# Is there an unique triangle?

In geometry, Euler's theorem states that the distance d between the circumcenter and incenter of a triangle is given by $d^2=R(R-2r)$ where $R$ and $r$ denote the radii of the circumscribed circle and inscribed circle, respectively.

QUESTION: Given the values of $d(=\sqrt{R^2-2Rr}),R,r$, is there an unique triangle that satisfies the conditions? Note by Inquisitor Math
3 months, 3 weeks ago

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No. In general we need three independent parameters to construct a unique triangle. Here we only have two independent parameters because d is derived from R and r.

- 3 months, 3 weeks ago

Hi again! If that is the case, are there infinitely many triangles that satisfy the conditions?

- 3 months, 3 weeks ago

I guess so. Scale the triangle so r=1. Side lengths of a,b and c then R=abc/2p where p=the perimeter, a+b+c. If R is 1.5 for example, a=b=c=3 is a solution but there are infinitely more.

- 3 months, 3 weeks ago

Hmm... I’m still not totally convinced there are infinitely many solutions.

Btw your example doesn’t satisfy $R\geq2r$.

- 3 months, 3 weeks ago

Yes, sorry, a lazy answer. There is indeed an infinite family of triangles. I looked at it a few months back. There's a thing known as Poncelet's Porism that asserts this. There's a community discussion on Brilliant about it with the question: how to parameterize this family of triangles and which of them has greatest area.

- 3 months, 3 weeks ago

Can you provide a link of the discussion plz? I’m trying to find it :)

- 3 months, 3 weeks ago

- 3 months, 3 weeks ago

Hi again! FYI here is an interesting equation I want to share. (Not sure whether the minus sign works too)

$s•x=x^2+2Rr \pm r\sqrt{4R^2-x^2}$

where $s=\dfrac{a+b+c}2$ and $x=a,b,c$

Idk how yet... but seems like this can yield some results.

- 3 months, 2 weeks ago

Just in case you are interested, I hope you can share your thoughts too :)

- 3 months, 2 weeks ago

Thanks for the equation. Where does it come from?

I can derive something similar but I'm not sure how accurate it is.

Looking again at this infinite family of triangles with the same incircle and circumcircle: If we still scale things so that r=1, the formula for R is abc/4s which implies 2R = abc/(a+b+c). The area of a triangle A is rs or, since r=1, just s. So area is maximised when a+b+c (=2s) is maximised and minimised when a+b+c is minimised.

If we substitute 3u=a+b+c and w^3 =abc (see https://brilliant.org/wiki/the-uvw-method/#tejs-theorem), then 2R = w^3/3u and so w^3=6Ru. A is maximised/minimised when u (or w^3) is maximised/minimised.

Tejs' theorem says that this happens when two of a,b,c are equal (or one of a,b,c is zero which is not a proper triangle). WLOG, we could say a=b: this yeilds an isoscoles triangle whenever area is minimised or maximised for a given R.

If a=b, we can substitute and remove b from the equations and say that 2R = (a^2)c/(2a+c) or (a^2)c-4Ra-2Rc=0. Solving for a using the quadratic formula yields a=(2R+/-sqrt(4R^2+2Rc^2))/c. This is starting to look like a version of your equation. I suspect that using the plus sign results in the "narrow" isoscoles triangle and the minus sign results in the "fat" isoscoles triangle that maximise and minimise the area.

- 3 months, 2 weeks ago

Since you are interested here’s how it’s derived: (like I first assumed, values of R and r are given)

notice that $a,b,c$ are roots of the following polynomial $x^3-2s•x^2+(s^2+4Rr+r^2)x-4Rrs=0$

(actually, at first I used $p=a+b+c$ to express the eq but s seems nicer)

equivalently, $s=\dfrac{x^2+2Rr\pm \sqrt{(x^2+2Rr)^2-x^2(x^2+4Rr+r^2)}}x$

therefore $s•x=x^2+2Rr \pm r\sqrt{4R^2-x^2}$

- 3 months, 2 weeks ago

This polynomial is really interesting. I see now how you derived that equation. It's hard for me to investigate it further because x "encodes" a, b and c as its roots and s also depends on a,b and c. So both x and s are variables that depend on a,b and c. This equation probably only works if s is constant for a given R and r which I don't think is the case.

Looking at the cubic polynomial has been interesting: Cubics have three real roots if and only if the discriminant of the cubic is positive. We want positive, real roots for a, b and c so under which conditions is the discriminant positive? Setting r to 1 again, the discriminant is -256R^3+16R^2s^2-192R^2+80Rs^2-48R-4s^4-8s^2-4. Although this is a long expression, it will be zero when s is at maximum or minimum. It has real roots when R=2 and s=3sqrt(3): an equilateral triangle with minimum R when r=1. In fact, for a given R, it simplifies into the form of As^4+Bs^2+C for some A,B and C that depend only on R. This is a quadratic in s^2 . Solving using the quadratic formula gives s^2=2R^2+10R-1+/-2sqrt(R(R-2)^3). Taking the square root of s^2 gives an expression for the maximum and minimum semiperimeter for any given R. It is interesting that this results in a complex number and no real solutions when I use my bad example above where R =1.5 (i.e. less than the minimum of 2). Also interesting but expected that when R=2, the expression under the square root is zero meaning that the maximum and minimum semiperimeter are identical. When R>2, the semiperimeter has different maximum and minimum values indicating that there is an infinite family of non-congruent triangles with different but continuously varying semiperimeters/areas.

- 3 months, 2 weeks ago

Yes I found this eq interesting too! (As you mentioned, the fact this eq involves the complex roots as well is noteworthy.) I’m glad I shared it with you:) I was hoping to go further but since there are not enough conditions, I guess that’s as far as we can go.

Btw is the minus part of the eq valid?

- 3 months, 2 weeks ago

I'm not sure but it would be interesting to look at it. In my reply above, there is an expression for s^2 of the form s^2=A+/-B. This gives not two but four possible values for s: a large s, a small s, a large negative s and a small negative s. Negative s could be thought of as a length/ distance in the opposite direction. Perhaps the same us true of the minus part of your equation?

- 3 months, 2 weeks ago