# Can a quadrilateral polygon have 3 obtuse angles?

I was messing around with quadrilaterals trying to draw one that has three obtuse angles. I couldn't create one because with 3 obtuse angles the shape would "open up too much".

I have finished high school math with a perfect score but hadn't messed around with university level math yet (just to give context to what I would understand easily and what would probably require a bit more explanation)

I can understand "visually" why there can't be 3 obtuse angles, but when I tried to prove it all I got was this:

Let there be a quadrilateral with 4 angles marked a, b, c, d.

Assuming 3 obtuse angles:

a > 90, b > 90, c > 90 therefore: a + b + c > 270

The sum of the internal angles in a quadrilateral is 360:

a + b + c + d = 360 => d = 360 - (a + b + c) therefore: d < 360 - 270 => d < 90

All I managed to prove here is that d must acute, but I couldn't figure out a mathematical proof why 3 is impossible.

I apologize in advance for any mistaken terms/conventions. I never learned math in English so I just used the terms that made sense to me.

• Try with the first three angles being $91°$, you'll easily find that is works :-) Commented Sep 4 at 14:55
• Surely you have proved here algebraically that it is possible to have three obtuse angles in a quadrilateral, i.e. a, b, c > 90 deg and d < 90 deg. Working on a bigger page do a geometrical proof for yourself. Commented Sep 4 at 14:55
• My Maylay kite seems to have one obtuse angle, two right angles, and an acute angle at the tail. Open those right angles to 91 degrees, and you have your counterexample. Commented 2 days ago

You can't prove that it's impossible for a quadrilateral to have 3 obtuse angles because it isn't impossible for a quadrilateral to have 3 obtuse angles. Here's one such counterexample:

All quadrilaterals have their internal angles sum to 360 degrees. That 3 of the angles are greater than 90 degrees simply means the last one is less than 90 degrees. This exactly the finding of your proof - that the final angle $$d$$ is less than 90 degrees, not that it can't exist.

The 3 obtuse angles must also sum to less than 360 degrees themselves, or else the remaining "angle" doesn't actually exist (since $$d>0$$). You can't use any set of obtuse angles to make a quadrilateral - it is possible to have the shape "open up" too much to close it with one angle, but that's not true of all sets of obtuse angles. If $$a+b+c\geq360$$, then the angles taken together are indeed "too obtuse" to form a quadrilateral, and no final angle $$d$$ will be able to close it. Here is an example of three obtuse angles which sum to more than 360 degrees and can't form the basis of a quadrilateral - the rays could extend infinitely but will never meet to form a closed angle. It would take at least 2 more angles to close this shape.

• If your two "dangling" edges went the other way (a long way down on your screen), you'd actually have a triangle with 3 acute angles and one obtuse angle which seems worth mentioning in the context of a question about 3 obtuse angles and one acute angle. Commented Sep 7 at 1:55

Construct an equiangular pentagon. Extend any two nonadjacent sides to their point of inrersection, forming a quadrilateral with three $$108°$$ angles from the original pentagon.

In the diagram above, the quadrilateral $$ABCD$$ has 3 obtuse angles:

1. $$\angle XAY$$ is a right angle: $$\vec{AX}\cdot\vec{AY}=(-1,1)\cdot (1,1)=0$$. Therefore $$\angle BAC$$ is obtuse.

2. $$\angle ABZ$$ is a right angle: $$\vec{BA}\cdot\vec{BZ}=(2,-1)\cdot (2,4)=0$$. Therefore $$\angle ABD$$ is obtuse.

3. $$\angle ACZ$$ is a right angle: $$\vec{CA}\cdot\vec{CZ}=(-2,-1)\cdot (-2,4)=0$$. Therefore $$\angle ACD$$ is obtuse.

• Mathematically correct, but a rather odd choice to use a not-to-scale diagram that depicts 3 visually acute angles. Commented Sep 4 at 13:42
• @NuclearHoagie I agree. Here is a Desmos graph showing the kite drawn to scale with angles marked: desmos.com/geometry/opqpw1msi8 Commented Sep 4 at 13:53
• @NuclearHoagie True, but your answer already has a nice visually accurate example. Also Oscar's answer has a very clever elegant construction that clearly works. What I wanted to add was a simple brute force way of writing down co-ordinates, where you can be certain that the required angles are obtuse by comparison with right angles. The odd diagram is probably down to me being a topologist by profession. Ask me to draw a coffee cup and you will get ...
– tkf
Commented Sep 4 at 14:16
• We should use the same scaling on the x and y axes here. Not doing so distorts the angles, making the figure presentation hard to interpret in tge mathematically proper sense. Commented Sep 4 at 23:17
• OK I have bowed to the consensus and rescaled :)
– tkf
Commented Sep 4 at 23:53