the included angle'), because AC = BC, CD = CE, and angle ACD = 120° = angle BCE . The two angles marked are therefore equal.
In the third diagram the coloured triangle CPD and the shaded triangle CQE are congruent ('two angles and the included side'), because angle PCD = 60° = angle QCE, angle PDC = = angle QEC and side CD = side CE.
Consequently CP = CQ which was to be shown.
Here's a nice problem I found on math stack exchange (question 1182471).
ACE is a straight line and triangles ABC and CDE are equilateral. Prove that CP = CQ.
If you have a solution do put in the comments. I imagine there are a number of different solutions. I'll post mine this Monday coming.
I saw this problem on Mind Your Decisions.
At left is an equilateral triangle. From an arbitrary point in its interior we draw line segments to its vertices, making angles , and as shown. If now we construct a second triangle (right) whose sides are equal in length to these three line segments, as indicated by the tick marks—What will the angles of the new triangle be?
I have never seen this before and have not viewed the solution. But I have worked out what the answer must be, just not proved it yet. And it's a truly beautiful result.
I wonder if it can be generalised to non-equalateral triangle? Or to a square?
Sangaku were geometrical puzzles from the 18th, 19th and early 20th centuries, painted on wooded tablets and hung in Japanese temples. Here is a problem I came across which is either a Sangaku or inspired by that tradition. It is very simple to state.
Inside a circle another smaller circle is drawn which is tangent to the bigger circle and to a diameter of the bigger circle.
An even smaller circle is then drawn which is tangent to the diameter and to both the other circles, as shown in Figure 1.
Figure 1. The green circle is tangent to the red circle, the diameter and the enclosing blue circle.
I had a lot of fun solving this problem which turns out to have a really nice answer. I'll post my solution, which I am pretty comfident is correct, at the end of the week.
Given an equilateral triangle LMN, let X lie on ML extended, and Y lie on MN, such that LX = NY.
Show that the point P where XY and LN intersect is the midpoint of XY.
This is a Sangaku-like puzzle (see https://learn1.open.ac.uk/mod/oublog/viewpost.php?post=230691)
We have a square inscribed in a triangle of known base b and height h, as shown. What is the length s of the square's side?
I called this a Samurai puzzle because many of the original sangaku were the work of Samurai.
(Solution in Comments.)
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