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This is from another question which I started answering but which has been closed before I could finish my answer. Right now there is still a gap in my answer, so it is now my turn to ask about this problem in an attempt to fill that gap.

We have a ring of six circles with radii $a,b,c,a,b,c$, each of them touching both the previus and the next one from the outside. They also all touch the inside of a circle of radius $r$. Show that $a+b+c=r$.

Configuration

The original question described this as a sangaku, so I'll keep that tag. But I don't know a source where this came from. (Update: See comment from brainjam for a source.) Below I'll show my approach for how to prove this fact, but there is one gap that I don't have a good handle for. Also, my approach used a lot of computer algebra, so there is a high chance someone has a more intuitive approach which would be better suited to a paper and pencil computation, or perhaps even avoid computations altogether.

The whole configuration has to be point symmetric with respect to the center of the circumcircle. Let's use $O$ to denote that center, $A,B,C$ for the centers of the first three circles, and $A',B',C'$ for those of their point reflections in $O$. Define $\alpha:=\measuredangle AOB, \beta:=\measuredangle BOC, \gamma:=\measuredangle COA'$. Due to the point symmetry, $\alpha+\beta+\gamma=\pi$. As we want to use the cosine rule eventually, let's first use angle sum formulas to rewrite this in terms of trigonometric functions.

\begin{align*} \alpha+\beta+\gamma&=\pi \\ \cos(\alpha+\beta+\gamma) &= -1 \\ \cos(\alpha+\beta)\cos\gamma-\sin(\alpha+\beta)\sin\gamma &= -1 \\ \cos\alpha\cos\beta\cos\gamma-\sin\alpha\sin\beta\cos\gamma -\sin\alpha\cos\beta\sin\gamma-\cos\alpha\sin\beta\sin\gamma &= -1 \end{align*}

Now the sines are in the way here. So let me combine $\sin^2\alpha+\cos^2\alpha=1$ with the equation above to eliminate $\sin\alpha$, then do the same with $\beta$ and $\gamma$. I used Sage and resultants for this elimination. The result I got was some polynomial of total degree $12$ in the cosines of the angles, but factoring it returned just a single factor with exponent $4$ which represents the following equivalent equation:

$$2\cos\alpha\cos\beta\cos\gamma + \cos^2\alpha+\cos^2\beta + \cos^2\gamma = 1$$

Now we can apply the cosine rule to find a relation between these cosines of the angles and the radii of the circles. That is because of the triangles seen in the figure above, the edges of which are sums or differences of radii.

\begin{align*} (a+b)^2 &= (r-a)^2 + (r-b)^2 - 2(r-a)(r-b)\cos\alpha \\ (b+c)^2 &= (r-b)^2 + (r-c)^2 - 2(r-b)(r-c)\cos\beta \\ (c+a)^2 &= (r-c)^2 + (r-a)^2 - 2(r-c)(r-a)\cos\gamma \end{align*}

I can use the first of these to eliminate $\cos\alpha$, the second to eliminate $\cos\beta$ and the third to eliminate $\cos\gamma$, again using resultants. When I factor the resulting degree $12$ polynomial I get:

$$ (a + b + c - r)\cdot r^2\cdot (a - r)^2 \cdot (b - r)^2 \cdot (c - r)^2 \cdot (4abc - ar^2 - br^2 - cr^2 + r^3) = 0 $$

The first of these factors is the condition $a+b+c=r$ we are trying to prove. So we need to show that all the other factors are non-zero to conclude that the first one has to be zero. This will hopefully boil down to some reasonable non-degeneracy assumptions.

$r\neq 0$ is reasonable to assume since otherwise we don't have a proper circumcircle anyway. $r\neq a, r\neq b, r\neq c$ follow from the assumption that the inner circles are actually smaller, so their radius can't be equal to that of the circumcircle.

But how do we know $4abc - ar^2 - br^2 - cr^2 + r^3\neq 0$? Why can we rule that out? Does that term have any intuitive geometric interpretation? Also, do you have any alternative approaches to suggest for this whole problem?

Update: Rewriting that last factor as $4abc - (a+b+c-r)r^2$ (as some of the proposed answers did) we can see that if the assumption holds, that parenthesis becomes zero and the whole term becomes positive. So we might want to prove that $4abc > (a+b+c-r)r^2$ holds without building that proof on the assumption $a+b+c=r$. Perhaps some clever chain of well-argued inequalities can achieve that. I'm convinced that $\frac r6\le a,b,c\le\frac r2$ but my reasoning for the lower bound currently involves the assumption we try to prove, and anyway doesn't lead to the desired inequality in an obvious way.

Update 2: After reading the answers it becomes clear that $a+b+c=r$ can be concluded if we assume that the six inner circles don't overlap, don't have any inner points common to two of them. That condition obviously holds in the example figure, but hasn't been explicitly stated in my original wording of the problem statement.

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8 Answers 8

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By symmetry, $A'C$ is parallel and congruent to $AC'$, $BC$ is parallel and congruent to $B'C'$ and $AB$ is parallel and congruent to $A'B'$. So $A'CBAC'B'A'$ is a hexagon with parallel and equal opposite sides. Such hexagon is centrally symmetric, the center of symmetry being the point of intersection of diagonals $BB'$, $CC'$ and $AA'$. Also by symmetry $BC$ is parallel to the diagonal $AA'$ etc. As a consequence, $A'CBO$ is a parallelogram. Its opposite sides are congruent: $a+b=r-c$, which is the claim ($a$ is the radius of the blue circle, $b$ is the radius of the green one and $c$ is the radius of the red one.

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  • $\begingroup$ It seems you are using merely the point symmetry as the basis to assume the parallelism between edges and diagonals. But that's not the case in general, as this counterexample shows. So while $BC\parallel AA$ holds in this specific configuration, the symmetry alone is not enough for that. If someone could show that parallelism by taking the circumcircle into account as well, then we'd have a nice alternate answer. $\endgroup$
    – MvG
    Commented Dec 7, 2023 at 7:20
  • $\begingroup$ You are absolutely right. An additional argument is needed, will think about it. $\endgroup$
    – GReyes
    Commented Dec 9, 2023 at 2:38
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Set up a coordinate system centered at $O$ and with $OA$ as $x$-axis. We have then $A=(r-a,0)$, $A'=(-r+a,0)$ and the conditions of tangency for the circles at $B$ and $C$ give a system of five equations: $$ \cases{ x_B^2 + y_B^2 = (r - b)^2\\ \\ x_C^2 + y_C^2 = (r - c)^2\\ \\ (x_B - x_C)^2 + (y_B - y_C)^2 = (b+c)^2\\ \\ (x_B - r + a)^2 + y_B^2 = (a + b )^2\\ \\ (x_C + r - a)^2 + y_C^2 = (a + c )^2 } $$ I solved this for $x_B, y_B,x_C,y_C,c$ with the help of Mathematica: the system is apparently of tenth degree, but there are only four solutions. And only two of them are really different, the other two solutions being the same but with reversed signs of $y_B$ and $y_C$. The first solution is the "good" one: $$ c = r - a - b\\ x_B = r - {r + a\over r - a}b\\ x_C = a - {r + a\over r - a}b\\ y_B=y_C = 2{\sqrt{r a b (r - a - b)}\over(r - a)} $$ The second solution, instead, gives: $$ y_C=-y_B {r (r- 2a) \over r^2 - 4 a b} $$ and implies $y_B$ and $y_C$ to have opposite signs, because $r>2a$ and $r>2b$. But this is not geometrically acceptable, because some of the inner circles would intersect (see figure below), hence this solution must be discarded.

enter image description here

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  • $\begingroup$ +1. That $y_B=y_C$ is a surprise. $\endgroup$
    – brainjam
    Commented Dec 10, 2023 at 23:26
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    $\begingroup$ Can you describe more about the solution process? Was that just a Mathematica solve call, or a more manual / guided work? Initially I assumed using coordinates instead of trigonometric functions to be the key distinguishing factor of your approach, but even with that if I eliminate $x_B,y_B,x_C,y_C$ I'm pretty much back with the expressions I had, except for different exponents and omission of $(b-r)(c-r)$. I can see how to get the equation for $y_B$ by eliminating $c,x_B,x_C,y_C$, but resultants give me additional terms for $y_C$ and I can't read your beautiful second solution from that yet. $\endgroup$
    – MvG
    Commented Dec 11, 2023 at 18:06
  • $\begingroup$ @MvG I got the result with Mathematica. And I was wrong: we get only two linear equations, not three. Hence there must be some other cancellation I haven't been able to track down. $\endgroup$ Commented Dec 11, 2023 at 21:17
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This is an answer to the question of what to do with the factor $4abc - (a+b+c-r)r^2$. Rather than try to show that it is non-zero, let's see what happens when it is zero. Solving for $c$, we get $$ c=\dfrac{r^2(a + b - r)}{4a b - r^2} $$

This describes the red circle $(E)$ in the figure below. It fits the bill of being tangent to circles (B) and (A') and the outer circle $(O)$, but, like four other factors in OP's 12 degree polynomial it has a property we don't like .. the tangency with $(O)$ is on the wrong side. So we can assume this factor is non-zero, leaving $(a + b + c - r)$ as the zero factor which describes circle $(C)$.

enter image description here

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  • $\begingroup$ Nice! I wish I could accept both your answer and that from intelligenti pauca since in yours I better understand where it's coming from, but it has a lingering feeling of “proof by example” which the sign analysis in that other answer avoids. However, I think we can argue that there will always be two circles touching the three like in your figure, that out of these two we always want the smaller one, and that this smaller one always corresponds to the factor without the $4abc$. With this I'm convinced your argument holds in general. $\endgroup$
    – MvG
    Commented Dec 12, 2023 at 7:06
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The aim of our answer here is to show that it is possible to find three different values for $r$ which make the factor in question equal zero. We obtained these expression for $r$ by solving the cubic equation, $$r^3-\left(a+b+c\right)r^2+4abc=0.$$

All three roots of this equation are real and they are, $$r_1 = \dfrac{a+b+c}{3}\left(1-2\cos\left(\dfrac{1}{3}\cos^{-1}\left(\dfrac{54abc}{\left(a+b+c\right)^3}-1\right)\right)\right),\qquad\qquad\qquad$$ $$r_2 = \dfrac{a+b+c}{3}\left(1-2\cos\left(\dfrac{1}{3}\left(\pi+\cos^{-1}\left(\dfrac{54abc}{\left(a+b+c\right)^3}-1\right)\right)\right)\right), \quad\text{and}$$ $$r_3 = \dfrac{a+b+c}{3}\left(1-2\cos\left(\dfrac{1}{3}\left(\pi-\cos^{-1}\left(\dfrac{54abc}{\left(a+b+c\right)^3}-1\right)\right)\right)\right).\qquad\quad$$

We also know that, $$r_1+r_2+r_3=a+b+c,\qquad\quad\tag{1}$$ $$r_1r_2+r_2r_3+r_3r_1=0, \qquad\text{and}\tag{2}$$ $$r_1r_2r_3=-4abc.\qquad\quad\qquad\qquad\tag{3}$$

Using (3), we can state that exactly one or all three roots are negative. However, in order to hold (1) and (2) valid, we cannot have more than one negative root. For instance, for $a=1$, $b=2$, and $c=3$, we get, $$r_1 = 2-4\cos\left(20^o\right)=-1.75877,\qquad$$ $$r_2 =2+4\cos\left(80^o\right)=2.69459, \quad\text{and}$$ $$r_1 = 2+4\cos\left(40^o\right)=5.06418.\qquad\quad$$

We actually hoped to find three negative roots or one negative real root and two complex roots, with which we could discard the mentioned factor.

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    $\begingroup$ Descartes' Rule of Signs tells us right away that there's exactly one negative root: substituting $r\to-r$ gives a polynomial with a single coefficient sign change. $\endgroup$
    – Blue
    Commented Dec 9, 2023 at 22:36
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The idea is to fix the black circle, the two blue circles, and the green circle, and somehow construct the red circle.

Denote the tangency points $D, P, Q, R$ as in the picture. Let $S$ be the second intersection of $PD$ with the black circle and $T$ be the second intersection of $PD$ with the green circle.

Note that $\angle CTD = \angle TDC = \angle PDA' = \angle A'PD$, hence $CT \parallel PQ$.

It follows that $\angle CRT = \angle RTC = \angle RQO = \angle ORQ$, which shows that $R, T, Q$ are collinear.

Let $V$ be the intersection of $CT$ and $OS$. Note that $\angle VTS = \angle OPS = \angle PSO = \angle TSV$, hence $VT=VS$. Moreover $\angle TDC = \angle CTD = \angle VTS = \angle TSV$, hence $VS \parallel CD$.

Since $CT\parallel PQ$ and $VS \parallel CD$, quadrilateral $OVCA'$ is a parallelogram. Since $\vec{OA}=\vec{A'O}=\vec{CV}$, quadrilateral $CVAO$ is a parallelogram as well. This shows $RO \parallel AV$.

Let $U$ be the intersection of $RQ$ and $AV$. Note that $\angle AUQ = \angle ORQ = \angle RQO = \angle UQA$, hence $AU=AQ$ and $U$ lies on the blue circle.

Note that $\angle VUT = \angle AUQ = \angle UQA = \angle UTV$, hence $VU=VT$.

We have shown that $VS=VT=VU$, hence $U, S, T$ lie on a circle with center $V$. Since $CTV$, $OVS$, and $VUA$ are straight lines, this circle is tangent to the green circle, black circle and right blue circle. As there exists only one circle externally tangent to the green circle and to the right blue circle and internally tangent to the shorter arc $RQ$ of the black circle, it follows that the circle with center $V$ passing through $U, S, T$ coincides with the red circle. Therefore $V=B$ and $BT=b$.

To finish the proof simply note that

$$a+b+c=AQ+CT+TV=AQ+CV=AQ+OA=OQ=r.$$

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  • $\begingroup$ Nice construction, +1! However, I think your argument is slightly different from what I was asking for. You explain that if we start with (a construction that will imply) $a+b+c=r$, then we get the described configuration of circles. My question was if we have such a described configuration of circles, then $a+b+c=r$ must follow. In other words, your answer is missing the part about how there can't be any other way to get a red circle that satisfies the configuration but doesn't follow from your construction. We need an argument to discuss that other circle and declare it invalid for a reason. $\endgroup$
    – MvG
    Commented Dec 14, 2023 at 10:13
  • $\begingroup$ @MvG Not sure if I understand what you mean. Are you asking why the circle I constructed coincides with the red circle given in the problem statement? This follows from the observation that there exists a unique circle tangent internally to the black circle and tangent externally to the green and blue circle. $\endgroup$
    – timon92
    Commented Dec 14, 2023 at 14:09
  • $\begingroup$ As the answer from brainjam and the answer from Intelligenti pauca show, there are in general two such circles with those tangentiality constraints. The second one is undesirable because it overlaps its own reflection / lies on the wrong side / touches the big circle in the wrong order, but without further argument it is not clear that your construction leads to the desirable one and not that other one. $\endgroup$
    – MvG
    Commented Dec 14, 2023 at 14:22
  • $\begingroup$ @MvG oh, I see. $\endgroup$
    – timon92
    Commented Dec 14, 2023 at 14:28
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What you find below is an alternative approach to prove $a+b+c=r$. To solve this seemingly difficult problem, we resort to inversive geometry heeding the advice of Jacobi, who once said, “You must always invert”$^\bf{1}$.

CircleInversion

For the circle inversion configuration shown in $\mathrm{Fig.\space 1}$, we have, $$O_{\small{B}}P\centerdot O_{\small{B}}\bar{P} = R^2.\qquad\space\space \tag{1}$$ $$\bar{\bar{r}}=\dfrac{R^2r}{O_{\small{B}}O^2 – r^2},\qquad \text{and}\tag{2}$$ $$r=\dfrac{R^2\bar{r} }{O_{\small{B}}\bar{O}^2-\bar{r}^2}.\qquad\quad\space \tag{3}$$

All three formulae come in handy in obtaining the required result for the scenario shown in $\mathrm{Fig.\space 2}$. After letting the radii of four of the seven circles mentioned in the problem statement $\Gamma_0$, $\Gamma_1$, $\Gamma_2$, and $\Gamma_4$ equal $r$, $a$, $b$, and $a$ respectively, we are looking for the hitherto unknown radius $c$ of a missing fourth circle $\Gamma_3$, which is tangent to $\Gamma_0$, $\Gamma_2$, and $\Gamma_4$.

Sangaku6Circles

To put it in a nutshell, our aim is to use (3) to calculate radius $c$ of the circle $\Gamma_3$. This requires radius $\bar{c}$ of the inverse circle of $\Gamma_3$ and the distance $O\bar{O}_3$. The former can be found through a construction and we hope to find the latter by applying Pythagoras theorem to $\triangle \bar{O}_3GO$.

As for the execution of the mentioned construction, we select a base circle $\Gamma$ of radius equal to $2r$, center of which is locate at the point of tangency $O$ of $\Gamma_0$ and $\Gamma_4$. This is because the inversions of $\Gamma_0$ and $\Gamma_4$ yield two straight lines Line-$0$ and Line-$4$ perpendicular to the line $OO_0$. These two lines meet the extended $OO_0$ at $A$ and $\bar{P}$. Since $\bar{P}$ is the inverse of $P$, we can find the perpendicular distance $A\bar{P}$ between the these two lines using (1) as shown below. $$A\bar{P}=O\bar{P}-OA=\dfrac{4r^2}{OP}-2r =\dfrac{4r^2}{2a}-2r=\dfrac{2r\left(r-a\right)}{a}.$$

Since the circle $\Omega_3$, which is the inversion of the circle $\Gamma_3$, is tangent to both $\Omega_0$ and $\Omega_4$, its diameter is equal to $A\bar{P}$. In other words, its radius $\bar{c}$ is given by, $$\bar{c}=\dfrac{A\bar{P}}{2}=\dfrac{r\left(r-a\right)}{a}.\tag{4}$$

Using (2), we can deduce the radius $\bar{a}$ of the circles $\Omega_1$, which is the inversion of the circle $\Gamma_1$. $$\bar{a}=\dfrac{4r^2a}{OO_1^2-a^2}=\dfrac{4r^2a}{\left(2r-a\right)^2-a^2}=\dfrac{ra}{r-a}\tag{5}$$

In a similar vein, using (2), we can express the radius $\bar{b}$ of the circles $\Omega_2$, which is the inversion of the circles $\Gamma_2$. $$\bar{b}=\dfrac{4r^2b}{OO_2^2 – b^2}\tag{6}$$

Consider $\triangle O_2O_0O$. Lengths of two of its sides are known, i.e., $O_0O=r$ and $O_2O_0=r-b$. Let $\angle O_1O_0O_2=\phi$. According to the law of cosines, we can write, $$OO_2^2-b^2=r^2+\left(r-a\right)^2+2r(r-a)\cos\left(\phi\right)-b^2=2r\left(r-b\right)\left(1+\cos\left(\phi\right)\right).\tag{7}$$

All three sidelengths of $\triangle O_1O_0O_2$ are known, i.e., $O_1O_0=r-a$, $O_0O_2=r-b$, and $O_2O_1=a+b$. Therefore, using cosine rule once again, we obtain, $$1+\cos\left(\phi\right)=1+\dfrac{\left(r-a\right)^2+\left(r-b\right)^2-\left(a+b\right)^2}{2\left(r-a\right)\left(r-b\right)}= \dfrac{2r\left(r-a-b\right)}{2\left(r-a\right)\left(r-b\right)}.\tag{8}$$

Now, using (7) and (8), we can express $OO_2^2-b^2$ in terms of $r$, $a$, and $b$. $$OO_2^2-b^2=2r\left(r-b\right)\dfrac{2r\left(r-a-b\right)}{2\left(r-a\right)\left(r-b\right)}=\dfrac{4r^2\left(r-a-b\right)}{r-a}\tag{9}$$

Substituting (9) in (6) yields, $$\bar{b}=4r^2b \dfrac{r-a}{4r^2\left(r-a-b\right)}=\dfrac{b\left(r-a\right)}{r-a-b}.\tag{10}$$

Since we now know $\bar{a}$, $\bar{b}$, and $\bar{c}$ from (5), (10), and (4), we can easily locate the positions of $\bar{O_1}$, $\bar{O_2}$, and $\bar{O_3}$ using appropriate geometric constructions. However, as mentioned above, we are still need to determine the length of $O\bar{O}_3$ algebraically, if we are to use (3) to find $\bar{c}$ as shown below. $$c= \dfrac{4r^2\bar{c}}{O\bar{O}_3^2-\bar{c}^2}, \quad\text{where}\quad O\bar{O}_3^2= OG^2+ G\bar{O}_3^2\tag{11}$$

So, before applying Pythagoras theorem to the right angled triangle $\bar{O}_3GO$ to determine $O\bar{O}_3$, we must find $OG$ and $G\bar{O}_3$. $OG$ is easy to find.

$$OG = OA+AG=2r+\bar{c}\qquad\rightarrow\qquad OG^2=4r^2+4r\bar{c}+\bar{c}^2\tag{12}$$

By applying Pythagoras theorem to the pair of right angled triangles $\bar{O}_2F\bar{O}_3$ and $\bar{O}_1A\bar{O}_2$, we can obtain $G\bar{O}_3$. $$E\bar{O}_2=2\sqrt{\bar{a}\bar{b}}\qquad\text{and}\qquad\bar{O}_2F= 2\sqrt{\bar{b}\bar{c}}.$$ $${\large\bf{{\therefore}}} \quad G\bar{O}_3=E\bar{O}_2+\bar{O}_2F= \sqrt{\bar{b}}\left(\sqrt{\bar{a}} +\sqrt{\bar{c}}\right)$$

Replacing $\bar{a}$, $\bar{b}$, and $\bar{c}$ with (5), (10), and (4), we get, $$\small{G\bar{O}_3=2\left(\sqrt{\dfrac{ra}{r-a}\dfrac{b\left(r-a\right)}{r-a-b}}+\sqrt{\dfrac{b\left(r-a\right)}{r-a-b}\dfrac{r\left(r-a\right)}{a}}\right)=2r\left(\sqrt{\dfrac{rb}{a\left(r-a-b\right)}}\right)}.\tag{13}$$

Using (12) and (13), we shall write, $$\small{O\bar{O}_3^2-\bar{c}^2= OG^2+ G\bar{O}_3^2-\bar{c}^2=4r^2+\dfrac{4r^2\left(r-a\right)}{a}+ \dfrac{4r^3b}{a\left(r-a-b\right)}=\dfrac{4r^3\left(r-a\right)}{ a\left(r-a-b\right)}}.\tag{14}$$

Finally, we substitute (4) and (14) in (11) to express $c$ in terms of $r$, $a$, and $b$. $$c=\dfrac{4r^2\bar{c}}{O\bar{O}_3^2-\bar{c}^2} =\dfrac{\dfrac{4r^3\left(r-a\right)}{a}}{\dfrac{4r^3\left(r-a\right)}{a\left(r-a-b\right)}}=r-a-b$$

$\small{{\bf{^1}}\text{ Men of Mathematics by E.T. Bell, vol II, page 355}}$

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  • $\begingroup$ I see you like, as I do, inversion transform ; see my recent answer there. $\endgroup$
    – Jean Marie
    Commented Jan 8 at 20:02
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enter image description here

Solution using the cosine rule, would help if there were a diagram, but i cannot obtain one.

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  • $\begingroup$ The original question had this image for illustration. The symbols seem to match your solution here, with the $O_i$ as centers of the six circles. Where did you get that solution from? Please cite your sources. $\endgroup$
    – MvG
    Commented Dec 10, 2023 at 12:20
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    $\begingroup$ Ah, I see this now as a part of the document linked in brainjam's comment. Your answer should cite that source. There is a figure at the bottom of that page. I'm a bit surprised that their factorisation has $16abc$ instead of the $4abc$ that I got. Also, it's not clear to me why they decide that they can ignore that factor. $\endgroup$
    – MvG
    Commented Dec 10, 2023 at 12:29
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    $\begingroup$ @MvG I checked your solution with Mathematica and it is correct, so $16$ is definitely wrong. $\endgroup$ Commented Dec 10, 2023 at 17:50
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I am not sure if I understood properly. A simple algebraic calculation is done here. Given that

$$(a + b + c - r)\cdot r^2\cdot (a - r)^2 \cdot (b - r)^2 \cdot (c - r)^2 \cdot (4abc - ar^2 - br^2 - cr^2 + r^3) = 0 \tag 1 $$

The 12th degree polynomial above proved by @MvG has six factors, two factors (first and last) need only be considered to take it further from here as the rest are positive. It is given that

$$(a + b + c - r) \cdot (4abc - ar^2 - br^2 - cr^2 + r^3) = 0 $$

There are two possibilities:

Case 1)

If first factor $(a+b+c-r)=0 , ~$ is given then show second factor

$$ 4abc -r^2(a+b+c-r)~ \ne 0 \tag 2 $$

This is easy to see as $(a,b,c)$ are positive, $4abc>0$.

Case 2)

If second factor $$ 4abc -r^2(a+b+c-r)= 0 ~ \tag 3 $$

is given, then show first factor

$$(a+b+c-r)\ne ~0 $$

In fact from 3) $$ (a+b+c-r) = \frac{4abc}{r^2} \ne ~0, \tag 4 $$ because $ (a,b,c,r) $ are all non-zero positive.

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  • $\begingroup$ We want to exclude the second case. Show that while it is algebraically sounds it doesn't fit the described geometry. $\endgroup$
    – MvG
    Commented Dec 10, 2023 at 12:22
  • $\begingroup$ How so? Given $ a1=a2=a , b\to 0~\text{or}~c\to 0~$ then circle $b$ does not touch circle $ a$ which violates given condition of tangency. Any attempt to make up a simple geometric sketch would not fit in as packed circles. $\endgroup$
    – Narasimham
    Commented Dec 10, 2023 at 14:41
  • $\begingroup$ It looks like you have not properly understood the question posed at the end the problem statement. After successfully arguing and discarding the six roots given by the factors $\left(a-r\right )^2$, $\space\left(b-r\right)^2$, and $\space\left(c-r\right)^2$, OP wants to know whether the last factor $r^3-\left(a+b+c\right)r^2+4abc$ is non-zero for $\bf{\text{reasonable}}$ positive real values of $r$. $\endgroup$
    – YNK
    Commented Dec 10, 2023 at 15:36

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