# Crazy fact(?) about circles drawn on base of triangle between cevians: they always fit, no matter what their order?

Take any triangle, and draw any number of cevians from the top vertex to the base, with any spacing between the cevians.

In each sub-triangle thus formed, inscribe a circle. Now rearrange the order of the circles from left to right (but don't change their radii).

Adjust the cevians so that each cevian touches two neighboring circles. The circles always seem to fit perfectly in the sub-triangles, with no gaps, no matter what their order!

Is this true, and if so, why?

I tested this with different size triangles and circles, and it always seems to work. I doubt it's just a coincidence.

At first I thought there must be a simple explanation, but I haven't found one. (I asked a coworker who is quite good at geometry about this, and he didn't know what to think.)

Notice that the angle at the top vertex subtended by say the blue circle, is different in the different arrangements.

Possibly useful: the inradius of a triangle is $$\frac{2\times \text{area}}{\text{perimeter}}$$.

• If you can prove it for two circles, then I believe that the rest should follow from induction.
– Blue
Jun 1 at 16:24
• Great suggestion Blue!
– Moti
Jun 1 at 19:57
• A similar issue here Jun 1 at 21:08
• @JeanMarie: I was reminded of the same thing. The result is known as "The Japanese Theorem (for cyclic polygons)".
– Blue
Jun 1 at 22:46
• I think this may even be reduced to a problem for one circle and one triangle: show that horizontally translating the base of the triangle, while varying the points of tangency between sides of the triangle and the circle, still allows the same circle to be inscribed within the triangle while keeping the base also tangent to the circle. Kind of reminds me of a a piece of food sliding within chopsticks :)
– Amit
Jun 3 at 10:20

Here's a proof for the case $$n=2$$: as suggested by Blue the other cases follow by swapping pairs of adjacent circles.

Suppose then we have a triangle $$ABC$$ with $$AC=a$$, $$BC=b$$, $$\angle A=2\alpha$$, $$\angle B=2\beta$$ and two inscribed circles of radii $$R$$ and $$r$$ (see figure below). From $$AH=R\cot\alpha$$ and $$CH=R\cot\alpha'$$ we get: $$R\cot\alpha+R\cot\alpha'=a$$ and analogously: $$r\cot\beta+r\cot\beta'=b.$$ These equalities can be rewritten as: $$\cot\alpha'={a\over R}-\cot\alpha \quad\text{and}\quad \cot\beta'={b\over r}-\cot\beta.$$ But $$\alpha+\beta+\alpha'+\beta'=90°$$, hence $$\cot\beta'=\tan(\alpha+\beta+\alpha')$$ and $$\tan(\alpha+\beta+\alpha')={b\over r}-\cot\beta. \tag1$$ We can then expand $$\tan(\alpha+\beta+\alpha')= {\tan(\alpha+\beta)+\tan\alpha'\over1-\tan(\alpha+\beta)\tan\alpha'} ={\cot\alpha'\tan(\alpha+\beta)+1\over \cot\alpha'-\tan(\alpha+\beta)}$$ and insert here the expression for $$\cot\alpha'$$ found above. Equation $$(1)$$ becomes then $${(a/R-\cot\alpha)\tan(\alpha+\beta)+1\over a/R-\cot\alpha-\tan(\alpha+\beta)} ={b\over r}-\cot\beta,$$ that is, after some passages: $${ab\over rR}+\cot\alpha\cot\beta+ (\cot\alpha+\cot\beta)\tan(\alpha+\beta)=\\ ={b\over r}\big(\cot\alpha+\tan(\alpha+\beta)\big) +{a\over R}\big(\cot\beta+\tan(\alpha+\beta)\big)+1. \tag2$$ From equation $$(2)$$ one can solve for $$r$$ given $$R$$, and vice versa. To finish the proof we must show that $$(2)$$ is invariant under the swap $$r\leftrightarrow R$$.

The LHS of $$(2)$$ is invariant at sight. Let's consider the first term on the RHS, we have: $${b\over r}\big(\cot\alpha+\tan(\alpha+\beta)\big)= {b\over r}\left({\cos\alpha\over\sin\alpha}+ {\sin\alpha\cos\beta+\cos\alpha\sin\beta\over \cos\alpha\cos\beta-\sin\alpha\sin\beta}\right)\\ ={1\over r}{b\over\sin\alpha\cos\alpha} {\cos\alpha\cos\beta\over\cos(\alpha+\beta)}.$$ But $${b\over\sin\alpha\cos\alpha}= {2b\over\sin2\alpha}=2d,$$ where $$d$$ is the diameter of the circle circumscribed to $$ABC$$. The second term on the RHS of $$(2)$$ can also be rewritten in an analogous way, hence equation $$(2)$$ can be recast as $${ab\over rR}+\cot\alpha\cot\beta+ (\cot\alpha+\cot\beta)\tan(\alpha+\beta) =\left({1\over r}+{1\over R}\right) 2d{\cos\alpha\cos\beta\over\cos(\alpha+\beta)}+1,$$ which is clearly invariant under $$r\leftrightarrow R$$, as it was to be proved. EDIT.

Different forms of equation $$(2)$$ relating $$r$$ and $$R$$ can be found here. One of them is worth mentioning because it is much simpler and also invariant under $$r\leftrightarrow R$$: $${1\over R}+{1\over r}={r_i\over rR}+{2\over h},$$ where $$r_i$$ is the inradius of triangle $$ABC$$ and $$h$$ its altitude corresponding to base $$AB$$.

• Great! I will elaborate on your statement "the other cases follow by swapping pairs of adjacent circles". Say for example the original order of the circles from left to right is $C_1, C_2, C_3, C_4, C_5$ and you want to rearrange them as $C_4,C_5,C_1,C_3,C_2$. First take $C_4$ and switch it with its left neighbor, then switch it with its new left neighbor, and so on, until it occupies the first position on the left. Then, in a similar fashion, move $C_5$ so that it occupies the second position from the left. And so.
– Dan
Jun 1 at 23:03
• I've just found a page about the same topic, with some relations which should be equivalent to the one I proved above: artofproblemsolving.com/community/… Jun 1 at 23:18
• I am wondering if a proof can take benefit of the property illustrated by this Wolfram animation Jun 2 at 16:46
• @Intelligentipauca: The AoPS formula can be written $$(h-2r)(h-2R)=h(h-2r_i)$$ (The right-hand is constant; in fact, it's the case of the left-hand w/ $r=0$ and $R=r_i$.) This suggested to me that the distances from the incircles to the line at distance $h$ from the base are significant. The realization that such a distance is a simple product using half-angle tangents led me to a proof that accommodates the permutation aspect without induction (except insofar as the commutativity of arbitrary products of reals is itself inductively derived :).
– Blue
Jun 3 at 1:49
• (+1) I have corrected a typo: the $\alpha'$ argument of the cotangent term in the denominator of the LHS of the equation prior to eqn 2 should (I believe) be just $\alpha$. Jun 3 at 6:18

Consider a triangle with base $$PP'$$ and a parallel "top-line" through the third vertex. Let $$\theta$$ and $$\theta'$$ be, respectively, the interior half-angle at $$P$$ and the exterior half-angle at $$P'$$; and let $$h$$ and $$h'$$ be, respectively, the distances from the top-line to $$P$$ and to the triangle's incircle. Then

Fun Fact: $$h \tan\theta = h' \tan\theta' \tag{\star}$$

(Proof of the Fun Fact appears below.)

Now, we can interpret a configuration of cevians and incircles (say, three of the latter) as defining a sequence of collinear points $$A_0$$, $$A_1$$, $$A_2$$, $$A_3$$, half-angles $$\alpha_0$$, $$\alpha_1$$, $$\alpha_2$$, $$\alpha_3$$ at these points and each on the "same side" of the segment joining the point to a common point $$O$$, and distances $$h_0$$, $$h_1$$, $$h_2$$, $$h_3$$ from the "top-line" through $$O$$ to $$A_0$$ (actually, any $$A_i$$) and the successive incircles. By the Fun Fact, we have \left.\begin{align} h_0 \tan\alpha_0 &= h_1 \tan\alpha_1 \\ h_0 \tan\alpha_1 &= h_2 \tan\alpha_2 \\ h_0 \tan\alpha_2 &= h_3 \tan\alpha_3 \end{align}\quad\right\}\qquad\to\qquad \tan\alpha_3 = \frac{h_0^3\tan\alpha_0}{h_1h_2h_3} \tag{\star\star}

The "crazy" observation here is that, for a fixed $$h_0$$ and $$\alpha_0$$, any permutation of $$h_1$$, $$h_2$$, $$h_3$$ yields the same $$\alpha_3$$ in $$(\star\star)$$, because —as you may have heard— multiplication is commutative. This guarantees that the sequence of incircles is always contained within a particular "bounding triangle" (namely, the triangle with height $$h_0$$, interior half-angle $$\alpha_0$$, and exterior half-angle $$\alpha_3$$). That is,

The (bounding) triangle can be divided by cevians into sub-triangles having the given incircles in any order.

For instance, defining $$\beta_0:= \alpha_0$$ and $$(k_0,k_1,k_2,k_3) := (h_0,h_3,h_1,h_2)$$, we get this figure ... ... such that

\left.\begin{align} k_0 \tan\beta_0 &= k_1 \tan\beta_1 \\ k_0 \tan\beta_1 &= k_2 \tan\beta_2 \\ k_0 \tan\beta_2 &= k_3 \tan\beta_3 \end{align}\right\}\;\to\; \tan\beta_3 = \frac{k_0^3 \tan\beta_0}{k_1 k_2 k_3} =\frac{h_0^3 \tan\alpha_0}{h_1 h_2 h_3}=\tan\alpha_3 \;\to\; \beta_3 = \alpha_3

Proof of Fun Fact. Using different notation, and trading the exterior half-angle for the corresponding interior half-angle for conceptual symmetry ...

Consider $$\triangle ABC$$ with inradius $$r$$ and height $$h$$ relative to base $$\overline{AB}$$. Take $$A'$$ and $$B'$$ on sides $$\overline{AC}$$ and $$\overline{BC}$$ such that $$\overline{AB}\parallel \overline{A'B'}$$ and $$\overline{A'B'}$$ is tangent to the incircle. Then $$h':=h-2r$$ is the height of $$\triangle A'B'C$$ relative to base $$\overline{A'B'}$$. Defining half-angles $$\alpha := \frac12\angle A$$ and $$\beta := \frac12\angle B$$, and noting that $$\triangle ABC\sim\triangle A'B'C$$, we have

$$\frac{h'}{h}=\frac{|A'B'|}{|AB|}=\frac{r\tan\alpha+r\tan\beta}{r\cot\alpha+r\cot\beta}=\frac{\sin(\alpha+\beta)}{\cos\alpha\cos\beta}\cdot\frac{\sin\alpha\sin\beta}{\sin(\alpha+\beta)}=\tan\alpha\tan\beta$$ which is equivalent to the Fun Fact. $$\square$$

I suspect that there's a route that avoids cancelling-out $$\sin(\alpha+\beta)$$, but that's what I have.

(The first part of orangeskid's answer is a very nice alternative proof.)

• Elegant! So this geometrical fact is only as crazy as the fact that multiplication is commutative.
– Dan
Jun 3 at 2:11
• Beautiful! (+1) Jun 3 at 7:55
• From now on I'll call lemmas "fun facts" and theorems "crazy facts". Jun 3 at 9:49
• @Amit: Bear in mind that my $\theta$ and $\theta'$ are half-angles at $P$ and $P'$. (I make consistent reference to "half-angle" in my discussion, but I admit I could've made this clearer in the diagram.) For a proof of the Lemma/Fun-Fact, see, for instance, the first part of orangeskid's answer.
– Blue
Jun 11 at 22:07
• @Blue Oh!! Thanks a lot, I actually didn't want to read on before I prove the Lemma, and I missed that little half-angle line I guess :) I'll try to do it again now that I know that... but thanks for the reference to the derivation too! Much appreciated.
– Amit
Jun 11 at 22:38

Consider a triangle $$ABC$$, with radius of inscribed circle $$r$$ We have

$$\tan\frac{B}{2}\cdot \tan \frac{C}{2} = \frac{r}{p-b}\cdot \frac{r}{p-c} = \frac{r^2}{(p-b)(p-c)}$$ but $$r = \frac{S}{p} = \frac{\sqrt{p(p-a)(p-b)(p-c)}}{p}$$ so $$\tan \frac{B}{2}\cdot \tan\frac{C}{2} =\frac{p-a}{p}= 1- \frac{a}{p} = 1 - \frac{2 r}{h_a}$$

Let us denote the above expression associated to the triangle $$\Delta ABC$$ (with base $$BC$$) by $$\rho$$.

Fact ( easy) if a triangle $$\Delta= ABC$$ is divided into two triangles $$\Delta' = ABM$$, and $$\Delta'' = AMC$$ then

$$\bbox[5px,border:2px solid yellow]{\rho = \rho' \cdot \rho''}$$

(use the fact that $$\tan \alpha \cdot \tan (\frac{\pi}{2}-\alpha) = 1$$)

The conclusion is that

$$\bbox[5px,border:2px solid red]{ 1- \frac{2 r}{h} = \left(1-\frac{2r'}{h}\right) \cdot \left(1-\frac{2 r''}{h}\right) }$$

It is easy now ( and we get more: no matter how we combine triangles with some inside radiuses, the radius of the composite is given by a formula ...).

$$\bf{Added:}$$

1. The quantity $$- \log ( 1- \frac{2 r}{h})$$ behaves like the measure of the segment $$BC$$. It would be interesting find the density function ( could be a classical thing, like a potential integral). In the solution of @blue: where $$\rho$$ is a quotient, that would work better, need to take the derivative of $$\log \tan \frac{\phi}{2}$$ where $$\phi$$ is the argument of $$i - t$$, $$t\in \mathbb{R}$$. Perhaps write $$t = \tan \frac{u}{2}$$? Yes, the formula below from Kellogg holds.

2. A similar thing works in non-euclidean geometries ( spherical and hyperbolic) Tried, does not work, see details below.

3. Can try to do this for other divisions, also in higher dimensions. ?? Unlikely, the analogue problem for the potential of a triangular plaque is hard to solve.

From my calculations, the line density is $$\frac{dt}{\sqrt{h^2 + t^2}}$$. This means we are dealing with the potential function of a linear uniform charge on the line. The potential at a point $$A$$ due to a uniform charge on the segment $$BC$$ is (proportional to) $$\log \frac{1}{1- \frac{2r}{h}}$$ This problem would have been known to the ones preparing for the Cambridge tripos some 150 years ago. One wonders if we have similar results for a planar charge on the base of a triangular pyramid.

Looking at the book of Kellogg, Foundations of potential theory ( potential of a homogeneous straight wire segment) we note the formula

$$\log \frac{1}{1-\frac{2r}{h}} = 2 \operatorname{arccotanh}\frac{b+c}{a}$$

both being equal to $$\log \frac{p}{p-a}$$.

$$\bf{Added:}$$ Wanted to see if the results holds in spherical geometry . Start with some relevant formulas. We are dealing with a spherical triangle of sides $$a$$, $$b$$, $$c$$ on a sphere of radius $$1$$. If the sphere had radius $$R$$, substitute everywhere $$a\to \frac{a}{R}$$ $$\ldots$$.

If we start from the fundamental formula

$$\cos a = \cos b \cos c + \sin b \sin c \cos A$$ we get

$$\begin{eqnarray} \sin \frac{A}{2} & = & \sqrt{\frac{\sin(p-b)\sin(p-c)}{\sin b \sin c}} \\ \cos \frac{A}{2} & = & \sqrt{\frac{\sin p \sin (p-a)}{\sin b \sin c}} \end{eqnarray}$$

and so $$\begin{eqnarray} \tan\frac{A}{2} = \sqrt{\frac{\sin(p-b)\sin(p-c)}{\sin p \sin (p-a)}} \end{eqnarray}$$

Also we have in a right angle triangle ($$A = \frac{Pi}{2}$$)

$$\begin{eqnarray} \tan B = \frac{\tan b}{\sin c} \end{eqnarray}$$

Now consider the inscribed circle in the triangle $$ABC$$. Looking at one of the small triangles formed we get from the above

$$\begin{eqnarray} \tan \frac{A}{2} = \frac{\tan r}{\sin (p-a) } \end{eqnarray}$$

and so

$$\tan r = \frac{ \sqrt{ \sin p \sin (p-a) \sin (p-b) \sin (p-c) }}{ \sin p}$$

( the numerator on RHS can be called "fake area", denoted $$\mathcal{S}$$ ) so $$\tan r = \frac{\mathcal{S}}{\sin p}$$.

Also, we have the height from $$A$$, $$h_a$$

$$\frac{1}{2} \sin a \cdot \sin h_a = \mathcal{S}$$

Also, from the above

$$\tan \frac{B}{2} \cdot \tan \frac{C}{2} = \frac{\sin(p-a)}{\sin p}$$

Now, we have $$\frac{2 \tan r}{\sin h_a} = \frac{\sin a}{\sin p}$$. However, there is no formula to connect this fraction with $$\frac{\sin(p-a)}{\sin p}$$. Therein lies the problem.

I've checked for a triangle with sides $$\frac{\pi}{2}$$, $$\frac{\pi}{3}$$, $$\frac{\pi}{4}$$ that the connection between the two in-radiuses of the triangles obtained by dividing the base $$\frac{\pi}{2}$$ is Not a symmetric relation. That means an analogue result $$does not hold$$ in spherical geometry.

• Thanks, I have a few questions if you don't mind. $p$ is half the perimeter of triangle $ABC$, right? How did you get your first equation (is it a known theorem)? In your last equation, what does $h$ mean?
– Dan
Jun 3 at 3:17
• @Dan: If you draw a picture with the center of the inscribed circle at the intersection of the angle bisectors from $B$ and $C$ then it is not that difficult. Also, $p$ is the half-perimeter of the triangle. And $h= h_a$, constant for all of the triangles with vertex $A$ and opposite side on a fixed line. Jun 3 at 4:18