Probably the HARDEST Geometry problem

                "Consider $n$ angles around a point, any two of which have distinct 

                 measures. Expressed in degrees, the measures of the $n$ angles are, 

                 any two, coprime natural numbers. Find the maximum value of $n$."

         Problem appeared in GMB 11/2024, page 648, proposed for 6th grade(E:17038), author Traian PREDA from Bucharest.

I don't have the ANSWER

I found examples with  ${\color{Green}{n=15}}$ 

and I would be happy if this was the maximum required.

          A few heuristic estimates will give us a first picture of the phenomenon.

         1) Because the angles around a point add up to $360^{\circ}$ and

$$1+2+3+\dots+26=351<360<378=1+2+3+\dots+26+27$$

we have the condition

$$3\leqslant n \leqslant 26 \tag{1}$$

          2) At most one of the angles has an even number of degrees as its measure.

          3) If we want as many angles as possible, then their sizes should be as small as possible.

So let's try prime numbers.

$1^{\circ}+2^{\circ}+3^{\circ}+5^{\circ}+7^{\circ}+11^{\circ}+13^{\circ}+17^{\circ}+19^{\circ}+23^{\circ}+29^{\circ}+31^{\circ}+37^{\circ}+41^{\circ}+$

$$+43^{\circ}+47^{\circ}+53^{\circ}=382^{\circ}$$

The result is $22^{\circ}$  more than it should be. We can eliminate $3^{\circ}+19^{\circ}$ or $5^{\circ}+17^{\circ}$ from the above sum and obtain

$$1^{\circ}+2^{\circ}+5^{\circ}+7^{\circ}+11^{\circ}+13^{\circ}+17^{\circ}+23^{\circ}+29^{\circ}+31^{\circ}+37^{\circ}+41^{\circ}+$$

$$+43^{\circ}+47^{\circ}+53^{\circ}=360^{\circ}$$

or

$$1^{\circ}+2^{\circ}+3^{\circ}+7^{\circ}+11^{\circ}+13^{\circ}+19^{\circ}+23^{\circ}+29^{\circ}+31^{\circ}+37^{\circ}+41^{\circ}+$$
$$+43^{\circ}+47^{\circ}+53^{\circ}=360^{\circ}.$$

Both writings have $n=15$ terms.

        By mistake, I wrote the terms (forgetting 47) but intentionally leaving out the even number 2

$$1^{\circ}+3^{\circ}+5^{\circ}+7^{\circ}+11^{\circ}+13^{\circ}+17^{\circ}+19^{\circ}+23^{\circ}+29^{\circ}+31^{\circ}+37^{\circ}+$$

$$+41^{\circ}+43^{\circ}+53^{\circ}=333^{\circ}$$

We would still have to add  $27^{\circ}$ to $360^{\circ}$. But then we would have to eliminate $3^{\circ}$ and instead of  $1^{\circ}+3^{\circ}$ we put  $4^{\circ}$. Therefore

$$4^{\circ}+5^{\circ}+7^{\circ}+11^{\circ}+13^{\circ}+17^{\circ}+19^{\circ}+23^{\circ}+$$

$$+27^{\circ}+29^{\circ}+31^{\circ}+37^{\circ}+41^{\circ}+43^{\circ}+53^{\circ}=360^{\circ}$$

So again a sum of  $n=15$  terms.

           As if with even more magic, I obtained another option. I see that

$$1^{\circ}+2^{\circ}+3^{\circ}+5^{\circ}+7^{\circ}+11^{\circ}+13^{\circ}+17^{\circ}+19^{\circ}+23^{\circ}+29^{\circ}+$$

$$+31^{\circ}+37^{\circ}+41^{\circ}+43^{\circ}+47^{\circ}=329^{\circ}$$

We would still need to add  $31^{\circ}$  to  $360^{\circ}$. We have the term $31^{\circ}$, but since $31+31=62$, close to 64 (!!) I will eliminate the term $31^{\circ}$ and replace it with $64^{\circ}$. We obtained a result with 2 more 360, so we will also eliminate the term $2^{\circ}$. Therefore

$$1^{\circ}+3^{\circ}+5^{\circ}+7^{\circ}+11^{\circ}+13^{\circ}+17^{\circ}+19^{\circ}+23^{\circ}+29^{\circ}+37^{\circ}+$$

$$+41^{\circ}+43^{\circ}+47^{\circ}+64^{\circ}=360^{\circ}$$

again a sum of  $n=15$  terms.

<last edit Ian, 23, 2025>

<will it continue ???>