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#1 h Jul 2, 2012, 3:13 AM • 2 Y

Y by Adventure10, Mango247

Circles $\Omega$ and $\omega$ are internally tangent at point $C$ . Chord $AB$ of $\Omega$ is tangent to $\omega$ at $E$ , where $E$ is the midpoint of $AB$ . Another circle, $\omega_1$ is tangent to $\Omega, \omega,$ and $AB$ at $D,Z,$ and $F$ respectively. Rays $CD$ and $AB$ meet at $P$ . If $M$ is the midpoint of major arc $AB$ , show that $\tan \angle ZEP = \tfrac{PE}{CM}$ .

Ray Li.

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yunxiu

#2 h Jul 2, 2012, 8:13 AM • 7 Y

Y by Wizard_32, Durjoy1729, MathbugAOPS, srijonrick, Adventure10, Mango247, poirasss

Denote $CD \cap {\omega _1} = G$ , then $\angle GDF = \angle CDM = 90^\circ$ , so $FG$ is the diameter of ${\omega _1}$ , hence $P$ is the homothetic center of ${\omega _1}$ and $\omega$ , so the centers of ${\omega _1}$ and $\omega$ are all in the line $PZ$ .
Because $D$ is the homothetic center of ${\omega _1}$ and $\Omega$ , so $M,D,F$ are collinear. Because $\angle CEF = 90^\circ$ , $\angle FDC=\angle MDC =90^\circ$ , $D,F,E,C$ are concyclic, so $MD \cdot MF = ME \cdot MC$ , $MZ$ is the radical line of ${\omega _1}$ and $\omega$ , so $MZ \bot ZP$ , hence $\angle CZM = \angle EZP$ , and we have $\Delta CZM \sim \Delta EZP$ , so $\tan \angle ZEP = \tan \angle ECZ = \frac{{EZ}}{{CZ}} = \frac{{PE}}{{MC}}$ .

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Sardor

#3 h Dec 16, 2014, 5:44 PM • 2 Y

Y by Adventure10, Mango247

Here my solution:
Obviously, $P$ is exsimilicenter of $w$ and $w_1$ , hence $P,O,O_1,Z$ are collinear, where $O ,O_1$ be the centers of $w,w_1$ , respectively.We have easily that $\angle PEZ=\angle ZCE$ , and $\angle MZC=\angle PZE$ (because we know that $MZ$ is common internal tangent of $w$ and $w_1$ ).Thus the triangles $PZE$ and $CZM$ are relatively similar.Hence $tan \angle ZEP=tan \angle ECZ= \frac{ZE}{CZ}=\frac{PE}{CM}$ , so we are done !

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#4 h May 15, 2016, 9:39 PM • 2 Y

Y by Adventure10, Mango247

Homothety that takes $w$ to $\omega$ takes to midpoint of arc $AB$ so we have $C-E-M$ so $CE$ is perpendicular on $AB$ and analogous $D-F-M$ .Trivialy $\tan \angle ZEF=\frac{ZE}{ZC}$ so for $\triangle EZP \sim \triangle CZM$ we only need $MZ$ perpendicular $PZ$ and we have that $\angle ZPE=\angle ZCM$ , now Mongue de Alambert theorem we have $C$ as center homothety $w$ to $\omega$ , $D$ as center of homothety $w_{2}$ to $\omega$ so center of homothety $w$ to $w_{1}$ lies on $CD$ ,so because $FE$ is common tangent we have $P-O_{1}-Z-O_{2}$ .Now we need to prove that $MZ$ is common tangent of $w_1$ , $w_{2}$ which is true because $CDEF$ is cyclic so $M$ is on radical axis which is common tangent.

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Delray

#5 h Apr 18, 2017, 10:14 PM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

Denote $O_\Omega$ , $O_w$ and $O_{w_1}$ as the centers of circles $\Omega$ , $w$ and $w_1$ . It is well known that the power of $M$ , the midpoint of major arc $BC$ , with respect to a circle tangent to $BC$ and minor arc $BC$ is equal to $MA^2=MB^2$ . It follows that M is on the radical axis of $O_{w_1}$ and $O_w$ , implying that $MZ \perp O_{w_1}O_w$ . Since $\angle{PEM}=90^{\circ}$ , we have that quadrilateral $PZEM$ is cyclic. This means that $\angle{ZPE}=\angle{ZME}$ , and since $\angle{ZEP}=\angle{ZCE}$ , we have that $\triangle{PZE}$ is similar to $\triangle{ZMC}$ , implying that $\frac{PE}{ZE}=\frac{MC}{ZC}$ . Rearranging we have that $\frac{PE}{CM}=\frac{ZE}{ZC}$ . Since $\angle{CZE}=90^{\circ}$ , we have that $\tan{\angle{ZCE}}=\frac{ZE}{ZC}=\frac{PE}{CM}$ . However, since $\angle{ZCE}=\angle{ZEP}$ , we obtain $\tan{\angle{ZEP}}=\frac{PE}{CM}$ as desired. $\square$

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#7 h Aug 2, 2017, 6:12 PM • 2 Y

Y by Adventure10, Mango247

Pure Angle Chasing

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yayups

#8 h Dec 29, 2017, 4:16 AM • 3 Y

Y by Wizard_32, Adventure10, Mango247

Computational Solution:
Diagram
Solution

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#9 h Dec 29, 2017, 6:42 PM • 2 Y

Y by Adventure10, Mango247

Dear yayups and MLs,

Very nice solution, in the style of the Great Yetti.

Happy New Year to all !

M.T.

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#10 h Nov 30, 2018, 6:38 PM • 1 Y

Y by Adventure10

A question, are $C,Z,F$ collinear?
PS: Okay, never mind, I got my doubt clarified myself.

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#11 h Mar 3, 2019, 7:19 PM • 1 Y

Y by Adventure10

Since $\tan \angle ZEP = \tan \angle ZCE = \tfrac{ZE}{ZC}$ , we will prove that $\tfrac{ZE}{ZC}=\tfrac{PE}{CM}$ by showing that $\triangle{ZEP}\sim\triangle{ZCM}$ .

By a homothety at $D$ sending $\omega_1$ to $(ABC)$ , we know that $D$ , $F$ , and $M$ are collinear. In addition, $ME\cdot MC=MB^2=MF\cdot MD$ by considering similar triangles $MEB/MBC$ and $MFB/MBD$ . Hence, $M$ is on the radical axis of $\omega$ and $\omega_1$ . This also means that $\triangle{ZCM}\sim\triangle{EZM}$ .

By Monge's Theorem with circles $(ABC)$ , $\omega$ , and $\omega_1$ , $C$ , $D$ , and $P'$ are collinear, where $P'$ is intersection of the common external tangents of $\omega$ and $\omega_1$ . Since $P'$ is on $\overleftrightarrow{CD}$ and $\overleftrightarrow{EF}$ , we must have $P'=P$ . Then $\angle PZM = 90^{\circ}=\angle PEM$ , so $PZEM$ is cyclic. We have that $\angle{EMZ}=\angle{EPZ}$ and $\angle{ZEP}=\angle{EZM}=\angle{ECZ}$ , so $\triangle{EZM}\sim\triangle{ZEP}$ (this is actually a congruence).

Therefore, $\triangle{ZEP}\sim\triangle{EZM}\sim\triangle{ZCM}$ and $\tan \angle ZEP = \tan \angle ZCE = \tfrac{ZE}{ZC}=\tfrac{PE}{CM}$ , as desired.

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#12 h Aug 5, 2019, 9:23 PM • 1 Y

Y by Adventure10

Let $O_2$ be the center of $\omega$ and $O_3$ be the center of $\omega_1$ .

We first note that because $E$ is the midpoint of $\overline{AB}$ and $M$ is the midpoint of arc $\overline{AB}$ , we have that $\angle MEB = 90^\circ$ . By the properties of circles inscribed within segments, $M$ , $E$ , and $C$ are collinear, so $\angle CEB = 90^\circ$ . Since $\angle CEB = 90^\circ$ and $\omega$ is tangent to $AB$ at $E$ , $CE$ is a diameter of $\omega$ .

Since $\angle ZEP = \angle ZCM$ , it suffices to show that $\angle EPZ = \angle CMZ$ .

Since $\omega$ and $\omega_1$ intersect at just $Z$ , their radical axis is the line going through $Z$ that is tangent to both $\omega$ and $\omega_1$ .

By the properties of circles inscribed in segments, $\mbox{Pow}_{\omega_1}(M) = MB^2 = \mbox{Pow}_{\omega}(M)$ . Thus, $M$ is on the radical axis of $\omega$ and $\omega_1$ and $MZ$ is tangent to both $\omega$ and $\omega_1$ .

Let the other intersection of $PC$ with $\omega$ be $C'$ . This intersection exists because we would otherwise have that $PC$ is perpendicular to a diameter of $\omega$ , $CE$ , while $PE$ is also perpendicular to $CE$ , making $P$ not exist on the Euclidean plane. By Power of a Point, $PC' \cdot PC = PE^2$ , meaning that $\frac{PE}{PC} = \frac{PC'}{PE}$ and $\triangle PEC \sim \triangle PC'E$ . Since $\angle CEP = \angle CEB = 90^\circ$ , we then have that $\angle EC'P = 90^\circ$ . Since $MC$ is a diameter of $\Omega$ , $\angle MDC = 90^\circ$ . By the properties of circles inscribed in a segment, $M$ , $D$ , and $F$ are collinear, so $\angle FDC = 90^\circ$ .

Now, consider the homothety about $P$ taking $E$ to $F$ , $\mathcal{H}$ . Since $\angle FDP = 90^\circ = \angle EC'P$ , $\mathcal{H}(C') = D$ . Since $\omega$ goes through $C'$ and is tangent to $EP$ at $E$ and $\omega_1$ goes through $D$ and is tangent to $FP$ at $F$ , we must have that $\mathcal{H}(\omega) = \omega_1$ . Thus, we then have that $\mathcal{H}(O_2) = O_3$ , meaning that $O_2$ , $O_3$ , and $P$ are collinear. Since $\omega$ and $\omega_1$ are tangent to each other at $Z$ , we have that $O_2$ , $Z$ , and $O_3$ are collinear. Thus, $P$ , $O_3$ , and $Z$ are collinear. Since $O_3Z$ is perpendicular to $MZ$ , we have that $PZ$ is perpendicular to $MZ$ , and so $\angle PZM = 90^\circ$ . Since $\angle PZM = 90^\circ$ and $\angle PEM = \angle MEP = \angle MEB = 90^\circ$ , we have that $PZEM$ is a cyclic quadrilateral.

Now, we have that $\angle EPZ = \angle EMZ = \angle CMZ$ . Because $\angle ZEP = \angle ZCM$ , we have that $\triangle ZEP \sim \triangle ZCM$ . Then, we have that $\frac{EZ}{CZ} = \frac{PE}{CM}$ . Since $\angle CZE = 90^\circ$ , we have that $\frac{EZ}{CZ} = \tan \angle ZCE = \tan \angle ZEP$ . Thus, we have that $\tan \angle ZEP = \frac{PE}{CM}$ , and we are done. $\fbox{}$

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#13 h Sep 1, 2019, 6:23 PM • 1 Y

Y by Adventure10

Let $O_1$ and $O_2$ be the centers of $\omega$ and $\omega_1$ . First of all note that by Archimedes + Shooting Lemma we have $CE \cap DF = M$ and $MA^2 = MB^2 = ME \cdot MC = MF \cdot MD$ so $EFDC$ is cyclic.

Claim: $O_1,Z,O_2,P$ are collinear.
Proof: Let $E'$ and $F'$ be the reflections of $E$ and $F$ across $O_1ZO_2$ (which we know to be collinear). By PoP we have $PF \cdot PE = PF' \cdot PE' =PD \cdot PC$ hence $E'F'DC$ is cyclic. By definition we know $EFF'E'$ is isosceles trapezoid so it is cyclic. We use radical axis theorem on the three circles $(EFF'E'), (EFDC), (E'F'DC)$ to obtain $E'F', EF, CD$ concur at $P$ . Obviously $E'F', O_1O_2, EF$ must concur so $O_1O_2$ passes through $P$ as desired. $\square$

Since $E$ is the midpoint of $AB$ we know $CE$ is the diameter of $\omega$ . Since $AB$ tangent to $\omega$ , we know $\angle ZEP = \angle ZCM$ and since $CEDF$ is cyclic, $M$ lies on the tangent at $Z$ . Therefore, by angle chasing with $MZ$ tangent to $\omega$ , we get $\angle CZM = 180^{\circ} - \angle CEZ = 180^{\circ} - \angle O_1ZE = \angle EZP$ since $O_1, Z, P$ collinear from our claim.

Therefore, by AA similarity we have $\triangle CZM \sim \triangle EZP$ so therefore $\tan{\angle ZEP} = \tan{\angle ZCM} = \frac{EZ}{ZC} = \frac{PE}{CM}$ as desired. $\blacksquare$

reeeeeeee

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#16 h Nov 24, 2019, 12:22 PM • 2 Y

Y by Adventure10, Mango247

I have used Yayups' diagram
Diagram
Definitions

Claim 1 : $\overline {C - E - M}$ and $\overline {D - F - M}$
Proof : This follows from Shooting Lemma.

Claim 2 : $CEFD$ is a cyclic quadrilateral .
Proof : $MZ$ is the radical axis of $\omega_1$ and $\omega$ . Now it is just Radical Axis Theorem.

From Claim 2 $\angle CEP = \angle CDF = 90^{\circ} \Longrightarrow CM$ is the diameter of $\Omega$ and $FG$ is the diameter of $\omega_1$ . So, there is a homothety centred at $P$ mapping $\omega_1 \mapsto \omega \Longrightarrow \overline{P - Z - T}$
Thus, $\angle PZM = 90^{\circ} \Longrightarrow PEZ\sim MCZ \Longrightarrow \tan \angle ZEP = \frac{EZ}{ZC} = \frac{PE}{CM}$

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#17 h Nov 27, 2019, 5:46 PM • 4 Y

Y by lilavati_2005, Pi-is-3, Adventure10, Mango247

Note that $\angle ZEP=\angle ZCE$ and $PE=CE\tan\angle DCE$ . As $\angle MDC=\frac{\pi}{2}$ , we have $\tan\angle DCE=\frac{MD}{DC}$ . Hence $PE=CE\frac{MD}{DC}$ .
Now consider an inversion $f$ with centre at $C$ and radius 1. The resulting diagram is attached.
Clearly, $|f(Z)f(E)|=|f(D)f(M)|$ . Hence $\tan\angle ZCE=\tan\angle f(E)Cf(Z)=\frac{|f(Z)f(E)|}{|f(E)C|}=\frac{|f(D)f(M)|}{|f(E)f(M)|+|f(M)C|}=\frac{\frac{MD}{CM\cdot CD}}{\frac{EM}{CE\cdot CM}+\frac{1}{CM}}=\frac{\frac{MD}{DC}}{\frac{CM}{CE}}=\frac{\frac{PE}{CE}}{\frac{CM}{CE}}=\frac{PE}{CM}$ .

QED

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Reason: latex error

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#18 h Mar 25, 2020, 3:19 AM

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Let $O_1$ be the center of $\Omega$ , $O_2$ be the center of $\omega$ , and $O_3$ be the center of $\omega_1$ . Also, let $G$ be the point on $\omega$ diametrically opposite to $F$ .

Lemma: $O_2, Z, O_3, P$ are collinear, $C, Z, F$ are collinear, $E, Z, G$ are collinear, and $D, F, M$ are collinear.

Proof: It is well-known that $D, F, M$ are collinear (quick sketch of the proof: the homothety about $D$ sending $\omega_1$ to $\Omega$ sends $F$ to $M$ ). $C, Z, F$ are collinear and $E, Z, G$ are collinear because the negative homothety about $Z$ sending $O_2$ to $O_3$ sends $C$ to $F$ and $e$ to $G$ . It is obvious that $O_2, O_3, Z$ are collinear, and the homothety about $P$ sending $\triangle{PFG}$ to $\triangle{PEC}$ sends $O_2$ to $O_3$ (since they are midpoints of $GF, CE$ respectively), so $P, O_2, O_3$ are collinear, and we have proven all four parts of the lemma. End lemma.

Now note that $\angle{ZCE}=\angle{ZEP}$ , so it suffices to show that $\tan{\angle{ZCE}}=\frac{PE}{CM}$ . But $\tan{\angle{ZCE}}=\frac{EZ}{CZ}$ , so thus it suffices to prove that $\frac{EZ}{PE}=\frac{CZ}{CM}$ . But note since $\angle{ZCE}=\angle{ZEP}$ , this is equivalent to proving $\triangle{PEZ}~\triangle{MCZ}$ , which in turn is equivalent to proving $\angle{CMZ}=\angle{EPZ}$ . But note that we have $\angle{ECZ}=\angle{O_2ZC}=\angle{PZF}$ , and we also have $\angle{ECZ}=\angle{ZEP}=\angle{EZM}$ , meaning that $\angle{EZM}=\angle{PZF}$ , so thus $\angle{PZM}=90$ because $\angle{EZF}=90$ . But since $\angle{PEM}=90$ , we have $PZEM$ is cyclic, so thus $\angle{EMZ}=\angle{CMZ}=\angle{EPZ}$ , and hence we are done.

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#19 h Feb 1, 2021, 7:07 PM • 1 Y

Y by Mango247

It's well known $DFM$ collinear, and it's clear this line is perpendicular to $CP$ , hence $F$ is the orthocenter of $\triangle CMP$ . Considering the homothety at $Z$ sending $\omega_1$ to $\omega_2$ we see $CZF$ are collinear, and this line is perpendicular to $PM$ by our previous observations. Since $CEDF$ is cyclic by radical axis we have $MZ$ is tangent to $\omega_1,\omega_2$ , so $\angle EZM = \angle ZEP$ . In addition, $EZ \perp CZ \perp MP$ so $EZPM$ is an isosceles trapezoid. To finish, we have $\tan(\angle ZEP) = \tan(\angle ECF) = \tan(90^\circ - \angle CMP) = \frac{EM}{EP}$ so it suffices to show $\frac{EM}{EP}=\frac{PE}{CM}$ , which follows since $EM \cdot CM = MZ^2 = PE^2$ .

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#20 h May 13, 2021, 5:17 PM • 4 Y

Y by centslordm, Mango247, Mango247, Mango247

Quite nice, although it's a bit tricky to pick which length ratio to focus on (I originally tried to do $\frac{PF}{PE}=\frac{EM}{CM}$ which failed). Needed a few hints to solve.

Since $\angle CEF=\angle EZC=90^\circ$ , we have $\triangle CZF \sim \triangle CEF \sim \triangle EZF$ , so $\tan\angle ZEP=\tan\angle ZEF=\tan\angle ECZ=\frac{EZ}{CZ}$ . Hence we just have to prove:
$\frac{EZ}{CZ}=\frac{PE}{MC}.$ Now let $G \neq D$ denote the point where $\overline{CP}$ intersects $\omega_1$ . We then have $\angle GDF=\angle CDM=90^\circ$ , hence $\overline{FG}$ is a diameter. This implies that the homothety sending $\omega_1$ to $\omega$ is centered at $P$ .
Observe that the homothety sending $\omega$ to $\Omega$ sends $E$ to $M$ and is centered at $C$ , hence $C,E,M$ are collinear. Similarly, $D,F,M$ are collinear. Also, as $\angle CEF=\angle CDF=90^\circ$ , we have $C,D,E,F$ concyclic. Thus $ME\cdot MC=MF \cdot MD$ , so $M$ lies on the radical axis of $\omega$ and $\omega_1$ . As the perpendicular to the radical axis (passing through the centers) also passes through $P$ , we obtain $\overline{PZ} \perp \overline{ZM}$ . Since $\angle CZE=90^\circ$ , we have $\angle CZM=\angle PZE$ . We can also obtain that $\angle PEZ=\angle MCZ$ from the fact that $\triangle CEF$ is right with $Z$ as the foot of the $E$ -altitude, hence $\triangle PZE \sim \triangle MZC$ , from which we obtain $\frac{EZ}{CZ}=\frac{PE}{MC}$ as desired. $\blacksquare$

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#21 h Jul 10, 2021, 6:31 AM

Y by

i am definitely not sleep-deprived while writing this. also had the same issue as @above, focused on the wrong ratio at first
Since $\angle ZEP=\angle ZCE$ , it suffices to prove
$\begin{align*} \frac{PE}{CM}=\frac{ZE}{CM}\implies\frac{PE}{ZE}=\frac{CM}{CZ}, \end{align*}$ or $\triangle PEZ\sim\triangle MCZ$ . Manipulating ratios, we have
$\begin{align*} \frac{PZ}{ZE}=\frac{MZ}{CZ}\implies\frac{PZ}{MZ}=\frac{ZE}{CZ}, \end{align*}$ which means it is the same to prove $\triangle MZP\sim\triangle CZE$ . However, this is also the same as proving $MEZP$ is cyclic, as if so, then $\angle MZP=\angle MEP=90^\circ=\angle CZE$ and $\angle ZEP=\angle ZEB=\angle ECZ$ . Moreover, proving that $MZ$ is tangent to $\omega$ finishes, as if so, $\angle MZP=90^\circ=\angle MEP$ which implies the cyclicity as needed. Indeed, note that $M$ is on the radical axis of $\omega$ and $\omega_1$ since $MA^2=ME\cdot MC=MF\cdot MD$ by Shooting Lemma. $Z$ is trivially on the radical axis as well, and since $\omega$ and $\omega_1$ are externally tangent so we may conclude.

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#22 h Sep 12, 2021, 12:48 AM • 2 Y

Y by centslordm, hussien

Solution

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#23 h Jan 6, 2022, 5:29 AM

Y by

$C,E,M$ collinear
$D,F,M$ collinear
$C,Z,F$ collinear
Remaining sinus Law

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#24 h Jul 13, 2022, 7:54 PM • 1 Y

Y by hungrypig

Solution

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#25 h Jul 18, 2022, 10:04 AM • 2 Y

Y by Mango247, Mango247

Solution

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#26 h Aug 11, 2022, 1:21 PM • 1 Y

Y by Jndd

It is easy to see that $\angle ZEP \cong \angle ZCM$ . Then $\tan \angle ZEP = \tan \angle ZCM = \tan \angle ZCE = \frac{EZ}{ZC}$ . We need to show this is the same as $\frac{PE}{CM}$ . Notice that the power of $M$ w.r.t to $\omega_1, \omega_2$ is $MA^2=MB^2$ , so $M$ lies on the radical axis, along with $Z$ . Then $\angle MZO_1 = 90$ , so $\angle CMZ = \angle O_1 MZ = 90 - \angle ZO_1 M = 90 - \angle PO_1 E = \angle O_1 P E = \angle ZPE$ . Then by AA, $\triangle EZP \sim \triangle CZM$ , so $\frac{EZ}{ZC} = \frac{PE}{CM}$ , as desired.

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Jndd

#27 h Aug 23, 2022, 8:41 PM

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Let $O$ and $O_1$ be the centers of $\omega$ and $\omega_1$ respectively, and let $X$ be the intersection of $PC$ with $\omega_1$ with $X\neq D$ and $Y$ be the foot of the perpendicular from $X$ to $EC$ . $C, E, M$ are collinear and it's well known that $D, F, M$ are collinear. $C, Z, F$ are collinear because $\angle{O_1ZF}=\angle{O_1FZ}=\angle{ZCO}=\angle{OZC}$ because $FX\parallel EC$ . Since $\triangle{ZEF}\sim\triangle{ECF}$ , $\angle{ZEP}=\angle{ECF}$ . Call $H$ the orthocenter of $\triangle{CEX}$ . Because $\triangle{CHX}\sim\triangle{CFP}$ and $\triangle{CEH}\sim\triangle{CMF}$ , we get $\frac{CX}{CP}=\frac{CH}{CF}=\frac{CE}{CM}$ which means $\triangle{CEX}\sim\triangle{CMP}$ .
Hence, $\tan{\angle{ZEP}}=\tan{\angle{ECF}}=\frac{FE}{CE}=\frac{XY}{CE}=\frac{PE}{CM}$ , as desired.

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PNT

#29 h Dec 7, 2022, 1:24 PM

Y by

Sardor wrote:

Here my solution:
Obviously, $P$ is exsimilicenter of $w$ and $w_1$ , hence $P,O,O_1,Z$ are collinear, where $O ,O_1$ be the centers of $w,w_1$ , respectively.We have easily that $\angle PEZ=\angle ZCE$ , and $\angle MZC=\angle PZE$ (because we know that $MZ$ is common internal tangent of $w$ and $w_1$ ).Thus the triangles $PZE$ and $CZM$ are relatively similar.Hence $tan \angle ZEP=tan \angle ECZ= \frac{ZE}{CZ}=\frac{PE}{CM}$ , so we are done !

Why $P, O_1, Z$ and $P$ are collinear?

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#30 h Dec 15, 2022, 5:14 AM

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@above There is a homothety that takes $\omega _1$ to $\omega$ . Lilavati's solution explains it well.

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#31 h Sep 1, 2023, 11:04 AM • 1 Y

Y by Vahe_Arsenyan

Clarly $C-E-M$ by shooting lemma, so as $E$ is the midpoint of $AB$ we get that $CM$ is diameter of $\Omega$ and $CE$ is diameter of $\omega.$ Now notice that $\angle ZEP=\angle ZCE,$ and $\tan \angle ZCE=\frac{ZC}{CE}.$ We will show that $\triangle ZEP\sim \triangle ZCM,$ which will trivially imply that we are done. As $\angle ZEP=\angle ZCE$ we just need to show $\angle PZE=\angle MZC:$

For this invert at $M$ with radius $MA=MB.$ Clarly $\omega \to \omega, AB\to \Omega, \Omega\to AB.$ So this shows that $(DZF)\to(DZF),$ implying that $MZ$ is tangent to both $\omega, (DZF).$ We now claim that $\angle PZM=90^\circ.$ Notice that by the tangency, to show this is equivalent to showing that $PZ$ passes through the centers of $(DZF)$ and $\omega.$ But this is true by Monge d'Alembert on $(DZF), \omega, \Omega,$ as this shows that $P$ is the exscimillicenter of $(DZF), \omega.$ Now just end the problem by computing $\angle PZE=90^\circ+\angle MZE=\angle MZC,$ solving the problem.

This post has been edited 1 time. Last edited by Bigtaitus, Sep 1, 2023, 10:12 PM

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#32 h Apr 3, 2025, 10:40 PM

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Note first that the points $P$ , $D$ , $E$ , and $M$ are concyclic because $\angle{PDM}=\angle{PEM}=90$ .

Claim: $Z$ lies on $(PDEM)$ .

Proof: It suffices to show that $\angle{PZM}=90$ . Since $M$ is the midpoint of major arc $AB$ , $MZ$ is tangent to $\omega$ and $\Omega$ , and since $P$ is the center of a homothety mapping $\omega$ to $\Omega$ , $PZ$ passes through the centers of both circles.

Extend $CZ$ to point $P’$ such that the foot of the altitude from $P’$ to $CE$ is $M$ . Note that $\triangle CZE \sim \triangle CMP’$ , so $\frac{CZ}{CM}=\frac{CE}{CP’}\implies CZ(CP’)=CE(CM)$ . By power of a point, $Z$ , $P’$ , $M$ , and $E$ are concylic, meaning $P’$ lies on $(PDZEM)$ . Since $PM$ is a diameter, $\angle{PP’M}=90$ , implying $PP’ME$ is a rectangle. Finally, $\tan{ZEP}=\tan{ZCM}=\frac{P’M}{CM}=\frac{PE}{CM}$ , as desired.

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