# What proportion of triangles are acute?

An acute triangle is one in which all angles are acute, ie all angles $<90^\circ$. What proportion of triangles are acute?

I have two attempted answers which are different; I suspect this may be because the question is incoherent. I'd appreciate either an explanation of the correct answer or of why the question is insufficiently defined.

Answer 1: $\frac{1}{2}$.

Since no more than one angle in a triangle can fail to be acute, whether the triangle is acute is determined by the value of one of its angles. Let $T$ be a triangle and pick an angle $\theta(^{\circ})$. Since $T$ is a triangle we must have $0<\theta<180$.

Now $T$ is acute iff $0<\theta<90$, so $\frac{1}{2}$ of all triangles are acute.

Answer 2: $\frac{1}{4}$.

(I lack the graphics skills to properly illustrate this.) Consider 3D Cartesian axes with $0< x,y,z< 180$. Now the points on the plane $x+y+z=180$ represents all possible triples $(\alpha,\beta,\gamma)$ that could be angles of a triangle. Note that this plane (or its closure I suppose, if we're being picky) has the shape of an equilateral triangle with vertices at $(180,0,0)$, $(0,180,0)$ and $(0,0,180)$.

Now we restrict the axes to $0<x,y,z<90$. This restriction of $x+y+z=180$ forms an equilateral triangle with vertices at $(90,90,0)$, $(0,90,90)$ and $(90,0,90)$ (the midpoints of the sides of the first triangle). But this new triangle's area is $\frac{1}{4}$ of the area of the previous triangle so $\frac{1}{4}$ of all triangles are acute.

• This seems very close to Bertrand's paradox, at least at first glance. Nov 2, 2015 at 16:48
• How in the world can an acute triangle exist if each of the angles is less than 60 degrees? Guessing that $<60$ should actually be a $<90$. Nov 2, 2015 at 16:52
• I think of this question as a question in geometric probability. The answer will depend on the process by which "random" triangles are generated. Nov 2, 2015 at 16:52
• Your first paragraph says an acute triangle is one where all angles are less than $60$, is this a mistake? There are no triangles with all angles less than $60$. Nov 2, 2015 at 16:52
• As others have pointed out, you'll get different answers depending on exactly what you mean by "randomly chosen triangle". this link provides many reasonably natural sounding ways to do it (yielding a variety of answers): math.northwestern.edu/~diana/math/acute.pdf
– lulu
Nov 2, 2015 at 16:58

Inspired by the article on Bertrand's paradox as provided by @Arthur, I offer this thought experiment:

Consider taking the isosceles right triangle and constructing a perpendicular bisector to its hypotenuse. Take the two endpoints of the right triangle and any point along the perpendicular bisector as a new triangle, as shown:

Any point chosen between the base and the height of the right triangle will result in an obtuse triangle, and any point chosen along the line beyond the height will make an acute triangle. From this we can obviously see, there are infinitely many more acute triangles than obtuse.

• This is, hopefully, intended as a joke, since if I just pull the vertices on the $x$ axis farther out I can just as easily show there are more obtuse than acute triangles. Nov 2, 2015 at 17:32
• Yep, exactly. Your answer will depend on how you construct your randomly chosen triangles Nov 2, 2015 at 17:35

I'm going to offer this argument for the answer being $1/2$; Consider two points that lie on the unit circle, one is fixed and one moves around the circle clockwise. A "random" isosceles triangle can be constructed by connecting the fixed point, the mobile point, and the center of the circle. As the mobile point moves around the circle, the resultant triangle oscillates between acute and obtuse, the changeover occurs when the mobile point crosses an imaginary diameter of the circle perpendicular to the radius of the fixed point. This imaginary diameter bisects the area of the circle, therefore the triangle is acute half the time and obtuse the other half. Therefore half of all (isosceles) triangles must be acute.

One quarter.

A "circle" argument.

Pick points $$A, B, C$$ "at random". Now pick polar coordinates so that $$O$$ is $$ABC$$'s circumcentre and $$A$$'s phase is $$\theta=0$$. Let $$B$$'s and $$C$$'s phases be $$\beta, \gamma$$. Normalise phase to be in the interval $$[0, 2\pi)$$. Then $$ABC$$ is obtuse if any of the following conditions occurs:

• $$\beta<\pi$$ and $$\gamma<\pi$$
• $$\beta>\pi$$ and $$\gamma>\pi$$
• $$\beta>\gamma+\pi$$
• $$\gamma>\beta+\pi$$

These conditions are mutually exclusive.

Assuming that $$\beta$$ and $$\gamma$$ are chosen from the uniform distribution on $$[0, 2\pi)$$, the above conditions occur with probabilities $$1/4, 1/4, 1/8$$ and $$1/8$$. So $$ABC$$ is obtuse with probability $$3/4$$ and acute with probability $$1/4$$.

An "orthic system" argument.

Pick three points this way. First, pick an orthic system "at random". Then pick three of those four points to be your triangle's vertices, each combination of three being picked with probability $$1/4$$. With probability $$3/4$$, one of the three you picked was the interior point and your triangle is obtuse; with probability $$1/4$$, you picked the three outside points (the three that form the convex hull) and your triangle is acute.

One thing I like about this argument is that it doesn't matter what distribution the orthic system is picked from. And every triangle has a unique orthocentre, and is thus in an orthic system, and thus has a chance of getting picked.