# Simple proof of "maximum number of right angles in a convex $n$-gon is 3 for $n\geq 5$" for a 8th grade student?

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?

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 $$90^{\circ}$$.
• 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. Nov 22, 2020 at 18:38
• 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. Nov 22, 2020 at 19:47