# Three externally touching circles have their centers on the same line and have radii $a$,$b$ and $c$ (where $a<b<c$).Prove that $b^2=ac$.

Three circles have their centers on the same line and have radii $$a$$,$$b$$ and $$c$$ (where $$a).The circle with radius $$b$$ touches the other two circles but circles with radii $$a$$ and $$c$$ do not touch each other.The three circles also have a common tangent.Prove that $$b^2=ac$$. I solved it with good deal of calculations. Is there an elementary way to do this problem

• Why should this be true? What's to keep me from, say, making the right hand circle as large as I want? Usually these sort of problems assume that the various circles have a common tangent or something like that. – lulu Sep 21 '20 at 11:14
• I forgot to draw the common tangent. Editing. – Maverick Sep 21 '20 at 11:16
• If there is a common tangent somewhere in the picture, then these problems tend to come down to some simple application of the Pythagorean Theorem. Just try to find some right triangles. – lulu Sep 21 '20 at 11:17
• Above image is by player3236 – Maverick Sep 23 '20 at 11:50

## 4 Answers

Elementary solution :

The top two right triangles in your diagram are similar. So

$$\frac{height}{base} = \frac{b-a}{b+a} = \frac{c-b}{c+b}$$

By componendo and dividendo, \begin{align} \frac{b}{a} & = \frac{c}{b} \\ b^2 & = ac \end{align}

This aside, @Somos' answer is great. Since all circles are similar, one can get any circle by dilating or contracting any other circle. The ratio of dilation is just the ratio of radii (radius is the only parameter of a circle).

In your diagram, add other direct common tangent to these three circles. The two tangents meet at P, which clearly is center of similarity (homothety).

Fun fact :

Whenever there are tangent circles inscribed in an angle like given, the radii of circles form geometric progression.

When the number of circles is any odd number, the circle exactly in the middle has radius equal to the geometric mean of radii of two circles at the end.

Do you see why this is obvious? :)

We are given a circle $$A$$ with radius $$a$$ externally tangent to a circle $$B$$ with radius $$b>a$$. The two common tangents $$T_1,T_2$$ of $$A$$ and $$B$$ meet in point $$P$$. On the other side of $$B$$ from $$A$$ is the circle $$C$$ with radius $$c>b$$ externally tangent to $$B$$.

Let $$T$$ be a similarity (or homothetic) map with center at $$P$$ which maps the center of $$A$$ to the center of $$B$$. The two common tangents $$T_1,T_2$$ are fixed by $$T$$ since they each contain $$P$$. Since $$A$$ and $$B$$ are externally tangent, $$T$$ must map $$B$$ to a circle externally tangent to $$B$$ and with tangents $$T_1,T_2$$. Thus that circle must be $$C$$.

Because ratios are unchanged by similarities, the ratios $$b/a$$ and $$c/b$$ must be equal, hence $$b^2=ac$$.

• What do you mean by 'which maps the circle with radius a to the circle with radius b'. – Maverick Sep 21 '20 at 14:50
• @Maverick: If you have a picture with only the two smaller circles (radius $a$ and radius $b$) and the two tangent lines in it, and then enlarge the picture by a factor $b/a$, then the small circle with radius $a$ becomes $a$ circle with radius $b$. The other circle scales up from radius $b$ to radius $b*b/a$. Because there is only one possible circle that touches the two tangent lines and touches the radius $b$ circle, you get that this large circle of radius $b*b/a$ must also have radius $c$. – Jaap Scherphuis Sep 21 '20 at 15:49 Let $$AK=x$$. By similar triangles $$\triangle AHE \sim \triangle AID \sim \triangle AJB$$:

$$\frac {x+a}a=\frac {x+2a+b}b=\frac{x+2a+2b+c}c$$

Subtracting $$1$$ from each fraction:

$$\frac xa = \frac{x+2a}b=\frac {x+2a+2b}c=:y$$

We have $$\frac{ay+2a}b=y = \frac{by+2b}c$$

$$acy+2ac = b^2y+2b^2$$

Factoring out $$y+2$$ gives the result.

Say common tangent length between circle A and B (IJ) is $$x$$, B and C (JK) is $$y$$ and so between A and C (IK) is $$(x + y)$$. Applying Pythagoras, between circles of radius a and b,

$$x^2 = (a+b)^2 - (b-a)^2 = 4ab$$

Applying Pythagoras, between circles of radius b and c,

$$y^2 = (b+c)^2 - (c-b)^2 = 4bc$$

Applying Pythagoras, between circles of radius a and c,

$$(x+y)^2 = (a+2b+c)^2 - (c-a)^2 = 4(b^2+ab+bc+ca)$$

i.e $$x^2 + y^2 + 2xy = 4b^2+4ca+x^2+y^2$$

i.e $$4b^2+4ca = 2xy = 8b\sqrt{ac}$$

i.e $$b^2 - 2b\sqrt{ac} + ca = 0$$

i.e $$b^2 = ca$$