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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.

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?

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.)