Which Triangle is More Equilateral?
Two consecutive isosceles triangle days (10/10/11 and 10/11/11) got me thinking: Which of the following triangles is more equilateral, the 10-10-11 triangle, or the 10-11-11 triangle?
Or perhaps I should say equilateraler?
[Update: here is my first attempt to answer this question in a mathematically sensible way.]
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35 Comments
John Golden · October 10, 2011 at 10:50 am
Are you proposing a distance metric for triangles? Love it! What would that be like?
MrHonner · October 10, 2011 at 11:04 am
I guess I am proposing something like a measure of equilateralism. I’ve got some ideas; I’ll try to write them up tomorrow, on our next isosceles triangle day!
Geoff · October 10, 2011 at 12:43 pm
Love this problem. I’m adding to my Geometry curriculum map.
Robert L. Tuva (@rltuva) · October 10, 2011 at 1:01 pm
Both have the same standard deviation of side lengths.
MrHonner · October 10, 2011 at 1:09 pm
Glad you like it, Geoff! Please feel free to link to your curriculum map.
Robert, I also thought about SD of side length. An interesting approach.
Tao Wang (@MathLaoshi) · October 10, 2011 at 3:26 pm
How about the distance between the incenter and the circumcenter? I constructed the two triangles in GeoGebra. The incenter and circumcenter of the 10-11-11 are slightly closer together than those of the 10-10-11.
MrHonner · October 10, 2011 at 5:43 pm
I like that idea, Tao. My two initial thoughts are: (1) is there an obvious reason that closing that distance makes a triangle more equilateral; (2) we probably want our quantification of equilateralism to be invariant under dilation.
James Cleveland · October 10, 2011 at 7:02 pm
They don’t have the same standard deviation of angles, though. The SD of the angles of the 10-11-11 triangle is a bit smaller than for the 11-10-10 one.
MrHonner · October 10, 2011 at 8:00 pm
James, I thought of this, too, when Robert pointed out that the SD of the sides is the same. This may be a good approach. However, It should be noted that the question, as posed, is about equilateralism, not equiangularism.
Tao Wang (@MathLaoshi) · October 10, 2011 at 7:20 pm
(1) To me, the “obvious” reason is that in an equilateral triangle, the incenter and the circumcenter coincide. (2) Hm, hm… that’s a good point. It didn’t occur to me since the areas of the triangles are so similar. A 20-22-22 triangle shouldn’t be any less equilateral than a 10-11-11. I’m eagerly awaiting what you have for your next post.
MrHonner · October 10, 2011 at 8:00 pm
I’ll try to come up with something good. 🙂
Curmudgeon · October 10, 2011 at 7:47 pm
Which fraction is closer to unity: 5/6 or 6/7?
This is the same kind of question.
MrHonner · October 10, 2011 at 8:06 pm
I like the intuition, but I’m not sure it agrees with what I believe is the more equilateral of the two.
James Cleveland · October 10, 2011 at 8:45 pm
I see your point, though if we are sticking Euclidean geometry equiangular triangles are always equilateral. AND it solves the dilation issue, since the angles would stay the some.
But still, nothing to do with side lengths. I wonder if we could use the compass & straightedge construction as a way to evaluate this, by looking at the variation in the arc and circle, but I’m not certain how to go about that.
mrgaffey · October 10, 2011 at 9:38 pm
An equilateral triangle can be circumscribed by a circle. I wonder which triangle has vertices furthest away from the edge of the circle.
JBL · October 11, 2011 at 10:32 am
Riffing on Tao Wang’s comments, here are two possible metrics:
1) the equilateral triangle is the triangle of largest area with a given circumcircle. Thus, we can define the equilateralness of a triangle to be the ratio of its area to the area of its circumcircle.
2) the equilateral triangle is the triangle of smallest area with a given incircle. Thus, we can define the equilateralness of a triangle to be the ratio of the area of the incircle to the area of the triangle.
3) the equilateral triangle is the triangle with the largest area given its perimeter. Thus, we can define the equilateralness of a triangle to be the ratio of the area to the square of the perimeter. (Note the squaring to make units cancel; this leaves equilateralness unchanged under dilation.)
Question: is there a pair of triangles such that the first is more equilateral under one of these definitions while the second is more equilateral under the other two? Are there triangles that tie for some but not others?
MrHonner · October 11, 2011 at 6:27 pm
These are the measures I’ve been playing around with, but I’ve heard lots of great ideas, from both commenters and students.
Your questions at the end really get to the heart of mathematical exploration. Once we settle on one (a few?) measures of equilateralness, then the real fun begins: we can start exploring the consequences of that measure by asking question like “What kinds of triangles have the same equilateralness?” Or, “Are there some triangles that create a conflict between our intuition and our definition?”
This simple problem creates a lot of opportunity.
Christopher · October 11, 2011 at 2:11 pm
Lovely question. It’s a pretty abstract thing you’re trying to measure, of course. But it’s an activity that cuts to the heart of what it means to measure abstractly.
How will we know when we’ve got the measure right? Is it intuitively clear which one is more equilateral than the other?
Hard to sort out with this example. What about a 1-1-2 vs. a 2-2-1? Is it more transparent there? Perhaps a Twitter poll is in order to see whether you’re trying to measure something that people can even perceive?
I took on a similar challenge with some preservice secondary teachers a few years back. We tried to measure the cubeyness of various prisms. This sort of thing can lead to some interesting mathematical places.
MrHonner · October 11, 2011 at 6:31 pm
I love the question “How will we know when we’ve got the answer right?” It’s a big leap for students to realize that the answer is not “When the teacher tells you.”
JBL · October 11, 2011 at 9:44 pm
Well, 1-1-2 vs. 2-2-1 isn’t a close thing: the first is degenerate (i.e., not actually triangular) and the second is a perfectly respectable triangle with positive area.
MrHonner · October 11, 2011 at 9:55 pm
‘Degenerate’ is such a strong word.
In all seriousness, it’s worth exploring what happens with our proposed metrics on these extreme cases. For example, Heron’s formula works perfectly well on the 1-1-2 ‘triangle’; maybe we shouldn’t be so quick to discount it.
Christopher · October 12, 2011 at 9:29 am
Drat! Of course!
Darn Triangle Inequality!
Always sneaking in there to foil my ill-conceived examples.
I retract my degenerate-triangle example and suggest 2-2-3 v. 3-3-2 instead. And again I ask the question that was the point of the example. Absent an “equilateral” measure, can we intuitively agree on which one is more equilateral? I wanna say the 3-3-2 is, but it has nothing to do with circumcenters; it’s about angles. 3-3-2 is acute, where 2-2-3 is obtuse (quickly making GeoGebra sketch before posting to make sure mental image matches reality).
nateharman1234 · October 18, 2011 at 8:45 am
Here is an idea:
Among triangles, an equilateral triangle uniquely minimizes perimeter for a given area. I think that from this perspective the ratio A/P^2 is a good thing to look at. It is dimensionless, maximized for the equilateral triangle, and can be generalized to polygons with more sides.
MrHonner · October 18, 2011 at 9:13 am
Nate, this is essentially the approach I took here: https://mrhonner.com/2011/10/12/which-triangle-is-more-equilateral-part-ii/.
I like it for just the reasons you said–it’s dimensionelss, and it generalizes. easily and naturally.
Although it strikes me that it’s more a measure of regularness than equilateralness.
Sue VanHattum · October 22, 2011 at 10:12 am
You might like this similar exercise, which asks about the squareness of rectangles.
Thanks for reminding me of it. I think I might use it at my next math salon.
I’m headed out to an Escape From the Textbook meeting. We played with the rectangles last time, so I’ll point people to the post for an extension.
MrHonner · October 22, 2011 at 12:14 pm
Hi Sue-
The “How square are these rectangles?” problem you’ve shared is a great lead-in to this problem. All the interesting questions are still there–what does ‘squareness’ mean? How can we measure it? Which measure should we choose? Why? Which measures are equivalent?–and the subsequent math is a bit easier to handle.
I can definitely see using squareness as an introduction to equilateralness. Thanks for sharing!
Cody Johnson · October 29, 2011 at 11:55 am
It is the 10-11-11 triangle because:
The angles formed inside the 10-11-11 triangle are calculated to be 62.964308, 62.964308, and 54.071384. The angles in the 10-10-11 triangle are 56.632987, 56.632987, and 66.734026.
Next, I calculated the absolute value of 60 minus the angle, added them together for each triangle, and divided by 3. Whichever number is lower is a more efficient triangle, therefore “equilateraler.”
| 60 – 62.964308 | = 2.964308
| 60 – 62.964308 | = 2.964308
| 60 – 54.071384 | = 5.928616
(2.964308 + 2.964308 + 5.928616) / 3 = 3.952411
| 60 – 56.632987 | = 3.367013
| 60 – 56.632987 | = 3.367013
| 60 – 66.734026 | = 6.734026
(3.367013 + 3.367013 + 6.734026) / 3 = 4.489351
3.952411 < 4.489351, therefore the 10-11-11 triangle is "equilateraler."
Cody Johnson · October 29, 2011 at 12:00 pm
Piece of cake.
MrHonner · October 29, 2011 at 12:11 pm
Very nice! Essentially you are looking at which set of angles has the lower mean deviation from 60, which is a very intuitive and natural approach.
However, one interesting consequence of this is that, if triangle ABC has angles 58, 58, and 64, and triangle PQR has angles 62, 62, and 56, then these two triangles are equally equilateral. Does that make sense?
Cody Johnson · October 30, 2011 at 6:47 pm
But the angles are not 58, 58, and 64. For the left and right corners of the 11-10-10 triangle, the angles equal arccos(5.5 / 10) = 56.632987 degrees. Therefore, the other angle is 180 – 2arccos(5.5 / 10) = 66.734026 degrees. So the 11-10-10 triangle is 56.632987, 56.632987, and 66.734026 degrees. For the 10-11-11 triangle, the left and right corners are arccos(5 / 11) = 62.964308 degrees, and the other one being 180 – 2arccos(5 / 11) = 54.071384 degrees. The 10-11-11 triangle angle measurements are, respectively, 62.0964308, 62.0964308, and 54.071384 degrees. So the triangle’s angles are not as you say.
Cody Johnson · October 30, 2011 at 6:50 pm
Therefore the mean derivation from 60 is different in the triangles.
MrHonner · October 30, 2011 at 7:15 pm
I like the mean deviation method you’ve proposed for the 10-10-11 and 10-11-11 triangles, and I think it makes sense. What I suggested was that we consider how your method works on other triangles.
For example, if we have two triangles with angles {58,58,64} and {56,62,62}, your method will declare them equally equilateral. So the question is, do we want to say that those two triangles are indeed equally equilateral?
Cody Johnson · October 31, 2011 at 4:13 pm
OH! Now I understand. You were speaking hypothetically and I thought that you were talking about in this particular instance. Yes, certainly mean deviation is a better method in the case of two triangles being equally equilateral.
Cody Johnson · October 31, 2011 at 4:30 pm
Like in two triangles with angles {65,56,59} and {58,57,65} (I used this example to prove it’s not just only in isosceles or equilateral triangles).
MrHonner · October 31, 2011 at 7:24 pm
Right, good example. So in this case, the mean deviation method declares these two triangles to be equally equilateral. I feel like one of them should be more equilateral than the other.