I know a proof of "maximum number of right angles in a convex $n$-polygon is 3 for $n\geq 5$" as follows:

Suppose $k$ is the number of right angles. Then $180(n-2)-90k$ is the sum of other $n-k$ interior angles. Now we can perform these $n-k$ angles such that all have equal angle i.e. equal to their average $\frac{180(n-2)-90k}{n-k}=\frac{180(n-k)+90k-360}{n-k}=180+\frac{90(k-4)}{n-k}$ that $\frac{90(k-4)}{n-k}$ must be negative since each angle $<180$ so $90(k-4)<0$ and $k<4$.

But this proof is a bit cumbersome for 8$^{th}$ grade school student. (The bolded part is also dubious to me and hard to accept it! Let alone the 8$^{th}$ grade school student). Is there any more simple argument?


1 Answer 1


Sum of exterior angles of a convex n-gon is $360^{\circ} = 4\cdot 90^{\circ}$. We conclude for $n > 4$, at most $3$ right angles are allowed. In that case, sum of rest $n-3$ exterior angles is $90^{\circ}$.

  • 2
    $\begingroup$ For some reason I thought that 'exterior angle' was just $360^{\circ}$ minus the internal angle, not the $180^\circ$ complement. Yay for standard terminology. $\endgroup$ Nov 22, 2020 at 18:38
  • $\begingroup$ Yes. I was thinking the same until I was reminded of past usage. :) $\endgroup$
    – cosmo5
    Nov 22, 2020 at 18:39
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    $\begingroup$ very simple. Thanks. Now I can explain this to my student even to my grandpa! $\endgroup$
    – C.F.G
    Nov 22, 2020 at 18:53
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    $\begingroup$ One tiny note I would add to this, BTW, is that convexity is exactly the condition that all exterior angles are positive, so if we do exceed $360^\circ$ then there's no way of getting back without a concavity. $\endgroup$ Nov 22, 2020 at 19:47
  • $\begingroup$ @StevenStadnicki: We should call it "turning angle" instead of "exterior angle"! $\endgroup$
    – user21820
    Jan 30, 2021 at 3:45

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