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Source: 41-st Vietnamese Mathematical Olympiad 2003

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#1 h Aug 28, 2004, 4:18 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

The circles $C_{1}$ and $C_{2}$ touch externally at $M$ and the radius of $C_{2}$ is larger than that of $C_{1}$ . $A$ is any point on $C_{2}$ which does not lie on the line joining the centers of the circles. $B$ and $C$ are points on $C_{1}$ such that $AB$ and $AC$ are tangent to $C_{1}$ . The lines $BM$ , $CM$ intersect $C_{2}$ again at $E$ , $F$ respectively. $D$ is the intersection of the tangent at $A$ and the line $EF$ . Show that the locus of $D$ as $A$ varies is a straight line.

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#2 h Aug 28, 2004, 1:54 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

And that line is the perpendicular to the line of the centers from $M$ . $EF$ is the image of $BC$ through the homothety of center $M$ turning $C_1$ into $C_2$ , which means that the pole of $EF$ wrt $C_2$ is the image of $A$ through this homothety. Call this point $A'$ . Since $A,M,A'$ are collinear, we get that $AA'$ passes through $M$ , so the intersection of the polars of $A,A'$ wrt $C_2$ must lie on a line, which is the polar of $M$ wrt $C_2$ (duality principle), and this line is the tangnt from $M$ to $C_2$ .

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#3 h Aug 28, 2004, 3:29 PM • 2 Y

Y by Adventure10, Mango247

Okay, I have the same proof but in a different form and a bit more detailed, so I'm posting it...

In fact, the locus of the point D is the common tangent of the circles $C_1$ and $C_2$ at the point M. In order to prove this, I will show that the point D always lies on the common tangent of the circles $C_1$ and $C_2$ at the point M.

Here is how I prove it (sorry, the proof uses pole-polar relationship): Let m be the common tangent of the circles $C_1$ and $C_2$ at the point M. Then we have to show that the point D lies on the line m.

The polar of the point A with respect to the circle $C_1$ is the line BC, since the points B and C are the points where the tangents from A to $C_1$ touch $C_1$ . Now, since the line MA passes through the point A, the pole of the line MA with respect to the circle $C_1$ must lie on the polar of the point A with respect to $C_1$ . Hence we have obtained the fact that the pole of the line MA with respect to the circle $C_1$ lies on the line BC.

Now, since M is the point of tangency of the circles $C_1$ and $C_2$ , there exists a homothety h with center M mapping the circle $C_1$ to the circle $C_2$ . This homothety h must take the points B and C to the points E and F, respectively (since the points E and F lie on the circle $C_2$ and on the lines BM and CM, respectively). Hence, this homothety h takes the line BC to the line EF. Also, this homothety h leaves the line MA invariant (since this line passes through the center M of the homothety). Finally, polar relation is clearly invariant under homotheties. Hence, from the fact that the pole of the line MA with respect to the circle $C_1$ lies on the line BC, we can derive using our homothety h that the pole of the line MA with respect to the circle $C_2$ lies on the line EF. But the pole of the line MA with respect to the circle $C_2$ is the point of intersection of the tangents to $C_2$ at the points M and A. The tangent to $C_2$ at M is the line m that we have met before. Hence, we see that the point of intersection of the line m with the tangent to $C_2$ at A lies on the line EF. In other words, the point of intersection D of the tangent to $C_2$ at A with the line EF must lie on the line m. And this is exactly what we wanted to prove.

Darij

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#4 h Aug 29, 2004, 8:36 PM • 2 Y

Y by Adventure10, Mango247

The condition:...the radius of $C_2$ is larger than that of $C_1$ ...
It is not necessary.
$AFME$ performed in this way is harmonic quadrilateral, also: $AE.MF=AF.ME$
From the another view point: $AB,AC$ cut at $J,K$ to circle $C_2$ , we know:
$MB$ and $MC$ are external bisector of $AMJ$ and $AMK$ triangles.
Here the proof:
The line $AM$ cut at $L$ to circle $C_1$ .
Draw the line $T$ common tangent at $M$ for two circles $C_1$ and $C_2$ .
The tangent line $T$ cut at $H,I$ to $AB,AC$ ,then:
$<HMJ=<MAJ=\gamma$ , $<LMB=<MAB+<MBA=\gamma+\delta$ ,
$<MBA=<MBH=<HMB=\delta$ , so $<BMJ=<BML=\gamma+\delta$ then $MB$ is external bisector of $AMJ$ triangle.
$<KMI=<KAM=\alpha$ , $<CML=<CAM+<ACM=\alpha+\beta$ ,
$<ACM=<ICM=<IMC=\beta$ , so $MC$ is external bisector of $AMK$ triangle.
The another angular relations in the figure.
It is not hard prove that the tangent line $T$ at $M$ and the tangent at $A$ with $EF$ are concur at $D$ .
Therefore the locus is the tangent line $T$ at $M$ .

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yetti

#5 h Dec 17, 2006, 8:21 AM • 2 Y

Y by Adventure10, Mango247

Let $O_{1},\ O_{2}$ be centers and $r_{1},\ r_{2}$ radii of the circles $\mathcal C_{1},\ \mathcal C_{1}$ touching at M. These 2 circles can touch either externally or internally. Only in the case of internal tangency we need $r_{2}> r_{1},$ otherwise the tangents to $\mathcal C_{1}$ from $A \in \mathcal C_{2}$ would not exist. Assuming external tangency, M is the internal similarity center of the circles $\mathcal C_{1},\ \mathcal C_{2},$ hence $\mathcal C_{1}\sim \mathcal C_{2}$ are centrally similar with similarity center M and similarity coefficient $-\frac{r_{1}}{r_{2}}.$ The triangles $\triangle MBC \sim \triangle MEF$ are centrally similar with the same similarity center and coefficient, hence their corresponding sides $BC \parallel EF$ are parallel. Since BC is a polar of A with respect to $\mathcal C_{1},$ $BC \perp AO_{1}$ and consequently, $EF \perp AO_{1}$ as well. Let AM meet $\mathcal C_{1}$ at Z. Since AB is a tangent of $\mathcal C_{1}$ at B, $\angle ABM = \angle MZB = \angle MAE.$ Using the sine theorem for the triangles $\triangle ABM,\ \triangle AEM$ then yields

$\frac{EA}{EM}= \frac{\sin \widehat{AME}}{\sin \widehat{MAE}}= \frac{\sin \widehat{AMB}}{\sin \widehat{ABM}}= \frac{AB}{AM},\ \ \ EA^{2}= EM^{2}\cdot \frac{AB^{2}}{AM^{2}}$

Because of the central similarity of $\mathcal C_{1},\ \mathcal C_{2}$ with center M, $AZ = AM+MZ = AM \left(1+\frac{r_{1}}{r_{2}}\right)$ and $EB = EM+MB = EM \left(1+\frac{r_{1}}{r_{2}}\right).$ Power of A to $\mathcal C_{1}$ is then $AB^{2}= AM \cdot AZ = AM^{2}\left(1+\frac{r_{1}}{r_{2}}\right).$ Substituting this to the above equation yields $EA^{2}= EM \cdot EB,$ which means that E lies on the radical axis of the point A and the circle $\mathcal C_{1}.$ Since $EF \perp AO_{1},$ EF is their radical axis. On the other hand, the tangent $t_{A}$ of $\mathcal C_{2}$ at A is the radical axis of the point A and the circle $\mathcal C_{2}.$ The radical axes $EF,\ t_{A}$ meet at the radical center D of the point A and the circles $\mathcal C_{1},\ \mathcal C_{2},$ which lies on the radical axis $t_{M}$ of the circles $\mathcal C_{1},\ \mathcal C_{2},$ their single common internal tangent at M.

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yetti

#6 h Dec 19, 2006, 8:23 AM • 2 Y

Y by Adventure10, Mango247

And I could have made this simpler, too:

... Since AB is a tangent of $\mathcal C_{1}$ at B, $\angle ABM = \angle MZB = \angle MAE.$ The triangles $\triangle ABE \sim \triangle MAE$ with a common angle at the vertex E are then similar, having equal angles, hence

$\frac{EA}{AB}= \frac{EM}{AM},\ \ \ EA^{2}= EM^{2}\cdot \frac{AB^{2}}{AM^{2}}$

etc.

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7054 posts

#7 h Dec 19, 2006, 2:40 PM • 3 Y

Y by gemcl, Adventure10, Mango247

Very nice both the proposed problem and the Yetti proof !

Lemma. If the circles $c\ ,\ c'$ are tangent in the point $T$ and are given
the points $\{A,B\}\subset c$ , $\{A',B'\}\subset c'$ so that $T\in AA'\cap BB'\ ,$ then $AB\parallel A'B'\ .$

Another proof of the proposed problem (similarly with the Yetti's).
Denote the second intersection $L$ of the line $AM$ with the circle $C_{1}\ .$ Therefore,
$\begin{array}{ccccccc}BL\parallel AE & \Longrightarrow & \widehat{ABE}\equiv\widehat{BLA}\equiv\widehat{EAM}& \Longrightarrow & EAM\sim EBA & \Longrightarrow & EA^{2}=EM\cdot EB\\\ CL\parallel AF & \Longrightarrow & \widehat{ACF}\equiv\widehat{CLA}\equiv\widehat{FAM}& \Longrightarrow & FAM\sim FCA & \Longrightarrow & FA^{2}=FM\cdot FC\end{array}\|$
$\implies$ the line $EF$ is the radical axis between the circle $C_{1}$ and the point $A$ (null circle).

$D\in EF\cap AA$ and $AA$ is the radical axis of the circles $C_{2}$ and $A$ $\implies$ $D\in MM\ .$

This post has been edited 5 times. Last edited by Virgil Nicula, Dec 19, 2006, 7:26 PM

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#8 h Dec 19, 2006, 4:21 PM • 2 Y

Y by Adventure10, Mango247

Thank you very much. Very nice solution mr Virgil .
I remember I friend Nick Rapanos solved it by homothety ,but I don't remember the full solution .Could anyone find such one ?

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#9 h Dec 22, 2006, 9:34 PM • 1 Y

Y by Adventure10

We denote as $L,$ the intersection point of the circle $(O_{1}),$ from the segment line $AM.$ Also we denote as $D,$ $H,$ the intersection point of the common tangent line at $M$ $($ radical axis $),$ of $(O_{1}),$ $(O_{2}),$ at points $A,$ $L,$ respectively.

By applying the polar theory, because of the segment line $LM,$ as the polar of $H,$ with respect to the circle $(O_{1}),$ passes through the point $A,$ we have that the segment line $BC,$ as the polar of $A,$ with respect also to $(O_{1}),$ passes through the point $H.$

$($ We can also prove this result, without polar theory. If we denote as $K,$ $N,$ the intersection points of the segment line $LH,$ from the tangent lines $AB,$ $AC$ respectively, we have the configuration of the triangle $\bigtriangleup AKN,$ taken the circle $(O_{1})$ as it’s incircle and so, based on a well known $($ at least to me $)$ Lemma, we have that the segment lines $KN,$ $BC$ and the tangent line of $(O_{1})$ at $M,$ are concurrent. An elementary proof can be found, by Newton’s, Menelaus’s and Ceva’s theorems $).$

We will prove now, that the segment line $EF,$ passes through the point $D.$ From similar isosceles triangles $\bigtriangleup DAM\sim \bigtriangleup HLM,$ $\Longrightarrow$ $\frac{MD}{MH}= \frac{MA}{ML}= \frac{R_{2}}{R_{1}}$ $,(1)$ $($ because of $\bigtriangleup O_{1}LM\sim \bigtriangleup O_{2}AM$ $).$

It is easy to prove that $BC\parallel EF$ $($ from $\angle CBM = \angle CMH = \angle FMD = \angle FEM$ $).$

We denote as $D',$ the intersection point of $EF,$ $HM.$ From similar triangles $\bigtriangleup CMH\sim \bigtriangleup FMD'$ $\Longrightarrow$ $\frac{MD'}{MH}= \frac{MF}{MC}= \frac{R_{2}}{R_{1}}$ $,(2)$ $($ because of $\bigtriangleup O_{1}MC\sim \bigtriangleup O_{2}MF$ $).$

From $(1),$ $(2)$ $\Longrightarrow$ $MD' = MD$ $\Longrightarrow$ $D'\equiv D$ $,(3)$

From $(3),$ we conclude that the concurrency point of the segment line $EF,$ with the tangent line of $(O_{1}),$ at point $A,$ lies on the radical axis of $(O_{1}),$ $(O_{2})$ and the proof is completed.

Kostas Vittas.

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#10 h Apr 22, 2009, 12:17 PM • 2 Y

Y by Adventure10, Mango247

Let me restate the problem so that I can fit my solution:
"The circles $(C_1), (C_2)$ externally touches at $A$ ( $(C_2)>(C_1)$ ). Let $B$ be an any point on $(C_2)$ . From $B$ , let $BM,BN$ be the tangents wrt $(C_1)$ . $AM$ , $AN$ , respectively, intersects $(C_2)$ at $P,Q$ . $PQ$ intersects the tangents from $B$ of $(C_2)$ by the point $K$ . Prove that $K$ blongs to a fixed line."
Proof:
Consider the homothety with center $A$ , ratio $\frac {-r_1}{r_2}=k$ , we get:
$H(A,k): P\mapsto M$ , $Q\mapsto N$ . Hence, $H(A,k): PQ\mapsto MN$ . Also, the tagents from $B$ also maps to the tangents from $C$ . Let $H$ be the intersection of $MN$ to the tangent from $C$ . It is easy to see that $H(A,k): H\mapsto K$ . Now, we need to prove that $H$ belongs to the fixed line. But this is obviously true because the fact that $AMCN$ is the harmonic quadrilateral, therefore, $H$ belongs the the common internal tangent of $(C_1),(C_2)$ . The homothety center $A$ , ratio $k$ , which turns the tangent at $A$ of $(C_1),(C_2)$ into itself. Therefore, K is also a member of the common internal tangent of $(C_1),(C_2)$ , which is fixed.
Our proof is completed

p/s: Sorry if my solution is the same to anyone's

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vslmat

#11 h Aug 8, 2012, 2:16 PM • 1 Y

Y by Adventure10

Another solution using properties of symmedian:
Easy to show that $LO_{1} // AO_{2}$ and $CO_{1} // FO_{2}$ and $LC // FA$ , also $LB // AE$ , therefore $BC//EF$ .
$\Delta BLC \sim \Delta EAF$ . Since $LC // AF$ , point $K$ in $\Delta BLC$ corresponds to point $H$ in $\Delta EAF$ .
As $AB$ and $AC$ are two tangents to the circle $C_{1}$ , $AL$ , resp. $KL$ is the symmedian in $\Delta BLC$ , so $AH$ is the symmedian in $\Delta EAF$ .
Furthermore, $HE$ , resp. $DE$ is the symmedian in $\Delta MEA$ , so $D$ must be the intersection of the tangents at $A$ and $M$ to the circle $C_{2}$ .
With the solution, other interesting facts can be seen, e.g. the two tangents at $E$ and $F$ to the circle $C_{2}$ must intersect at a point on line $AL$ .

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#12 h Feb 22, 2016, 4:33 AM • 1 Y

Y by Adventure10

Sorry to revive an old thread, but can someone please tell me if the following proof is correct or not?

Suppose line $AM$ meets $C_1$ again at $G$ . Evidently, $MCGB$ is a harmonic quadrilateral, and so $M(MCGB)$ is a harmonic pencil. Intersecting this with $C_2$ , we conclude that $MFAE$ is a harmonic quadrilateral. If the tangent to $C_2$ from $D$ touches $C_2$ at $D'$ , then $D'FAE$ is harmonic as well. Thus $D'\equiv M\implies D\in$ the common tangent of $C_1$ and $C_2$ , which is a straight line.

Please let me know if it's wrong, since it appears to be too short to be true.

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#13 h Feb 24, 2016, 3:38 AM • 2 Y

Y by Adventure10, Mango247

Hello, can anyone please check the above proof?

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bonciocatciprian

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bonciocatciprian

#14 h Feb 24, 2016, 3:42 PM • 4 Y

Y by Ankoganit, PRO2000, Adventure10, Mango247

Ankoganit wrote:

Sorry to revive an old thread, but can someone please tell me if the following proof is correct or not?

Suppose line $AM$ meets $C_1$ again at $G$ . Evidently, $MCGB$ is a harmonic quadrilateral, and so $M(MCGB)$ is a harmonic pencil. Intersecting this with $C_2$ , we conclude that $MFAE$ is a harmonic quadrilateral. If the tangent to $C_2$ from $D$ touches $C_2$ at $D'$ , then $D'FAE$ is harmonic as well. Thus $D'\equiv M\implies D\in$ the common tangent of $C_1$ and $C_2$ , which is a straight line.

Please let me know if it's wrong, since it appears to be too short to be true.

It seems okay. Also, you should note that for a specific $\triangle ABC$ inscribed in $(O)$ (unrelated to our problem) there are three distinct points on the circle, each completing $\triangle ABC$ to a harmonic quadrilateral. What makes $D' \equiv M$ in your case is that they are both on the same arc determined by $E$ and $F$ .

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#15 h Feb 25, 2016, 3:40 AM • 1 Y

Y by Adventure10

@bonciocatciprian Thank you for your comments.

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jred

#16 h Dec 12, 2016, 1:39 PM • 1 Y

Y by Adventure10

Indeed,the locus of $D$ is the internally common tangent of $C_1$ and $C_2$ except $M$ .

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JackXD

#17 h Jun 6, 2017, 3:09 AM • 2 Y

Y by gemcl, Adventure10

Let $AM \cap C_1=J$ . Since BC is the polar of A w.r.t C1 the lines BM,BJ,BC,BA form a harmonic pencil $\implies (M,J;C,B)=-1$ .Seeing this from M we get that MM,MJ,MC,MB form a harmonic pencil.Projecting this onto C2 we get $(M,A;F,E)=-1$ .Thus AM,AD,AF,AE form a harmonic pencil $\implies AM$ is the polar of $D$ w.r.t C2.Therefore $D$ lies on the polar of M w.r.t C2 (duality principle) which is a straight line.

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#18 h Dec 2, 2018, 4:21 AM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

Symmedians!

Let $AM$ intersect $C_2$ again at $A'$ . Then, since $A'A$ is a symmedian, we have $-1=(M,A';B,C)\stackrel{M}{=}(M,A;E,F)$ Thus quadrilateral $MEAF$ is harmonic. Since $D$ lies on the intersection of tangent of $A$ , and the $EF$ , $DM$ must be tangent to $C_1$ , which means it's also tangent to $C_2$ . As, $A$ varies, $D$ traces the radical axis of $C_1$ and $C_2$ , which is a line, as needed.

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#19 h Sep 10, 2022, 6:43 AM • 1 Y

Y by Mango247

It is sufficient to prove $MD$ is tangent.

Let $N=AM\cap (MBC)$ . Since $AB, AC$ is tangent so $MBNC$ is a harmonic quadrilateral. A homothety centered at $M$ takes $MBNC$ to $MEAF$ . So $MEAF$ is also harmonic quadrilateral. Meaning $MD$ is tangent because $AD$ is tangent.

This post has been edited 1 time. Last edited by Tafi_ak, Sep 10, 2022, 6:44 AM

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