My friend gave me this puzzle:

What is the probability that a point chosen at random from the interior of an equilateral triangle is closer to the center than any of its edges?

I tried to draw the picture and I drew a smaller (concentric) equilateral triangle with half the side length. Since area is proportional to the square of side length, this would mean that the smaller triangle had $1/4$ the area of the bigger one. My friend tells me this is wrong. He says I am allowed to use calculus but I don't understand how geometry would need calculus. Thanks for help.

Parcly Taxel
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  • The vertices of the smaller triangle you drew are much closer to the edges of the original triangle than they are to the center. –  Mar 08 '16 at 19:57
  • @Rahul Yes I know it is wrong, they look closer to circles than straight lines .. – terrace Mar 08 '16 at 20:02
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    @N.S.JOHN: The body of the question specifies that it's an *equilateral* triangle, so I think "center" is reasonably well-defined. – ruakh Mar 10 '16 at 02:28
  • @ruakh please avoid sarcastic comments like this. This is an education site and just provide the answer. – N.S.JOHN Mar 10 '16 at 03:37
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    @N.S.JOHN: I did not intend my comment to be sarcastic, and I'm sorry if it came off that way. (I had assumed that you simply saw the phrase "inside a triangle" in the title of the question and didn't notice the additional word "equilateral" in the body of the question. So I didn't realize any further explanation was needed beyond pointing out that word.) To answer your question: "center" does not mean "circumcenter", no; but with an equilateral triangle specifically, the circumcenter, incenter, orthocenter, and centroid are all the same point, so it's reasonable to just call it "the center". – ruakh Mar 10 '16 at 04:11
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    [Problem B1 on the 50th Putnam (1989) is very similar](https://mks.mff.cuni.cz/kalva/putnam/putn89.html). – Zubin Mukerjee Mar 21 '16 at 20:04
  • FWIW: This subsequently appeared as a Problem of the Week by Cal Poly's Dept of Mathematics. [**Link**](http://www.calpoly.edu/~sherman1/puzzleoftheweek/pow_143soln.pdf) (pdf) with solution. – Benjamin Dickman May 23 '16 at 17:32
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    @BenjaminDickman Wow, shouldn't they credit Stack Exchange or something like that? – terrace Jun 05 '16 at 04:35
  • @terrace If they learned about it from Stack Exchange? Then yes, absolutely; I find the question somewhat natural, though (as mentioned above, the version for a square is Putnam B1/1989). You said that your friend gave you the puzzle; could you find out the source from your friend? I'd be interested to know! – Benjamin Dickman Jun 05 '16 at 05:18
  • I'm baffled nobody even mentions the question is not well defined if how the point is chosen is specified, wich is equivalent to set the measure space with the probability function. [Bertrand paradox applies here!](https://en.m.wikipedia.org/wiki/Bertrand_paradox_(probability)) – Santropedro Jan 19 '17 at 02:41
  • @Santropedro i think choosing chord at random doesnt make sense but choosing point is easier – terrace Feb 03 '17 at 00:56

4 Answers4


You are right to think of the probabilities as areas, but the set of points closer to the center is not a triangle. It's actually a weird shape with three curved edges, and the curves are parabolas.

The set of points equidistant from a line $D$ and a fixed point $F$ is a parabola. The point $F$ is called the focus of the parabola, and the line $D$ is called the directrix. You can read more about that here.

In your problem, if we think of the center of the triangle $T$ as the focus, then we can extend each of the three edges to give three lines that correspond to the directrices of three parabolas.

Any point inside the area enclosed by the three parabolas will be closer to the center of $T$ than to any of the edges of $T$. The answer to your question is therefore the area enclosed by the three parabolas, divided by the area of the triangle.

Let's call $F$ the center of $T$. Let $A$, $B$, $C$, $D$, $G$, and $H$ be points as labeled in this diagram:

Voronoi diagram for a triangle and its center

The probability you're looking for is the same as the probability that a point chosen at random from $\triangle CFD$ is closer to $F$ than to edge $CD$. The green parabola is the set of points that are the same distance to $F$ as to edge $CD$.

Without loss of generality, we may assume that point $C$ is the origin $(0,0)$ and that the triangle has side length $1$. Let $f(x)$ be equation describing the parabola in green.

By similarity, we see that $$\overline{CG}=\overline{GH}=\overline{HD}=1/3$$

An equilateral triangle with side length $1$ has area $\sqrt{3}/4$, so that means $\triangle CFD$ has area $\sqrt{3}/12$. The sum of the areas of $\triangle CAG$ and $\triangle DBH$ must be four ninths of that, or $\sqrt{3}/27$.

$$P\left(\text{point is closer to center}\right) = \displaystyle\frac{\frac{\sqrt{3}}{12} - \frac{\sqrt{3}}{27} - \displaystyle\int_{1/3}^{2/3} f(x) \,\mathrm{d}x}{\sqrt{3}/12}$$

We know three points that the parabola $f(x)$ passes through. This lets us create a system of equations with three variables (the coefficients of $f(x)$) and three equations. This gives

$$f(x) = \sqrt{3}x^2 - \sqrt{3}x + \frac{\sqrt{3}}{3}$$

The integral of this function from $1/3$ to $2/3$ is $$\int_{1/3}^{2/3} \left(\sqrt{3}x^2 - \sqrt{3}x + \frac{\sqrt{3}}{3}\right) \,\mathrm{d}x = \frac{5}{54\sqrt{3}}$$

This gives our final answer of $$P\left(\text{point is closer to center}\right) = \boxed{\frac{5}{27}}$$

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Zubin Mukerjee
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    Could you expand on how “by similarity” points G and H are spaced 1/3 from the ends? Points A and B are the intersections of the parabolas, but it's not obvious to me that they have any particular properties. – Kevin Reid Mar 09 '16 at 15:18
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    @KevinReid introduce a new point J that is the midpoint of CD, and draw FJ. CAG is a 30-60-90 triangle, as is CFJ. Since AG=AF (by definition of a parabola) and AC=2AG (because of the 30-60-90 triangle relationships), CAG is 2/3 the size of CFJ, and CG is 2/3 of CJ. – MattPutnam Mar 09 '16 at 15:37
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    @MattPutnam I see. I think that's sufficiently unobvious to be worth writing in the answer. – Kevin Reid Mar 09 '16 at 16:39
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    Isn't the result equal to $3^{-1} - 3^{-2} - 3^{-3}$? – Alecos Papadopoulos Mar 09 '16 at 22:29
  • Which is the third _easy_ point through which the parabola passes except $A$ and $B$? Nice @AlecosPapadopoulos. I wonder why it is. – woky Mar 24 '16 at 16:40
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    @woky the point whose $x$ coordinate is the same as $F$'s and whose $y$ coordinate is half of $F$'s – Zubin Mukerjee Mar 24 '16 at 17:38

In response to Benjamin Dickman's request for a solution without calculus, referring to dtldarek's nice diagram in Zubin Mukerjee's answer (with all areas relative to that of the triangle $FCD$):

The points $A$ and $B$ are one third along the bisectors from $F$, so the triangle $FAB$ has area $\frac19$. The vertex $V$ of the parabola is half-way between $F$ and the side $CD$, so the triangle $VAB$ has width $\frac13$ of $FCD$ and height $\frac16$ of $FCD$ and thus area $\frac1{18}$. By Archimedes' quadrature of the parabola (long predating the advent of calculus), the area between $AB$ and the parabola is $\frac43$ of the area of $VAB$. Thus the total area in $FCD$ closer to $F$ than to $CD$ is

$$ \frac19+\frac43\cdot\frac1{18}=\frac5{27}\;. $$

P.S.: Like Dominic108's solution, this is readily generalized to a regular $n$-gon. Let $\phi=\frac\pi n$. Then the condition $FB=BH$, expressed in terms of the height $h$ of triangle $FAB$ relative to that of $FCD$, is

$$ \frac h{\cos\phi}=1-h\;,\\ h=\frac{\cos\phi}{1+\cos\phi}\;. $$

This is also the width of $FAB$ relative to that of $FCD$. The height of the arc of the parabola between $A$ and $B$ is $\frac12-h$. Thus, the proportion of the area of triangle $FCD$ that's closer to $F$ than to $CD$ is

$$ h^2+\frac43h\left(\frac12-h\right)=\frac23h-\frac13h^2=\frac{2\cos\phi(1+\cos\phi)-\cos^2\phi}{3(1+\cos\phi)^2}=\frac13-\frac1{12\cos^4\frac\phi2}\;. $$

This doesn't seem to take rational values except for $n=3$ and for $n\to\infty$, where the limit is $\frac13-\frac1{12}=\frac14$, the value for the circle.

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The problem can be solved without calculus by using Archimedes Quadrature of the Parabola. To prove this Archimedes did, however, use techniques that are closely related to modern calculus.

Link to image

According to Archimedes Quadrature of the Parabola the area between the parabola and the edge EF is $4\over3$ the area of triangle EFG.

Let's assume that EK has length x, then due to similarity DI has length ${3\over2}x$ and DH has length ${1\over2}x$. Because point G is on the parabola we have that the length of DG is half the length of DI, hence the length of DG is ${3\over4}x$. Now the length of HG is ${3\over4}x-{1\over2}x={1\over4}x$. So the length of DH is twice the length of HG

From this it follows that the area of triangle EFG is half the area of triangle EFD. The area between EF and the parabola is ${4\over3}\cdot{1\over2}={2\over3}$ times the area of triangle EFD.

As the area of triangle EFD is equal to ${1\over27}$ the area of triangle ABC we now find that the area of points closer to the middle then to the edges is equal to $3\cdot1{2\over3}\cdot{1\over27}={5\over27}$ the area of triangle ABC. Hence P(point is closer to the center)=${5\over27}$.

Erik Jurriën
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I am just passing by here. Perhaps the idea is to have something elegant, but this is just straightforward: an integral in polar coordinates with the help of Mathematica. I reuse the picture of the solution of Zubin. Let J be the midpoint of CD. By symmetry, we can restrict the analysis to CFJ. Without loss of generality, let the length of FJ be $1$. The total area of CFJ is $\sqrt{3}/2$. We need the area above the curve in CFJ. Pick an arbitrary point L on the curve inside CFJ. Let $\theta$ be the angle between FJ and FL. Some trigonometry gives us that the length of FL is $x = (1+ \cos\theta)^{-1}$. (It helps to draw the straight line that extends FL and meets CJ at some point K and note that the length of FK is $1/\cos \theta$.) So, the area above the curve is the integral of $(1+\cos \theta)^{-2}/2$ from $0$ to $\pi/3$. Using mathematica, the primitive is $((2 + \cos\theta)\sin\theta)/(6(1 + \cos\theta)^2)$. So, the area above the curve is $((2 + \cos(\pi/3))\sin(\pi/3))/(6(1 + \cos(\pi/3))^2) = 5 \sqrt 3/ 54$ . So, the probability to be above the curve is $(5 \sqrt 3/ 54)/(\sqrt 3/2) = 5/27$.

This solution generalizes to any regular polygone. We integrate from $0$ to $\pi/n$ instead of from $0$ to $\pi/3$ and the total area of CFJ becomes $\tan(\pi/n)/2$ instead of $\sqrt 3/2$, every thing else stays the same.

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