Show that the radii of three inscribed circles are always in a geometric sequence

Question

A triangle is inscribed in a circle so that three congruent circles can be inscribed in the triangle and two of the segments. Each circle is the largest circle that can be inscribed in its region.

Keeping the shortest side (in red) fixed, move the vertex opposite the shortest side along the circle. As before, each inscribed circle is the largest circle that can be inscribed in its region. Do not let the triangle become degenerate.

Show that the radii of the inscribed circles are always in a geometric sequence.

---

I made a Geogebra page where you can move the vertex and observe the circles.

I will post my solution, which involves complicated algebra. I am looking for a more elegant solution.

Answer 1

I have a proof using trigonometry. I label the points as the following figure.

Let's denote the inradius by $r$ and the radii of the three circles by $t_{A}, t_{B}, t_{C}$.

The key formulas in this proof are: $$r = 4R\sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}$$ and $$t_{A} = R{\left(\sin\dfrac{A}{2}\right)}^{2}, t_{B} = R{\left(\sin\dfrac{B}{2}\right)}^{2}, t_{C} = R{\left(\sin\dfrac{C}{2}\right)}^{2}$$

I guess I should provide a proof for these formulas.

From the incenter-excenter lemma, $D$ is the circumcenter of triangle $IBC$. $$IB\cdot IC\cdot\sin\angle BIC = 2\cdot \text{area}(IBC) = r\cdot BC$$

On the other hand

$$\sin\angle BIC = \sin\left(\dfrac{\pi}{2} + \dfrac{A}{2}\right) = \sin\left(\dfrac{\pi}{2} - \dfrac{A}{2}\right) = \cos\dfrac{A}{2}$$

and

$$BC = 2R\sin A = 4R \sin\dfrac{A}{2}\cos\dfrac{A}{2}, IB = \dfrac{r}{\sin\dfrac{B}{2}}, IC = \dfrac{r}{\sin\dfrac{C}{2}}$$

so we obtain that

$$r = 4R \sin\dfrac{A}{2}\sin\dfrac{B}{2}\sin\dfrac{C}{2}.$$

According to the law of sine

$$DX = DB\cos\dfrac{\pi - A}{2} = 2R\sin\dfrac{A}{2}\sin\dfrac{A}{2}$$

so $t_{A} = R{\left(\sin\dfrac{A}{2}\right)}^{2}$.

Back to the original problem.

$$ \begin{align*} r^{2} = t_{B}t_{C} & \iff 16R^{2}{\left(\sin\dfrac{A}{2}\right)}^{2}{\left(\sin\dfrac{B}{2}\right)}^{2}{\left(\sin\dfrac{C}{2}\right)}^{2} = R^{2} {\left(\sin\dfrac{B}{2}\right)}^{2}{\left(\sin\dfrac{C}{2}\right)}^{2} \\ & \iff {\left(\sin\dfrac{A}{2}\right)}^{2} = \dfrac{1}{16} \\ & \iff \sin\dfrac{A}{2} = \dfrac{1}{4} \end{align*} $$

This implies the desired result.

Answer 2

The result generalizes a bit. Since other answers provide detailed proofs of the original result, I'll just hit the highlights.

Let $P$ be an interior point of $\triangle ABC$. Specifically, we can write $$P = \frac{u A + v B + w C}{u+v+w}$$ for positive real constants $u$, $v$, $w$ (that we'll consider fixed throughout). Let lines $PB$ and $PC$ meet the sides $AC$ and $AB$ at $B'$ and $C'$. Let circles tangent to $AC$ and $AB$ at $B'$ and $C'$ also be tangent to the circumcircle (while not overlapping the triangle itself). Let $b$ and $c$ be the radii of these circles, and let $i$ be the triangle's inradius.

Then (barring degeneracies) $b$, $i$, $c$ are in geometric progression (that is, $i^2 = bc$) when $$\sin^2 \frac12A = \frac{u^2vw}{(u+v)^2(u+w)^2}\tag{$\star$}$$

Since $(\star)$ only involves angle $A$, if one finds any instance where the geometric progression property holds (eg, when $b=i=c$), then the property holds so long as $\angle BAC$ is an inscribed angle of fixed size.

With some Mathematica-assisted coordinate bashing, I find that the tangent circles have radii given by $$ b = \frac{4 r uw \sin^2B_2}{(u + w)^2} \qquad c = \frac{4 r uv \sin^2C_2}{(u + v)^2} \tag1$$ where $r$ is the circumradius of $\triangle ABC$, and $X_2 := X/2$ (to save typing and reduce visual clutter). Moreover, we "know" that the inradius is given by $$i = 4r\sin A_2\sin B_2\sin C_2 \tag2$$ Thus, radii $b$, $i$, $c$ are in geometric progression iff $$0 = b c-i^2 = 16r^2\sin^2B_2\sin^2C_2\left( \frac{u^2vw}{(u+v)^2(u + w)^2} - \sin^2A_2 \right) \tag3$$ from which $(\star)$ follows (since $\sin B_2$ and $\sin C_2$ are non-zero in a non-degenerate triangle). $\square$

---

In the question as stated, we note that the tangent circles touch the midpoints of the triangle's sides. Consequently, in the notation of this answer, $BB'$ and $CC'$ are medians, making $P$ the centroid, whence $u=v=w$. In this case, $(\star)$ reduces to $\csc A_2 = 4$, echoing the result in @Duong Ngo's answer.

---

There's actually no requirement that $P$ be interior to the triangle. The result holds for configurations as shown:

---

It's a bit intriguing that we can re-write $(1)$ as $$ \frac{1}{b} = \frac{1}{4r}(u + w)\left(\frac1u+\frac1w\right)\csc^2B_2 \qquad \frac{1}{c} = \frac{1}{4r}(u+v)\left(\frac1u+\frac1v\right)\csc^2C_2 \tag{1'}$$ and $(\star)$ as $$\csc^2 A_2 = (u+v)\left(\frac1u+\frac1v\right)\,(u+w)\left(\frac1u+\frac1w\right) \tag{$\star'$}$$ I wonder if there's some geometric significance to these forms.

Update. There's this: Values $u$, $v$, $w$ can be considered the barycentric coordinates of point $P$, typically expressed as $u:v:w$ (since the proportion is all that matters). Given $P$, the point with bary-coordinates $\frac1u:\frac1v:\frac1w$ is called the isotomic conjugate of $P$. It's the point whose corresponding $B'$ is the reflection of $B'$ in the midpoint of $AC$; likewise for $C'$ (and the appropriately-defined $A'$). "Obviously", the tangent circles touching these points must be congruent to the respective tangent circles for $P$. Consequently, we should expect the formulas for their radii to be unchanged by the substitutions $(u,v,w)\to(1/u,1/v,1/w)$.

So, that helps explain the forms of $(1')$ and $(\star')$, albeit not their underlying geometry. (Perhaps it would be instructive to think of the product in, for instance, the $b$ formula as $(u+w)\csc B_2 \cdot \left(\frac1u+\frac1w\right)\csc B_2$.)

---

Another tidbit: We can eliminate angle references in the formulas for $b$, $c$, $i$, and the corresponding radius $a$ for a tangent circle along side $BC$. This gives a general relation among the four radii:

$$2 \;+\;\widehat{a} \;+\;\widehat{b} \;+\;\widehat{c} \;-\;\widehat{a}\,\widehat{b}\,\widehat{c} \;=\; 0 \tag{5}$$

where, leaning into my previous observation, I define cyclically $$ \widehat{a}:=\frac{a}{i}(v+w)\left(\frac1v+\frac1w\right) \qquad \widehat{b}:=\cdots\qquad \widehat{c}:=\cdots\tag6$$ Hmmm ... That $2$ seems out of place. Perhaps this form of the equation is better: $$ \left(1+\widehat{a}\right) \;+\;\left(1+\widehat{b}\right) \;+\;\left(1+\widehat{c}\right) \;=\;1+\widehat{a}\,\widehat{b}\,\widehat{c} \tag{5'}$$

Answer 3

Self-answering. This solution involves complicated algebra. I am looking for a more elegant solution.

---

Assume that the largest circle (in which the triangle is inscribed) is a unit circle.

The length of the fixed side of the triangle

Consider the diagram with the isosceles triangle.

Let:

• $a=$ half the length of the fixed side of the triangle
• $L=$ length of one of the longer sides of the isosceles triangle
• $r=$ radius of the three inscribed circles

Pythagorus gives

$$L=\sqrt{a^2+\left(\sqrt{1-a^2}+1\right)^2}=\sqrt{2+2\sqrt{1-a^2}}$$

We have

$$r=\frac{2\times(\text{area of triangle})}{\text{perimeter of triangle}}$$

$$\frac12\left(1-\sqrt{1-\left(\frac{L}{2}\right)^2}\right)=\frac{2a\left(\sqrt{1-a^2}+1\right)}{2a+2L}$$

which is satisfied by $a=\frac{\sqrt{15}}{8}$. (So the length of the fixed side of the triangle is $2a=\frac{\sqrt{15}}{4}$, and $L=\frac{\sqrt{15}}{2}$.)

A relationship between the other two sides

Now consider the variable triangle.

Let the two variable side lengths be $2x$ and $2y$, where $x\le y$ (so $0<x\le\frac{L}{2}$).

The angles at the centre of the circle subtended by the three sides are $2\arcsin x,\space 2(\pi-\arcsin y),\space 2\arcsin a$.

$$2\arcsin x+2(\pi-\arcsin y)+2\arcsin a=2\pi$$

$$y=\sin({\arcsin x+\arcsin a})=x\sqrt{1-a^2}+a\sqrt{1-x^2}$$

Expressions for the three radii

The radii of the circles in the segments are

$$R_1=\frac12\left(1-\sqrt{1-x^2}\right)$$

$$R_2 = \begin{cases} \frac12\left(1+\sqrt{1-y^2}\right) & \text{if }\space 0<x\le\frac78 \\ \frac12\left(1-\sqrt{1-y^2}\right) & \text{if }\space \frac78<x\le\frac{L}{2} \end{cases}$$

(There are two expressions for $R_2$ because when $x=\frac78$, the longest side of the triangle is a diameter of the circle, so when $x$ increases through $\frac78$, the circle in the segment bounded by the longest side changes from being in the major segment to being in the minor segment.)

The radius of the circle in the variable triangle is (again)

$$R_3=\frac{2\times(\text{area of triangle})}{\text{perimeter of triangle}}$$

Using Heron's formula,

$$R_3=\frac{\sqrt{(x+y+a)(-x+y+a)(x-y+a)(x+y-a)}}{x+y+a}=\sqrt{\frac{(-x+y+a)(x-y+a)(x+y-a)}{x+y+a}}$$

Showing that the radii are in a geometric sequence

Then we only need to show that $R_1R_2={R_3}^2$ for $0<x\le\frac{L}{2}$. This can be checked by a graph.

Answer 4

As Jean Marie noted, this geometric series relationship between $r_a, \; r$ and $r_c$ only exists for triangles standing on a unique chord of the circle, i.e. that chord upon which the congruent inscribed circles results. Refer to the diagram on the left above.

Determining the properties of the general case, we can easily deduce from the big circle geometry:

$$ r_a = \dfrac{R}{2} (1 - \cos{A}) = R \sin^2{\dfrac{A}{2}}$$

$$ r_b = \dfrac{R}{2} (1 - \cos{B}) = R \sin^2{\dfrac{B}{2}}$$

$$ r_c = \dfrac{R}{2} (1 - \cos{C}) = R \sin^2{\dfrac{C}{2}}$$

where $R$ is the outer circle's radius.

And $\;\;\; a = 2R\;\sin{A} \;\;\;\; b = 2R\; \sin{B} \;\;\;\; c = 2R \;\sin{C}$

Looking at the inscribed circle of radius $r$, we can write: $ \;\;a = r \cot{\dfrac{B}{2}} + r \cot{\dfrac{C}{2}} $

$$ \implies r = \frac{2R\sin{A}}{\cot{\dfrac{B}{2}} + \cot{\dfrac{C}{2}} } $$

$$ $$

The congruent situation first.

Clearly for $r_a = r_c = r\;$ we must have $ \;a = c \;$ and thus $ \; A = C$.

Thus $\; B = 180^{\text{o}} - 2A \;\;$ or $\;\; \dfrac{B}{2} = 90^{\text{o}} - A $.

$$ r_a = r \;\; \implies \;\; R \sin^2{\dfrac{A}{2}} = \frac{2R\sin{A}}{\cot{\dfrac{B}{2}} +\cot{\dfrac{C}{2}}} = \frac{4R\sin{\dfrac{A}{2}} \cos{\dfrac{A}{2}}}{\tan{A} +\cot{\dfrac{A}{2}}} $$

$$ \implies \tan{\dfrac{A}{2}} = \frac{4}{\tan{A} + \cot{\dfrac{A}{2}} } $$

$$ \implies \tan{\dfrac{A}{2}} \; (\frac{2 \tan{\dfrac{A}{2}}}{1 - \tan^2{\dfrac{A}{2}}}) + 1 = 4 $$

$$ \implies 2\; \tan^2{\dfrac{A}{2}} = 3 - 3 \; \tan^2{\dfrac{A}{2}} \; \implies \tan{\dfrac{A}{2}} = \sqrt{\dfrac{3}{5}} $$

$$ \implies \tan{A} = \dfrac {2\; \sqrt{\dfrac{3}{5}} } { (1 - \dfrac{3}{5}) } = \sqrt{15} \;\; \implies \; \cos{A} = \dfrac{1}{4} = 75.5225 ^{\text{o}} $$

Hence since $\sin{\dfrac{A}{2}} = \sqrt{\dfrac{3}{8}} \;\;$ we have $\;\; r_a = r_c = r = R \; \sin^2{\dfrac{A}{2}} = \dfrac{3R}{8} $

The general situation when $b$ is fixed at its congruent length.

Keeping $b$ fixed is effectively keeping angle $B$ fixed at its congruent value of $B = 180^{\text{o}} - 2 \cos^{-1}{\dfrac{1}{4}} = 28.955^{\text{o}} $. This means that changes to angles $A$ and $C$ come at the expense of each other, i.e.

$ A + C = 2A_o \;\; \implies \;\; \dfrac{C}{2} = A_o - \dfrac{A}{2} \;\; $ where $\;\; A_o = \cos^{-1}{\dfrac{1}{4}} $

$$ r = \dfrac {2R \sin{A}} { \cot{\dfrac{B}{2}} + \cot{\dfrac{C}{2}} } = \dfrac { 4R \sin{\dfrac{A}{2}} \cos{\dfrac{A}{2}} } { \sqrt{15} + \cot{(A_o - \dfrac{A}{2}}) } = \dfrac { 4R \sin{\dfrac{A}{2}} \cos{\dfrac{A}{2}} } { \sqrt{15} + \dfrac{ 1 + \sqrt{15} \tan{\dfrac{A}{2}} } { \sqrt{15} - \tan{\dfrac{A}{2}} } } $$

$$ \require{cancel} = 4R \sin{\dfrac{A}{2}} \cos{\dfrac{A}{2}} \; (\dfrac { \sqrt{15} - \tan{\dfrac{A}{2}} } { 15 - \cancel {\sqrt{15} \tan{\dfrac{A}{2}} } + 1 + \cancel {\sqrt{15} \tan{\dfrac{A}{2}} } } ) $$

$$ \implies \; r = \dfrac { 4R \sin{\dfrac{A}{2}} \cos{\dfrac{A}{2}} (\sqrt{15} - \tan{\dfrac{A}{2}} ) } {16} = \dfrac{R}{4} \sin{\dfrac{A}{2}} \cos{\dfrac{A}{2}} (\sqrt{15} - \tan{\dfrac{A}{2}} ) $$

$$ \boldsymbol{ \require{cancel} \dfrac{r_a}{r} } = \dfrac{ \cancel{R} \sin^\cancel{2}{\dfrac{A}{2}} } { \dfrac{\cancel{R}}{4} \cancel{\sin{\dfrac{A}{2}} } \cos{\dfrac{A}{2}} (\sqrt{15} - \tan{\dfrac{A}{2}} ) } = \boldsymbol{ \dfrac{4 \tan{\dfrac{A}{2}}} {\sqrt{15} - \tan{\dfrac{A}{2}}} } $$

$$ $$

$$ r_c = R\;\sin^2{\dfrac{C}{2}} = R\;(\sin{A_o} \cos{\dfrac{A}{2}} - \cos{A_o} \sin{\dfrac{A}{2}})^2 $$

Recalling that $\;\; \sin{A_o} = \dfrac{\sqrt{15}}{4} $ and $ \cos{A_o} = \dfrac{1}{4} $ we have:

$$ r_c = R\;\cos^2{\dfrac{A}{2}} (\dfrac{\sqrt{15}}{4} - \dfrac{1}{4}\;\tan{\dfrac{A}{2}})^2 = \dfrac{R}{16}\; \cos^2{\dfrac{A}{2}}\;(\sqrt{15} - \tan{\dfrac{A}{2}})^2 $$

$$ \require{cancel} \boldsymbol{\dfrac{r}{r_c}} = \dfrac{\dfrac{\cancel{R}}{4} \sin{\dfrac{A}{2}} \cancel{\cos{\dfrac{A}{2}} } \cancel{(\sqrt{15} - \tan{\dfrac{A}{2}})}} {\dfrac{\cancel{R}}{16}\; \cos^\cancel{2}{\dfrac{A}{2}}\;(\sqrt{15} - \tan{\dfrac{A}{2}})^\cancel{2}} = \boldsymbol{ \dfrac{4 \tan{\dfrac{A}{2}}} {\sqrt{15} - \tan{\dfrac{A}{2}}} } $$

So we have:

$$ \boldsymbol{ r_a:r \; = \; r:r_c \; = \; \dfrac{ 4 \tan{\dfrac{A}{2}} } {\sqrt{15} - \tan{\dfrac{A}{2}} } } $$

which corresponds to a geometric sequence $a, ar, ar^2, \dots \; $ if we have them in the ascending order $ r_c, r, r_a$.

The sequence ratio of

$$\; \dfrac{ 4 \tan{\dfrac{A}{2}} } { \sqrt{15} - \tan{\dfrac{A}{2}} }\;$$

collapses to unity (though still mathematically a geometric sequence with ratio $1$ !) at the congruent situation where $A = C = A_o $ and it rises to $\; +\infty \;$ as $A \; \rightarrow 2A_o = 151.045^\text{o} \;$ and then increases from $ -\infty $ after this.

I know that this is far from the elegant solution sought by Dan. I have a feeling that I am missing a geometrical perspective that might simplify things. I am also curious about the neglected small circle of radius $r_b$. But we must solve a problem elaborately before we find the elegant solution . . .

Answer 5

This may be long late answer, but one realizes that even if the question is about a geometric sequence, the real question would be the following: Given a circle of center $O$, radius $R$ and a chord $[AB]$; let $C$ be a point on some arc $\overparen{AB}$. If $M$ is the midpoint of $[AC]$ and $N$ the midpoint of $[BC]$ then $$\dfrac{(R-\overline{OM})(R-\overline{ON})}{r^2}$$ is a constant $k$ independent of the position of $C$, $r$ is the inradius of triangle $ABC$. The algebraic measure $\overline{ON}$ is negative if the angle at $A$ become obtuse and the same for $\overline{OM}$ at vertex $B$.

Here for the particular case, one find some configuration for wich the constant $k$ is $4$.

A proof is presumed computational because we have the inradius and circumradius in one expression, taking $R=1$, in a coordinates system, the coordinates are as follows: $A(-a;-\sqrt{1-a^2})$, $B(a;-\sqrt{1-a^2})$, $C(-b;\pm\sqrt{1-b^2})$, $a>0$ and $b\ge 0$. We will take $C(-b;\sqrt{1-b^2})$ (the same goes for the other case). Line $(AC)$ has the equation: $$(a-b)y-(\sqrt{1-a^2}+\sqrt{1-b^2})x-b\sqrt{1-a^2}-a\sqrt{1-b^2}=0$$

Line $(BC)$ has the equation: $$(-a-b)y-(\sqrt{1-a^2}+\sqrt{1-b^2})x-b\sqrt{1-a^2}+a\sqrt{1-b^2}=0$$ Now the distance of $O$ to $(AC)$ equals: $$\dfrac{a\sqrt{1-b^2}+b\sqrt{1-a^2}}{\sqrt{2-2ab+2\sqrt{(1-a^2)(1-b^2)}}}=d$$

The distance of $O$ to $(BC)$ equals: $$\dfrac{|-a\sqrt{1-b^2}+b\sqrt{1-a^2}|}{\sqrt{2+2ab+2\sqrt{(1-a^2)(1-b^2)}}}$$ Set $$h=\dfrac{a\sqrt{1-b^2}-b\sqrt{1-a^2}}{\sqrt{2+2ab+2\sqrt{(1-a^2)(1-b^2)}}}$$ $$r^2=\dfrac{4a^2\left(\sqrt{1-a^2}+\sqrt{1-b^2}\right)^2}{\left(2a+\sqrt{2-2ab+2\sqrt{(1-b^2)(1-a^2)}}+\sqrt{2+2ab+2\sqrt{(1-b^2)(1-a^2)}}\right)^2}=\dfrac{4S^2}{P^2}$$ $S$ is the area of $ABC$ and $P$ its perimeter. Finally check that $$\dfrac{(1-h)(1-d)}{r^2}$$ is independent of $b$. Taking $b=0$ one needs to show that the previous fraction is equal to $\dfrac{1+\sqrt{1-a^2}}{2a^2}$.

Notice that $\sqrt{2+2ab+2\sqrt{(1-b^2)(1-a^2)}}=\sqrt{(1+a)(1+b)}+\sqrt{(1-a)(1-b)}$ Doing some algebra the identity to prove is $$4\left(\sqrt{1-a^2}+\sqrt{1-b^2}\right)^3(1+\sqrt{1-a^2})=\left(2a+(\sqrt{1+b}+\sqrt{1-b})(\sqrt{1-a}+\sqrt{1+a})\right)^2\left(2\sqrt{1-a^2}+2\sqrt{1-b^2}+a^2-b^2+(b-a)(\sqrt{(1+b)(1+a)}+\sqrt{(1-b)(1-a)})-(a+b)(\sqrt{(1-b)(1+a)}+\sqrt{(1+b)(1-a)})\right) $$ which is true.

Also, to verify check that $$a^2-b^2=(-\sqrt{1-a^2}+\sqrt{1-b^2})(\sqrt{1-a^2}+\sqrt{1-b^2})$$ and that $$(a+b)(\sqrt{(1-b)(1+a)}+\sqrt{(1+b)(1-a)})=(\sqrt{1-a^2}+\sqrt{1-b^2})(\sqrt{(1+a)(1+b)}-\sqrt{(1-a)(1-b)})$$ Simplify by $(\sqrt{1-a^2}+\sqrt{1-b^2})$ to get:

$$4\left(\sqrt{1-a^2}+\sqrt{1-b^2}\right)^2(1+\sqrt{1-a^2})=\left(2a+(\sqrt{1+b}+\sqrt{1-b})(\sqrt{1-a}+\sqrt{1+a})\right)^2\left(2-\sqrt{1-a^2}+\sqrt{1-b^2}-\sqrt{(1+b)(1+a)}+\sqrt{(1-b)(1-a)}-\sqrt{(1-b)(1+a)}+\sqrt{(1+b)(1-a)}\right) $$ The factors are squares as $2(1+\sqrt{1-a^2})=(\sqrt{1-a}+\sqrt{1+a})^2$ and $$2\left(2-\sqrt{1-a^2}+\sqrt{1-b^2}-\sqrt{(1+b)(1+a)}+\sqrt{(1-b)(1-a)}-\sqrt{(1-b)(1+a)}+\sqrt{(1+b)(1-a)}\right)=\left((\sqrt{1+b}+\sqrt{1-b})-(\sqrt{1+a}-\sqrt{1-a})\right)^2$$

Taking square root, we prove: $$2\left(\sqrt{1-a^2}+\sqrt{1-b^2}\right)(\sqrt{1-a}+\sqrt{1+a})=\left(2a+(\sqrt{1+b}+\sqrt{1-b})(\sqrt{1-a}+\sqrt{1+a})\right)\left((\sqrt{1+b}+\sqrt{1-b})-(\sqrt{1+a}-\sqrt{1-a})\right) $$

Computations are checked with GeoGebra.