Hey there! Can someone help me out with this question -

What is the minimum number of acute exterior angles in a convex octagon?

(Note: a convex polygon is one in which none of the interior angles are greater than 180∘.)

Hey there! Can someone help me out with this question -

What is the minimum number of acute exterior angles in a convex octagon?

(Note: a convex polygon is one in which none of the interior angles are greater than 180∘.)

is the answer 4?

The sum of interior angles of an octagon is (8-2)x180° = 6x180° = 1080°

If we consider the ideal case of a regular octagon, then all interior angles will be 135°. And all exterior angles (extending one of the sides at each vertex) will be 45°. Notice that the internal and external angles add up to 180°. So we have all 8 exterior angles as acute angles.

Now consider, to reduce the number of acute exterior angles, we have to increase the angle measure from 45° to (at least) 90°. So if we increase one exterior angle from 45° to 90°, its corresponding interior angle also decreases from 135° to 90°. So to maintain that the sum of the interior angles of the octagon should remain 1080°, the 45° that was reduced should be added to another vertex, so that leads to a second vertex having a 180° angle (corresponding acute exterior angle of 0°). Notice that we cannot push the angle further because it would no longer remain convex. So we are done with 2 vertices. Repeat the same 3 more times. And boom, we started with 8 acute exterior angles, and we made it 4.

The catch is, *the figure we have ended up with at the end is a quadrilateral*. But we haven’t broken any rules though. each angle is still \leqslant 180°, and the exterior angles are (4x 90° => Not acute) and (4x 0° acute).

So the answer is actually 5. But your explanation does seem plausible.

I suppose you do not have to necessarily make one of the angles a complete 180 degree. I think that forgoes the property of an octagon, and hence maybe risky.

My logic (thanks to another fellow student as well) was like this:

The sum of exterior angles of a polygon will always be 360.

This implies that I cannot have more than 4 obtuse exterior angles since 4*90 = 360 and any angle more than that will admonish the above property mentioned.

This means my max obtuse exterior angles that I can take is 3 not more than that. Therefore, for acute angles, 8-3 = 5, which means I can have a minimum of 5 acute exterior angles.

(This property still holds true for interior angles since with 3 obtuse exterior angles imply that there are 3 acute interior angles and 5 acute exterior angles imply that there are 5 obtuse interior angles and these numbers can be balanced to make sure that they are 1080 degree.)

Let me know if this seems sound to you.

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