*This instalment is a Case Study concerning an unresolved elementary geometric problem which has recently come to light. It is another example of neglect of reasoning in maths. The problem was printed on the cover of the July 2020 issue of the *Mathematical Gazette*. It showed a triangle ***/\**** ABC residing within its circumcircle, and three special angles marked **θ

The backstory of this problem printed in the *Gazette* is that it was devised by Richard Beetham, an outstanding older UK maths teacher and problem-solver. He showed it to his lifelong friend Fred Hoyle, who in turn showed it to Prof. Jayant Narlikar a fellow cosmologist. The author of the recent article, Narlikar, describes how he had acquired two roundabout solutions from Beetham himself, one which used trigonometry and another —printed in the July 2020 *Gazette*—- which used ratios and Ceva’s theorem. These were comparatively efficient ways to get the answer… (θ = 50) but there was an unnoticed elephant in the room: *why was there no classic solution based on simple angle reasoning*?

We normally expect angle theorems to be provable using angle theorems.

But no one, it seemed, had succeeded in finding a way to solve the problem concerned using the classic angle theorems of Euclid! It was a problem which had been around, probably for fifty years, but which has remained unaccountably unsolved —even though it probably only required persistence in fielding the most elementary geometry.

It is, it turns out, peculiarly difficult to prove the result by this obvious, straightforward route. But many maths problems appear to be formidable on first acquaintance. It is the neglect of any sense of *urgency* and *attention* about the need to show that <<mathematical reasoning can resolve obscure logical situations painlessly>> which stands out most strikingly. It seems that no one among those who knew the challenge could be bothered to persevere with the inquiry.

The present author (Chris Ormell) has a particularly acute reason to be intrigued by the (for him, new) challenge, because he was (aged 16 at the time) taught by Richard Beetham in 1946 just after WW2. Beetham was by far the best of his early maths teachers, because he was young and enthusiastic: and he had a good repertoire of strategies for solving difficult mathematical puzzles. So, the present author has found himself suddenly confronted, in effect, with a challenge from beyond the grave! (Richard was aged about 31 in 1946. So he would be 106 if he were alive today.)

What has happened to the self-confidence of the mathematical community when numerous unsolved problems which stand out like sore thumbs, and have been around for decades, are left neglected, forgotten, and unexplored? We should, surely, be setting a better, more confident example to young mathematicians, especially as the world is now mathematicised up to its eyebrows, and is utterly dependent on the rigorous guardianship of those with effective mathematic skills.

The craven adoption of Zermelo-Fraenkel axioms in the 1920s unfortunately set a marker for defeatism in mathematics. It was a brazen attempt to try to make it *look as if* the contradiction implicit in Russell’s Paradox had been satisfactorily dealt-with, when it hadn’t. The Leadership of Mathematics, the heartland of all-but-certain truth, had stooped to fudging. It is hardly surprising that, as an unwitting consequence, maths went under a cloud of self-doubt and artificiality in the 1920s and 1930s… one which has remained to this day and has led to all sorts of missed opportunities.

Figure 1 shows the cover of the *Mathematical Gazette* for July 2020. The second diagram (Figure 2) shows the base angles *a, b, c* of the three isosceles sub-triangles which make up the triangle ABC. Clearly θ = *b+c* and *a*+*b*+*c*= 90. (*a, b, c* refer to the number of degrees of these angles.)

The angles of the triangle AEB are *b+c*, 2*b*+2*c* and *c*: so 4*c*+3*b*= 180. The latter result is illustrated in Figure 3, where the arcs BI and AP subtend 2*c* degrees at O and three equal arcs, IH, HG and GA subtend *b* degrees each at O. **Figure 1**

Figure 3 also shows the point A’ which is the mirror image of A in the line HC. K is the mid point of AE and the perpendicular bisector of AE, KD, also passes through A’. Given this *fact* —which is quite difficult to prove— it is easy to show that *a*= 2*b* and hence that *b*=20, *c*=30 and θ = 50. [The angles at D are 2*b*+*c*, *c*, 2*a*, 2*b*+*c*, 2*b*+*c* and *a*. Now the last three of these add up to 180, which tells us that *a*=2*b*, so, as *a*+*b*+*c* = 90, 3*b*+*c* = 90. Solving this with 4*c*+3*b* =180 gives the result.] **Figure 2**

It remains to prove that KD = passes through A’. This result can be obtained by considering the triangle /\AQF shown in grey in Figure 4. DK is extended to Q such that / AQD and /FQD are each *b* degrees. QA is extended to Q’. F is on the line AF such that the angle /Q’AF is 4*b* and /FEB is *b*/2 degrees. **Figure 3**

Now /Q’AF= /AQF + /AFQ. Hence 4*b* = 2*b* + /AFQ and /AFQ = 2*b* degrees. It follows that /Q”FA = 2*a*+2*c*. Also /CA’A is *a*+*c* degrees and /A’AF = 2*b* degrees, so /A’FA = *a+c* degrees =/A’FQ”. We now have AA’ as the angle bisector of /Q’AF and FA’ as the angle bisector of /Q”FA, so A’ is the excentre of /\ AQF *shown grey* and lies on the angle bisector QD. Thus QD passes through A’.QED.

**CHRISTOPHER ORMELL April 2021 **