Is there any geometrically simple reason why the inradius of a triangle should be at most half its circumradius? I end up wanting the fact for this answer.

I know of two proofs of this fact.

Proof 1:

The radius of the nine-point circle is half the circumradius. Feuerbach's theorem states that the incircle is internally tangent to the nine-point circle, and hence has a smaller radius.

Proof 2:

The Steiner inellipse is the inconic with the largest area. The Steiner circumellipse is the circumconic with the smallest area, and has 4 times the area of the Steiner inellipse. Hence the circumcircle has at least 4 times the area of the incircle.

These both feel kind of sledgehammerish to me; I'd be happier if there were some nice Euclidean-geometry proof (or a way to convince myself that no such thing is likely to exist, so the sledgehammer is necessary).

EDIT for ease of future searching: The internet tells me this is often known as "Euler's triangle inequality."

  • $\begingroup$ This is fairly old, but using Erdös-Mordell and Carnot's should work. $\endgroup$
    – Nairit
    Commented Jul 5, 2018 at 2:31

6 Answers 6


So Proof #1 can be modified to be completely elementary.

First, it is easy to show that the incircle is the smallest circle touching all 3 sides. The circle passing through the midpoints (the nine-point circle) obviously has circumradius half that of the larger circle. No need to invoke Feuerbach's theorem for this.



  • $\begingroup$ Good point. Thanks! $\endgroup$
    – Micah
    Commented Jul 14, 2012 at 21:20

Compute the area of a triangle (first method):

Consider the following diagram:

$\hspace{4.5cm}$enter image description here

The area of the green triangle is $$ A=\tfrac12ab\sin(\theta)\tag{1} $$ By the Inscribed Angle Theorem, the angle that $c$ subtends at the origin, $o$, is $2\theta$. Therefore, we get that $$ c=2R\sin(\theta)\tag{2} $$ Combining $(1)$ and $(2)$ yields $$ 4AR=abc\tag{3} $$ Compute the area of a triangle (second method):

Consider the following diagram:

$\hspace{4.5cm}$enter image description here

Note that the areas of the the red ($\color{#C00000}{\triangle iyz}$), green ($\color{#00A000}{\triangle izx}$), and blue ($\color{#0000FF}{\triangle ixy}$) triangles are $\frac12r$ times $a$, $b$, and $c$, respectively. Therefore, we get $$ 2A=r(a+b+c)\tag{4} $$

Compute $d$:

Translate the circumcenter, $o$, of $\triangle xyz$ to the origin. Then $$ |x|=|y|=|z|=R\tag{5} $$ Furthermore, using $a=|y-z|$, $b=|z-x|$, and $c=|x-y|$, we get $$ \begin{align} 2y\cdot z&=2R^2-a^2\tag{6a}\\ 2z\cdot x&=2R^2-b^2\tag{6b}\\ 2x\cdot y&=2R^2-c^2\tag{6c} \end{align} $$ Explanation:
$\text{(6a)}$: $a^2=(y-z)\cdot(y-z)=|y|^2+|z|^2-2y\cdot z$, then apply $(5)$
$\text{(6b)}$: $b^2=(z-x)\cdot(z-x)=|z|^2+|x|^2-2z\cdot x$, then apply $(5)$
$\text{(6c)}$: $c^2=(x-y)\cdot(x-y)=|x|^2+|y|^2-2x\cdot y$, then apply $(5)$

As mentioned above, the areas of the the red ($\color{#C00000}{\triangle iyz}$), green ($\color{#00A000}{\triangle izx}$), and blue ($\color{#0000FF}{\triangle ixy}$) triangles are proportional to $a$, $b$, and $c$, respectively. Thus, the barycentric coordinates of the incenter, $i$, are the mean of the vertices weighted by the lengths of the opposite sides: $$ i=\frac{ax+by+cz}{a+b+c}\tag{7} $$ and therefore, using $(3)$-$(7)$ yields that $d$, the distance between the incenter and circumcenter, satisfies $$ \begin{align} d^2 &=\frac{a^2R^2+b^2R^2+c^2R^2+2abx\cdot y+2bcy\cdot z+2caz\cdot x}{(a+b+c)^2}\tag{8a}\\ &=\frac{a^2R^2+b^2R^2+c^2R^2+ab(2R^2-c^2)+bc(2R^2-a^2)+ca(2R^2-b^2)}{(a+b+c)^2}\tag{8b}\\ &=\frac{(a+b+c)^2R^2-(a+b+c)abc}{(a+b+c)^2}\tag{8c}\\[3pt] &=R^2-\frac{abc}{a+b+c}\tag{8d}\\[6pt] &=R^2-2Rr\tag{8e}\\[12pt] &=R(R-2r)\tag{8f} \end{align} $$ Explanation:
$\text{(8a)}$: take the dot product of $(7)$ with itself and apply $(5)$
$\text{(8b)}$: apply $(6)$ to the dot products
$\text{(8c)}$: collect terms
$\text{(8d)}$: simplify
$\text{(8e)}$: apply $(3)$ and $(4)$
$\text{(8f)}$: factor

Since $d^2\ge0$, and $R>0$, we immediately get that $$ r\le\tfrac12R\tag9 $$

  • $\begingroup$ Thanks! I think this might strictly speaking be overkill as it's a lot more complex than the translational argument that Rofler didn't give, but it's great to have an expression for exactly how unequal the two quantities are... $\endgroup$
    – Micah
    Commented Jul 16, 2012 at 20:15
  • $\begingroup$ @Micah: I moved this part of my answer to this question since it applies more directly here. I reference it from the other answer. $\endgroup$
    – robjohn
    Commented Jul 16, 2012 at 20:47
  • $\begingroup$ @Micah: I didn't start from the 9-point circle. I hope this is simpler when starting from scratch. $\endgroup$
    – robjohn
    Commented Jul 16, 2012 at 20:55
  • $\begingroup$ To my mind the other argument is still simpler -- note that you don't actually need to know that it's the 9-point circle, just that it intersects the three midpoints. But I guess in some sense it's a question of algebraic vs. geometric complexity; visualizing the other proof is easier, but writing down an explicit set of translations that does the trick would probably require more work than what you did here. $\endgroup$
    – Micah
    Commented Jul 16, 2012 at 21:30
  • $\begingroup$ @Micah: Okay; I see. If we characterize the incircle as the smallest circle that touches all three sides, then yes, the circumcircle of the triangle constructed from the midpoints of the sides would be bigger than the incircle, and it would have half the radius of the original circumcircle. $\endgroup$
    – robjohn
    Commented Jul 17, 2012 at 0:33

I came up with the following (almost) purely algebraic proof for the simple reason that I'm not much of a geometry guy and it's too much mental work for me to keep track of the Nine-Point Circle.

We use the following formulae for the area of a triangle:

$A = \sqrt{s(s-a)(s-b)(s-c)} = rs = \dfrac{abc}{4R}$,

where $A$ is the area, $r$ is the inradius, $R$ is the circumradius and $s$ is the semiperimeter.


$R \geq 2r \iff \dfrac{abc}{4A} \geq \dfrac{2A}{s} \iff % \dfrac{abc}{8} \geq \dfrac{A^2}{s} \iff \dfrac{abc}{8} \geq (s-a)(s-b)(s-c)$.

To simplify the last inequality, use Ravi Substitution. i.e. Let $a = y + z$; $b = z + x$; $c = x + y$. (Note that this can always be done for any arbitrary triangle $ABC$. Also, $x,y,z$ are strictly positive. More information can be found if you google Ravi Substitution). It is obvious that $s = x + y + z$. Thus, last inequality is equivalent to

$\dfrac{(x+y)(y+z)(z+x)}{8} \geq xyz$.

This can be proved if we multiply the following three inequalities:

$\dfrac{x+y}{2} \geq \sqrt{xy}; \quad \dfrac{y+z}{2} \geq \sqrt{yz}; \quad \dfrac{z+x}{2} \geq \sqrt{zx}$.

(Note that $x,y,z$ are positive and $AM \geq GM$ can be applied safely).


enter image description here

It's enough to show that $OI^2=R(R-2r) \iff R^2-OI^2=2Rr$, where $O$,$I$,$R$, and $r$ is the circumcenter, incenter, circumradius, and inradius respectively. By power of a point, $R^2-OI^2=AI\times IL$, so it's enough to show that $AI\times IL =2Rr \iff \frac{AI}{r}=\frac{2R}{IL}$. This is trivial since $\triangle AFI \sim \triangle KBL$ and $LI=BL$ (By Incenter-Excenter lemma)


Expanding on Rofler's answer, just so there'll be a complete argument here:

Consider the circle which meets the midpoints of the three sides (this is secretly the 9-point circle, but we don't need to know that to finish this argument). It circumscribes a triangle which is similar to the reference triangle and scaled by $\frac{1}{2}$. Thus its radius is half the circumradius.

To show that the radius of the nine-point circle is larger than the inradius of the reference triangle, first translate it in your favorite direction until the first time it becomes tangent to a triangle side. Then slide it along that side of the triangle until the first time it becomes tangent to another triangle side. The result will be a circle which is tangent to two sides of the reference triangle and intersects the third, and any such circle must clearly be at least as large as the incircle.


enter image description here


The blue circle intersects all the sides of the green triangle. Now draw tangents to the blue circle parallel to the green sides. We have a magenta triangle, similar and larger than the green, for which the blue circle is inscribed. It follows that the radius of the blue circle is larger than the radius of the inscribed circle of the green triangle.

Now, when the blue circle passes through the midpoints of the green, we get the blue circle of radius $\frac{R}{2}$. Therefore, $\frac{R}{2}\ge r$.

This proof also works in $3$ d.


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