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$c_1$ centered at $A$ passing through $B$.

$BB′$ is a diameter of $c_1$.

$T$ a random point in segment $BB′$.

$c_2$ centered at $B′$ passing through $T$.

$c_3$ centered at $B$ passing through $T$.

$c_4$ tangent externally to $c_2$ and $c_3$ and internally tangent to $c_1$

$F$ is center of $c_4$ and $H,I$ are tangency points.

enter image description here

It is clear to me that $Z = HI \cap AF$ is the second homothety center of $c_1$ and $c_4$ and I would like to prove that it also lies in that line perpendicular to $AB$ through $T$.

important related result that you probably should know: Show these three circles share their external common tangent lines

This seems to be a general result about soddy circles

hellofriends
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2 Answers2

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  • Let common tangent at $T$ meet $AF$ at $Y$ and let perpendicular to $AB$ through $F$ meet $AB$ at $L$. enter image description here Then we calculate $y=LT$ by Pythagoras theorem: $$ B'F^2-B'L^2 = LF^2 =BF^2-BL^2$$ so $$ (b+c)^2-(b-y)^2 = (2a+b+c)^2-(2a+b+y)^2$$ and so we get $$y= {ac\over a+b}$$ so $${AY\over FY} = {AT\over LT} = {a\over y} = {a+b\over c}$$

  • On the other hand let $X$ be in $HI\cap AF$. enter image description here Homothety $H_1$ at $H$ and coefficient ${b\over c}$ takes $F$ to $B'$ and homothety $H_2$ at $G$ and coefficient ${a+b\over b}$ takes $B'$ to $A$, so composition $H_2\circ H_1$ takes $F$ to $A$ and has center at $FA\cap GH =X$. This composition has coefficent $${a+b\over b}\cdot {b\over c} = {a+b\over c}$$ so $X$ divides $AF$ in the same ratio as $Y$ and thus $X=Y$ and we are done.

nonuser
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  • -@Aqua, I request you to please intervene in this question and say whether the downvote on my answer is justified or not. Or mistake did I make? – SarGe Aug 21 '20 at 14:35
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    @Aqua, do you mean $H_1$ takes $F$ to $B'$ rather than $B$? – brainjam Aug 21 '20 at 17:07
  • @brainjam Yes, thanks. – nonuser Aug 21 '20 at 17:08
  • @Aqua, how do we know that $X$ is the center of the composition? – brainjam Aug 21 '20 at 17:29
  • since it takes F to A it must be on FA and since it is also on line between centers i.e. H and G it lies also on the line HG = line HI @brainjam – nonuser Aug 21 '20 at 18:48
  • @Aqua, thanks - that explains why $X$ is a fixed point (and thus center) of the composition. Hopefully fInal question: how do we know $HG=HI$? – brainjam Aug 21 '20 at 19:25
  • https://math.stackexchange.com/questions/3781074/show-these-three-circles-share-their-external-common-tangent-lines @brainjam – nonuser Aug 21 '20 at 19:28
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The argument in Aqua's answer can be shortened as follows. We use the same point names, but here $a,b,c$ are the radii of the circles centered at $A,B',F$ respectively (this changes the meaning of $a$). Let $LT:TA$ be $x$.

As described in Yiu's Triangle Geometry, pg 2, the internal homothetic center $X$ (aka internal center of similitude) of two circles $O(R),I(r)$ divides the segment $OI$ in the ratio $R:r$. Thus the internal homothetic point of $F(c),A(a)$ divides $FA$ in the ratio $c:a$.

Using Pythagoras' Theorem as in Aqua's answer we get

$$ (b+c)^2-(b-x(a-b))^2=(2a=b+c)^2-(2a-b+x(a-b))^2 $$

Solving for $x$ (using an online solver if we are lazy) we get $x=\dfrac{c}{a}$. Thus

$$ FY:YA = LT:TA = x = c:a, $$

so $Y$ is the internal homothetic center of $c_1,c_4$.

brainjam
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