12
$\begingroup$

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.

quadrilateral

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.

$\endgroup$
3
  • 9
    $\begingroup$ Try with the first three angles being $91°$, you'll easily find that is works :-) $\endgroup$
    – Dominique
    Commented Sep 4 at 14:55
  • 4
    $\begingroup$ 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. $\endgroup$
    – Trunk
    Commented Sep 4 at 14:55
  • $\begingroup$ 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. $\endgroup$
    – richard1941
    Commented 2 days ago

3 Answers 3

Reset to default
49
$\begingroup$

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:

enter image description here

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.

enter image description here

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – user121330
    Commented Sep 7 at 1:55
19
$\begingroup$

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.

$\endgroup$
1
4
$\begingroup$

enter image description here

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.

$\endgroup$
7
  • 5
    $\begingroup$ Mathematically correct, but a rather odd choice to use a not-to-scale diagram that depicts 3 visually acute angles. $\endgroup$
    – Nuclear Hoagie
    Commented Sep 4 at 13:42
  • $\begingroup$ @NuclearHoagie I agree. Here is a Desmos graph showing the kite drawn to scale with angles marked: desmos.com/geometry/opqpw1msi8 $\endgroup$
    – EphraimRuttenberg
    Commented Sep 4 at 13:53
  • $\begingroup$ @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 ... $\endgroup$
    – tkf
    Commented Sep 4 at 14:16
  • $\begingroup$ 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. $\endgroup$
    – Oscar Lanzi
    Commented Sep 4 at 23:17
  • $\begingroup$ OK I have bowed to the consensus and rescaled :) $\endgroup$
    – tkf
    Commented Sep 4 at 23:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .