Suppose $ABCD$ is a parallelogram. Consider circles $w_1$ and $w_2$ such that $w_1$ is tangent to segments $AB$ and $AD$ and $w_2$ is tangent to segments $BC$ and $CD$. Suppose that there exists a circle which is tangent to lines $AD$ and $DC$ and externally tangent to $w_1$ and $w_2$. Prove that there exists a circle which is tangent to lines $AB$ and $BC$ and also externally tangent to circles $w_1$ and $w_2$. Proposed by Ali Khezeli
Problem
Source: Iran TST 2012-Second exam-1st day-P3
Tags: geometry, parallelogram, geometric transformation, homothety, trigonometry, geometry proposed
paul1703
12.05.2012 16:28
Pretty sure $w_1, w_2$ are supposed to be tangent or some other condition....
goodar2006
12.05.2012 19:05
paul1703 wrote: Pretty sure $w_1, w_2$ are supposed to be tangent or some other condition.... Would you please show it on a picture if you think it's wrong?
paul1703
12.05.2012 19:18
here is the pic:D Edit: OK, i got it so basically you have to prove that the intersection point of a line and a hiperbola exists
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goodar2006
12.05.2012 19:46
paul1703 wrote: here is the pic:D What you didn't notice is that the fourth circle can be tangent to Lines $AB,BC$, which means it can also be on the other side of those two circles $w_1,w_2$
goodar2006
17.05.2012 09:17
Again I must say this: goodar2006 wrote: What you didn't notice is that the fourth circle can be tangent to Lines $AB,BC$, which means it can also be on the other side of those two circles $w_1,w_2$
yunxiu
17.05.2012 10:44
goodar2006 wrote: Again I must say this: goodar2006 wrote: What you didn't notice is that the fourth circle can be tangent to Lines $AB,BC$, which means it can also be on the other side of those two circles $w_1,w_2$ Thx.
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ArefS
25.05.2012 21:01
let $w_3$ be the circle that is tangent to $w_1,w_2,AD,DC$. let $R_1,R_2,R_3$ be the radius of three circles $w_1,w_2,w_3$. let $h_1,h_2$ be the altitudes of the parallelogram respective to sides $AB,AD$. let $M,N,P,Q,R,S,T,U$ respectively be: $w_1\cap w_3,w_2\cap w_3,AB\cap w_1,AD\cap w_1,BC\cap w_2,CD\cap w_2,AD\cap w_3,CD\cap w_3$ let $Q',S'$ be the diametrically opposite points of $Q,S$ wrt circles $w_1,w_2$. with some obvious homothety arguments we can see that triples $(P,M,U)$, $(Q',M,T)$, $(T,N,R)$, $(U,N,S')$ are collinear. let $l$ be the length of common external tangent of $w_1,w_2$. let $l_1,l_2$ be the length of tangents from $U,T$ to the circles $w_1,w_2$ respectively. it is obvious that: $TQ=\sqrt{TM\cdot TQ'},US=\sqrt{UN\cdot US'}$ and $l_1=\sqrt{UM\cdot UP},l_2=\sqrt{TN\cdot TR}$ because of Casey's Theorem the existence of the circle $w_3$ is equivalent to: $l\cdot TU+TQ\cdot US=l_1\cdot l_2$ so: $l\cdot TU+\sqrt{TM\cdot TQ' \cdot UN \cdot US'}=\sqrt{UM\cdot UP\cdot TN \cdot TR}$ using sin theorem we get that: $l\cdot 2R_3\cdot \sin (90-\frac{\angle {ADC}}{2})+2R_3\sqrt{US'\cdot\sin (\angle{NUS})\cdot TQ'\cdot \sin (\angle{MTQ})}=2R_3\cdot \sqrt{UP\cdot \sin (\angle{MUS})\cdot TR\cdot \sin (\angle{NTQ})}$ equivalently: $l\cdot \cos (\frac{\angle ADC}{2})+\sqrt{2R_1\cdot2 R_2}=\sqrt{h_1\cdot h_2}$. and this equation is also equivalent to existence of circle $w_4$ which is tangent to $w_1,w_2,AB,BC$. Hence we are done.
DVDthe1st
26.09.2013 04:39
Are there any synthetic ways to approach this beautiful problem?
jayme
26.09.2013 16:32
Dear Mathlinkers, I have this idea that perhaps it should be a link with Pedoe D., The most “elementary” theorem of Euclidean geometry, Math. Mag., 4 (1976) 40–42 If someone want to investigate this possibility, welcome... Sincerely jean-Louis
Stephen
05.04.2014 16:39
ArefS wrote: $l\cdot \cos (\frac{\angle ADC}{2})+\sqrt{2R_1\cdot2 R_2}=\sqrt{h_1\cdot h_2}$. and this equation is also equivalent to existence of circle $w_4$ which is tangent to $w_1,w_2,AB,BC$. Hence we are done. Could anyone explain why this last step works? Thanks.
guptaamitu1
25.09.2022 20:29
ArefS wrote: let $w_3$ be the circle that is tangent to $w_1,w_2,AD,DC$. let $R_1,R_2,R_3$ be the radius of three circles $w_1,w_2,w_3$. let $h_1,h_2$ be the altitudes of the parallelogram respective to sides $AB,AD$. let $M,N,P,Q,R,S,T,U$ respectively be: $w_1\cap w_3,w_2\cap w_3,AB\cap w_1,AD\cap w_1,BC\cap w_2,CD\cap w_2,AD\cap w_3,CD\cap w_3$ let $Q',S'$ be the diametrically opposite points of $Q,S$ wrt circles $w_1,w_2$. with some obvious homothety arguments we can see that triples $(P,M,U)$, $(Q',M,T)$, $(T,N,R)$, $(U,N,S')$ are collinear. let $l$ be the length of common external tangent of $w_1,w_2$. let $l_1,l_2$ be the length of tangents from $U,T$ to the circles $w_1,w_2$ respectively. it is obvious that: $TQ=\sqrt{TM\cdot TQ'},US=\sqrt{UN\cdot US'}$ and $l_1=\sqrt{UM\cdot UP},l_2=\sqrt{TN\cdot TR}$ because of Casey's Theorem the existence of the circle $w_3$ is equivalent to: $l\cdot TU+TQ\cdot US=l_1\cdot l_2$ so: $l\cdot TU+\sqrt{TM\cdot TQ' \cdot UN \cdot US'}=\sqrt{UM\cdot UP\cdot TN \cdot TR}$ using sin theorem we get that: $l\cdot 2R_3\cdot \sin (90-\frac{\angle {ADC}}{2})+2R_3\sqrt{US'\cdot\sin (\angle{NUS})\cdot TQ'\cdot \sin (\angle{MTQ})}=2R_3\cdot \sqrt{UP\cdot \sin (\angle{MUS})\cdot TR\cdot \sin (\angle{NTQ})}$ equivalently: $l\cdot \cos (\frac{\angle ADC}{2})+\sqrt{2R_1\cdot2 R_2}=\sqrt{h_1\cdot h_2}$. and this equation is also equivalent to existence of circle $w_4$ which is tangent to $w_1,w_2,AB,BC$. Hence we are done.
I think there's a small typo in the solution. It should
$$ \sin \left( \frac{\angle ADC}{2} \right) $$instead of
$$ \cos \left( \frac{\angle ADC}{2} \right) $$Reason is that
$$ TU = 2R_3 \cdot \sin \left( \frac{\angle ADC}{2} \right) $$and not
$$ TU = 2R_3 \cdot \sin \left( 90^\circ - \frac{\angle ADC}{2} \right) $$So instead of
$$ l\cdot \cos \left(\frac{\angle ADC}{2} \right)+\sqrt{2R_1\cdot2 R_2}=\sqrt{h_1\cdot h_2} $$our final equation should be
$$ \boxed{ l\cdot \sin \left(\frac{\angle ADC}{2} \right)+\sqrt{2R_1\cdot2 R_2}=\sqrt{h_1\cdot h_2} } $$
I have a doubt in the solution. In particular, I can't get how to prove converse direction. i.e. suppose we are given
I tried defining $\omega_3$ such that its tangent to three of the clines and then proving its also tangent to the fourth by invoking $(\star)$, but it didn't work.
There are some easier ways to conclude stuff like
\begin{align}
l_1 := p(U,\omega_1) &= \sqrt{2R_3 \cdot h_1} \\
l_2 := p(V,\omega_2) &= \sqrt{2R_3 \cdot h_2} \\
US &= 2 \sqrt{ R_3 \cdot R_2} \\
TQ &= 2 \sqrt{R_3 \cdot R_1}
\end{align}
Proof of $(1)$ : Since tangent to $\omega_3$ at $U$ is parallel to $AB$, so there exists a inversion $\Phi$ swapping $\omega_3$ and $AB$. Suppose $\ell_1,\ell_2$ are tangents from $U$ to $\omega_1$. Then $\Phi(\omega_1)$ is a circle tangent to all of
$$ \Phi(AB) = \omega_3 ~,~ \Phi (\omega_3) = AB ~,~ \Phi(\ell_1) = \ell_1 ~,~ \Phi(\ell_2) = \ell_2 $$it follows $\Phi(\omega_3) = \omega_3$, i.e. $\Phi$ fixes $\omega_1$. Now since the image of antipode of $U$ (wrt $\omega_3$) is the projection of $U$ onto $AB$, thus
$$ p(U,\omega_1) = \sqrt{ \text{squared radius of } \Phi} = \sqrt{2R_3 \cdot h_1} $$as desired.
Proof of $(2)$ : Analogous to proof of $(1)$.
Proof of $(3)$ : It is a known fact that the length of external common between two circles $\omega_2,\omega_3$ which touch each other externally equals
$$ 2 \sqrt{R_3 \cdot R_2} $$One way to prove this is to just note
\begin{align*}
\text{length of common external tangent} &= \sqrt{ (\text{distance between centers})^2 - (\text{difference between radi})^2} \\
&= \sqrt{(R_2 + R_3)^2 - (R_2 - R_3)^2} \\
&= 2 \sqrt{R_3 \cdot R_2}
\end{align*}Another (inversive) way to prove this is: Consider inversion $\Psi$ at $U$ with radius $US$. Clearly $\Psi$ fixes $\omega_2$. Now $\Psi(\omega_3)$ is a line parallel $US$ and tangent to $\omega_2$. Thus
$$ \text{dist}(U,\Psi(\omega_3)) = 2R_2 $$Now since $\Psi$ sends antipode of $U$ (wrt $\omega_3)$ to projection of $U$ onto $\Psi(\omega_3)$, so it follows
$$ US^2 = (\text{squared radius of } \Psi)^2 = 2R_3 \cdot 2R_2 $$as desired.
proof of $(4)$ : Analogous to proof of $(3)$.
Also, can someone please post a synthetic proof.
guptaamitu1
03.10.2022 17:55
There are two possible configurations! At first sight it seems the configuration in post #8 is the only one.