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Another circle tangent to circle problem

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Source: INAMO 2023 P7 (OSN 2023)

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#1 h Aug 30, 2023, 8:40 AM • 1 Y

Y by HWenslawski

Given a triangle $ABC$ with $\angle ACB = 90^{\circ}$ . Let $\omega$ be the circumcircle of triangle $ABC$ . The tangents of $\omega$ at $B$ and $C$ intersect at $P$ . Let $M$ be the midpoint of $PB$ . Line $CM$ intersects $\omega$ at $N$ and line $PN$ intersects $AB$ at $E$ . Point $D$ is on $CM$ such that $ED \parallel BM$ . Show that the circumcircle of $CDE$ is tangent to $\omega$ .

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#2 h Aug 30, 2023, 9:06 AM • 2 Y

Y by arifsulistianingsih, HWenslawski

$[asy] usepackage("tikz"); label("\begin{tikzpicture}[font=\small] \coordinate[label=below left:$C$] (C) at (0,0); \coordinate[label=above:$A$] (A) at (0,6); \coordinate[label=below:$B$] (B) at (5,0); \coordinate[label=above right:$O$] (O) at (2.5,3); \coordinate[label=below:$P$] (P) at (2.5,-2.08); \coordinate[label=right:$M$] (M) at (3.75,-1.04); \coordinate[label=above left:$N$] (N) at (3.09,-.86); \coordinate[label=above right:$E$] (E) at (4.06,1.13); \coordinate[label=below:$D$] (D) at (2.03,-.56); \draw[thick] (A)--(B)--(C)--cycle; \draw[thick] (C)--(P)--(B); \draw[thick] (D)--(E)--(P); \draw[dashed] (1.93,.94) circle (2.14); \draw[thick] (C)--(E); \draw[thick] (C)--(M); \draw[thick] (C)--(O)--(N); \draw[thick] (O) circle (3.91); \draw[thick] (B)--(N); \coordinate[label=below right:$N'$] (N') at (4.41,-1.22); \draw[thick] (P)--(N')--(B); \draw[thick] (M)--(N'); \draw[dotted] (2.5,.46) circle (2.54); \foreach \s in {A,B,C,O,E,D,M,N,P,N'}\filldraw (\s) circle (1.5pt); \end{tikzpicture}"); [/asy]$

Suppose $O$ is the centre of $(ABC)$ , it is clear that $O$ is the midpoint of $AB$ . Suppose $\angle CAB=\alpha$ , then $\angle CBP=90^\circ-\angle CBA=\alpha$ . Given the length $PB=PC$ , then $\angle BCP=\alpha$ and $\angle CPB=180^\circ-2\alpha$ . Suppose $N'$ is the reflection of point $M$ to point $N$ . From PoP to $(ABC)$ holds $MP\cdot MB=MB^2=MN\cdot MC=MN'\cdot MC$ , then $BN'PC$ is cyclic. Consider $BNPN'$ a parallelogram and $\angle COB = 2\angle CAB = 2\alpha$ and
$\angle CNE = \angle PNN' = \angle BN'N=\angle BN'C = \angle BPC= 180^\circ-2\alpha.$ Since $\angle BOE+\angle BNE=180^\circ$ , then $CNEO$ is cyclic. From this, it is obtained
$\angle NCB = \frac{\angle NOB}{2}=\frac{\angle NOE}{2} =\frac{\angle NCE}{2}\implies \angle NCB=\angle ECB.$ Then
$\begin{align*} \angle CED &= 180^\circ - \angle ECD - \angle EDC\\\\ &= 180^\circ-2\angle MCB - \angle BMC\\\ &= \angle MBC - \angle MCB\\\ &= \angle PBC - \angle MCB\\\ &= \angle PCB - \angle MCB\\\ &= \angle PCN, \end{align*}$ which is proven by Alternate Segment Theorem.

Remark. Looking at the midpoint will most likely pertain to PoP. After a few steps back, I needed to prove the CB bisect $\angle ECN$ . Seeing the midpoint condition also motivated me to construct the parallelogram, which killed it :p

This post has been edited 4 times. Last edited by Wildabandon, Aug 30, 2023, 9:21 AM
Reason: Add remark

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#3 h Aug 30, 2023, 11:16 AM • 1 Y

Y by arifsulistianingsih

INAMO 2023/7 wrote:

Given a triangle $ABC$ with $\angle ACB = 90^{\circ}$ . Let $\omega$ be the circumcircle of $ABC$ . The tangents to $\omega$ at points $B$ and $C$ meet at $P$ . Let $M$ be the midpoint of $PB$ . The line $CM$ intersects $\omega$ at $N$ , and the line $PN$ intersects $AB$ at $E$ . Point $D$ is on $CM$ such that $ED \parallel BM$ . Prove that the circumcircle of triangle $CDE$ is tangent to $\omega$ .

Reflect $N$ with respect to $AB$ to get point $F$ . Since $\angle ACB = 90^{\circ}$ , $AB$ is the diameter of the circle and hence $AB \perp PB$ . Since $FN \perp AB$ by definition, we obtain $FN \parallel PB$ .

Claim. $C, E, F$ are collinear.
Proof. We'll do this by phantom point. Let $E' = CF \cap AB$ . We'll prove that $P, N, E'$ are collinear. Note that by Power of Point, $MB^2 = MN \cdot MC$ and since $M$ is the midpoint of $BP$ , we obtain $MP^2 = MN \cdot MC$ , which implies $\triangle MPN \sim \triangle MCP$ . This implies $\angle MCP = \angle MPN$ . Therefore, we have
$\angle FNE' = \angle E'FN = \angle CFN = \angle NCP = \angle MCP = \angle MPN$ However, as $NF \parallel MP$ , we obtain the desired result.

To finish the problem, note that we have $DE \parallel PB \parallel NF$ , $C,D,N$ are collinear and $C,E,F$ are collinear. Thus, the homothety from point $C$ sends $(CNF) \equiv \omega$ to $(CDE)$ and hence $(CDE)$ is tangent to $\omega$ as desired.

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#4 h Aug 30, 2023, 11:22 AM • 2 Y

Y by Mhremath, MS_asdfgzxcvb

$PN$ intersects circle at $F$ . $NF \cap BC=G$ .
$-1=(F,N;G,P)=(CF\cap BP,M; B,P) \implies CF \parallel BP \parallel DE$ . If $CE \cap (ABC)=H$ , then $FE=EC, FC \perp AB \implies HN \parallel FC \parallel ED$ which is enough.

This post has been edited 2 times. Last edited by SerdarBozdag, Aug 30, 2023, 4:39 PM

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#5 h Aug 30, 2023, 4:00 PM

Y by

We remove $A$ ; all we need from the information about $A$ is that $E \in PN$ and $BE$ is perpendicular to $BP$ . Instead, we start with an isosceles triangle $BCP$ with $PB = PC$ , and construct $\omega$ as the unique circle tangent to $BP$ and $B$ and $CP$ at $C$ . (Note that $\omega$ only has to pass through $C$ ; the property $PB = PC$ would automatically ensure that $\omega$ is tangent to $CP$ at $C$ .) We replace the condition $ED || BM$ with $ED \bot BE$ , since they are indeed equivalent due to $BE \bot BP$ .

Now invert at $B$ , and you get the following very nice picture. The description changes as follows.

Let $BCP$ be an isosceles triangle with $BC = CP$ . Let $\omega$ be the line parallel to $PB$ passing through $C$ . Let $M$ be the point on the plane such that $P$ is the midpoint of $BM$ . The line $\omega$ and the circumcircle of $BCM$ intersects at $N$ . The circumcircle of $BPN$ intersects the line perpendicular to $BP$ passing through $B$ at $E$ . (In short, $E$ is the antipodal point of $P$ with respect to the circumcircle of $BPN$ .)

Now let $D$ be point on the circumcircle of $BCM$ such that $BD \bot DE$ (i.e. it is the second intersection of the circumcircle of $BCM$ and the circle with diameter $BE$ .) Show that the circumcircle of $CDE$ is tangent to the line $\omega$ .

Solution.
Let $F$ be the second intersection of the line $\omega$ (or $NC$ ) with the circumcircle of $BPN$ . It suffices to show that $\angle EDC = \angle ECF$ . Now notice that
$\angle EDC = \angle EDB - \angle CDB = \pi/2 - \angle CNB = \pi/2 - \angle FNB = \pi/2 - \angle FEB,$ with the second equation being true since the problem condition implies that $B, C, D, N, M$ lie in one circle. Since $\overline{BE} \bot \overline{BM} || \overline{CN} = \overline{CF}$ , we have $\angle ECF = \pi/2 - \angle CEB$ . So it remains to show that $\angle CEB = \angle FEB$ . It now suffices to show that $F$ is actually the reflection of $C$ with respect to the line $BE$ .

Since $CNMB$ is cyclic and $\overline{CN}$ is parallel to $\overline{BM}$ , $N$ is actually the reflection of $C$ with respect to the perpendicular bisector of $BM$ , which in turn is the line through $P$ perpendicular to $BM$ since $P$ is the midpoint of $BM$ . In particular, $PN = PC = CB$ .

With the same reasoning, since $FPBN$ is cyclic and $\overline{FN}$ is parallel to $\overline{BM}$ , $F$ is the reflection of $N$ with respect to the perpendicular bisector of $BP$ , which passes through $C$ since $BC = CP$ . In summary, the reflection of the perpendicular bisector of $PB$ brings $P$ to $B$ , $N$ to $F$ , and $C$ to itself. Thus $CB = CP = PN = BF$ . Since $\overline{BE}$ is perpendicular to $\overline{CF}$ , the rest follows.

For some reason, I've written too much for showing that $\overline{BE}$ is the perpendicular bisector of $CF$ . Seriously, I think that the inverted diagram speaks for itself.

When I tried to solve it, I noticed that maybe I could invert at $B$ , but I was also looking towards inverting at $C$ . The point $M$ was very annoying to me, maybe because I actually looked at $C$ instead of $B$ for a long time. Initially, I only actually try to invert at $B$ not because $M$ can be inverted nicely, but because I noticed that the circle $PCN$ is tangent to $PB$ at $P$ (after which I removed $M$ .) After solving it this way, I realized that inverting $M$ at $B$ adds no complication at all...

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#7 h Aug 30, 2023, 5:25 PM • 4 Y

Y by hi_how_are_u, centslordm, Sanjaya7114, MS_asdfgzxcvb

For homothety reasons it suffices to show that if $N'$ is the reflection of $N$ over $\overline{AB}$ and $E'=\overline{AB} \cap \overline{CN'}$ , then $E',N,P$ are collinear. This is straightforward by complex, with $A=-1$ and $B=1$ , so $c$ lies on the unit circle. First,
$P=\frac{2c}{c+1} \implies M=\frac{3c+1}{2c+2}.$ Then we can calculate
$N=\frac{c-\frac{3c+1}{2c+2}}{\frac{c(c+3)}{2c+2}-1}=\frac{2c^2-c-1}{c^2+c-2}=\frac{2c+1}{c+2}.$ In general we should have $E'=\tfrac{cn+1}{c+n}$ by complex intersection. Plugging in our expression for $N$ yields
$E'=\frac{\frac{2c^2+c}{c+2}+1}{c+\frac{2c+1}{c+2}}=\frac{2c^2+2c+2}{c^2+4c+1}.$ Now we want to show that
$\frac{\frac{2c^2+2c+2}{c^2+4c+1}-\frac{2c+1}{c+2}}{\frac{2c^2+2c+2}{c^2+4c+1}-\frac{2c}{c+1}} \in \mathbb{R} \iff \frac{(c+1)((2c^2+2c+2)(c+2)-(2c+1)(c^2+4c+1)}{(c+2)((2c^2+2c+2)(c+1)-2c(c^2+4c+1))} \in \mathbb{R}.$ It turns out that this second fraction simplifies nicely to
$\frac{(c+1)(-3c^2+3)}{(c+2)(-4c^2+2c+2)}=\frac{3}{4}\cdot\frac{(c-1)(c+1)^2}{(c-1)(c+2)(2c+1)}=\frac{3}{8}\cdot \frac{c+\frac{1}{c}+2}{c+\frac{1}{c}+\frac{5}{2}},$ which is evidently real. $\blacksquare$

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#8 h Aug 31, 2023, 1:09 AM • 2 Y

Y by IAmTheHazard, centslordm

You want to prove that if $CE$ intersect $\omega$ at $F$ , then $NF\parallel BP$ . But this is the same as if $C'$ is the reflection of $C$ about $AB$ then we want $C', N, P$ colinear.

But project from $C$ , $CC'$ is parallel to $PB$ , $CN$ bisects $BP$ , and $C'C$ is tangent to $\omega$ , so $(C', N; B, C)=-1$ , so $P, N, C'$ colinear. QED

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#9 h Aug 31, 2023, 11:32 AM • 1 Y

Y by arifsulistianingsih

Good problem. I have different sol. Let $AN\cap BC=L$ , and $E'$ be the foot of perpendicular through $L$ to $AB$ . Since $MB$ is tangent to $(N C B)$ we have $MN*MC=MB^2=MP^2$ , then $MP$ is tangent to $(C N P)$ . Then $\angle NPM=\angle PCM=\angle NBC=\angle NBL$ , since $(L E' B N)$ is cyclic, $\angle NPM=\angle NBL=\angle NE'L$ adding to this $E'L//PB$ (because $\angle LE'B=\angle PBE'=90$ ) gives that $P$ , $N$ , $E'$ are collinear which means that $E'=E$ . Then we know that $( C A E L)$ is cyclic, then $\angle LEC=\angle LAC=\angle NAC=\angle PCM$ which means that $PN$ is also tangent to $(C D E)$ , so done:).

This post has been edited 4 times. Last edited by X.Allaberdiyev, Sep 1, 2023, 4:42 PM
Reason: :)

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crocodilepradita

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#10 h Jun 30, 2024, 4:28 AM

Y by

Good problem. Here is my solution.

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Let $ED \cap CP=X$ and $O$ is the center of $\omega$ .

Claim. $ONXC$ is a kite
Proof.
Note that using Power of Point we have $MB^2=MN \cdot MC=MP^2$ , therefore by Alternate Segment Theorem $MP$ tangent to $(CNP)$ .

So, $\angle{MPN}=\angle{DEN}=\angle{XEN}$ . Thus, $\angle{XCN}=\angle{XEN} \implies$ $CENX$ cyclic.

Furthermore, because $ED\parallel BM$ thus $\angle{DEO}=\angle{XEO}=90^\circ$ . Because $\angle{XCO}=90^\circ$ we have $OEXC$ cyclic.

Therefore, $ONEXC$ is cyclic quadrilateral. So $\angle{ONX}=\angle{OEX}=90^\circ$ , and because $ON=OC$ we have $ONXC$ is a kite. $\square$

Therefore, we have $\angle{XON}=\angle{COX}=\angle{CEX}=\angle{CED}$ but we know that $\angle{XON}=\angle{XEN}=\angle{MPN}=\angle{PCD}$ .

By Alternate Segment Theorem, we have $(CDE)$ is tangent to $\omega$ . $\blacksquare$

This post has been edited 12 times. Last edited by crocodilepradita, Jun 30, 2024, 5:16 AM

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#11 h Mar 1, 2025, 3:55 AM

Y by

Nice problem :-D

:-D

, there are lots of solutions, here is the solution I found during the contest:

Let $F$ be the reflection of $P$ onto $B$ . The main idea is to prove that $C,E,F$ are collinear.
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.1454568863013947, xmax = 1.6563565995292908, ymin = -0.05935998820010408, ymax = 1.1097715542036222; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw(arc((-0.5419443881458441,0.),0.06153323907388036,0.,27.440649059739854)--(-0.5419443881458441,0.)--cycle, linewidth(1.5) + dtsfsf); draw(arc((-0.49838208602155787,0.5401907493436052),0.06153323907388036,-94.61049462177301,-67.16984556203316)--(-0.49838208602155787,0.5401907493436052)--cycle, linewidth(1.5) + dtsfsf); draw(arc((0.5419443881458441,0.),0.06153323907388036,152.55935094026015,180.)--(0.5419443881458441,0.)--cycle, linewidth(1.5) + dtsfsf); draw(arc((0.,0.2814052630635615),0.06153323907388036,512.5593509402601,540.)--(0.,0.2814052630635615)--cycle, linewidth(1.5) + dtsfsf); /* draw figures */ draw(circle((0.,0.5), 0.5), linewidth(2.) + green); draw((0.,1.)--(-0.49838208602155787,0.5401907493436052), linewidth(1.6) + wrwrwr); draw((-0.49838208602155787,0.5401907493436052)--(0.,0.), linewidth(1.6) + wrwrwr); draw((-0.49838208602155787,0.5401907493436052)--(-0.5419443881458441,0.), linewidth(1.6) + wrwrwr); draw((-0.5419443881458441,0.)--(0.,0.), linewidth(1.6) + wrwrwr); draw((0.,1.)--(0.,0.), linewidth(1.6) + wrwrwr); draw((-0.49838208602155787,0.5401907493436052)--(-0.27097219407292206,0.), linewidth(1.6) + wrwrwr); draw((0.,0.2814052630635615)--(-0.5419443881458441,0.), linewidth(1.6) + wrwrwr); draw((-0.389438385033842,0.2814052630635615)--(0.,0.2814052630635615), linewidth(1.6) + wrwrwr); draw((0.,0.)--(0.5419443881458441,0.), linewidth(1.6) + wrwrwr); draw((-0.49838208602155787,0.5401907493436052)--(0.5419443881458441,0.), linewidth(1.6) + wrwrwr); draw(circle((-0.194719192516921,0.5157026315317808), 0.3046486842341096), linewidth(2.) + green); /* dots and labels */ dot((0.,1.),linewidth(2.pt) + dotstyle); label("$A$", (-0.009143071403737501,1.0174716955928018), NE * labelscalefactor); dot((0.,0.),linewidth(2.pt) + dotstyle); label("$B$", (-0.005040855465478811,-0.04679780053968129), NE * labelscalefactor); dot((-0.49838208602155787,0.5401907493436052),linewidth(2.pt) + dotstyle); label("$C$", (-0.5442267115008499,0.5375124308165351), NE * labelscalefactor); dot((-0.5419443881458441,0.),linewidth(2.pt) + dotstyle); label("$P$", (-0.5931977629143075,-0.04679780053968129), NE * labelscalefactor); dot((-0.27097219407292206,0.),linewidth(2.pt) + dotstyle); label("$M$", (-0.27988932332881106,-0.04679780053968129), NE * labelscalefactor); dot((-0.3195799007083312,0.11546302250205742),linewidth(2.pt) + dotstyle); label("$N$", (-0.29424707911271647, 0.09267754136111412), NE * labelscalefactor); dot((0.,0.2814052630635615),linewidth(2.pt) + dotstyle); label("$E$", (0.023674656102332024,0.2749706107679791), NE * labelscalefactor); dot((-0.389438385033842,0.2814052630635615),linewidth(2.pt) + dotstyle); label("$D$", (-0.380393613816149,0.2893283665518845), NE * labelscalefactor); dot((0.5419443881458441,0.),linewidth(2.pt) + dotstyle); label("$F$", (0.5528605121377032,-0.04679780053968129), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Observe that $MP^2 = MB^2 = MN \cdot MC$ , so $MP$ is tangent to $(PNC)$ , so $\angle PCM = \angle FPE$ .
Observe that since $AB$ is a diameter of $\omega$ and $PB$ is tangent to $\omega$ at $B$ , we have $EB\perp PF$ . So, $\angle FPE = \angle PFE$ .

Claim: $C,E,F$ are collinear.
Proof: since $PC^2=PB^2 = \left(\frac{1}{2}PB \right) (2PB) = PM \cdot PF$ , hence $PC$ is tangent to $(CMF)$ , hence $\angle PFC = \angle PCM = \angle FPE = \angle PFE, \text{ proven } \square$
Finally, observe that $ED\parallel FP$ so $\angle DEC = \angle PFC = \angle PCM$ . By inscribed angle theorem, $PC$ is tangent to $(CDE)$ at $C$ . Since $PC$ is also tangent to $\omega$ at $C$ , we conclude that $(CDE)$ and $\omega$ are tangent at $C$ $\blacksquare$

This post has been edited 2 times. Last edited by GreenTea2593, Mar 1, 2025, 9:56 AM

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#12 h Mar 1, 2025, 1:01 PM

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$CE$ meets $(\omega)$ at $J, C$ . We will aim to show that $DE \parallel NJ$ which is enough. This is equivalent to $NJ \parallel BP$ or showing that $BA$ is the perpidencular bisector of segment $NJ$ . Since we already have $ON = OJ$ , proving $BA$ bisects $\angle JBN$ will suffice.
The rest is just a very long angle chase: Notice how $N$ is the $C$ -humpty point of $\triangle BCP$ , so:
$\angle ABJ = \angle ABN \Leftrightarrow \angle ACE = \angle ABN \Leftrightarrow \angle ACE + \angle ACM = 180^{\circ} \Leftrightarrow \angle ECB = \angle MCB$ Notice that, $\angle COE = 180^{\circ} - \angle CPB = 180^{\circ} - \angle CNE$ , so $C, O, E, N$ are concyclic. Thus:
$\begin{split} \angle ECB = \angle OCB - \angle OCE = \angle OCB - \angle ONE &= \angle OBC - \angle ONB + \angle BNE = \angle OBC - \angle OBN + \angle BNE \\&= \angle BNE - \angle CBN = 180^{\circ} - \angle BNP - \angle CBM + \angle NBM \\&= \angle BCP - \angle CBM + \angle BCM = \angle MCB \end{split}$ So we should be done...
Kudos to whoever read my crappy solution :wacko:

:wacko:

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#13 h Mar 4, 2025, 2:28 AM

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GreenTea2593 wrote:

Nice problem :-D

:-D

, there are lots of solutions, here is the solution I found during the contest:

Let $F$ be the reflection of $P$ onto $B$ . The main idea is to prove that $C,E,F$ are collinear.
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.1454568863013947, xmax = 1.6563565995292908, ymin = -0.05935998820010408, ymax = 1.1097715542036222; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw(arc((-0.5419443881458441,0.),0.06153323907388036,0.,27.440649059739854)--(-0.5419443881458441,0.)--cycle, linewidth(1.5) + dtsfsf); draw(arc((-0.49838208602155787,0.5401907493436052),0.06153323907388036,-94.61049462177301,-67.16984556203316)--(-0.49838208602155787,0.5401907493436052)--cycle, linewidth(1.5) + dtsfsf); draw(arc((0.5419443881458441,0.),0.06153323907388036,152.55935094026015,180.)--(0.5419443881458441,0.)--cycle, linewidth(1.5) + dtsfsf); draw(arc((0.,0.2814052630635615),0.06153323907388036,512.5593509402601,540.)--(0.,0.2814052630635615)--cycle, linewidth(1.5) + dtsfsf); /* draw figures */ draw(circle((0.,0.5), 0.5), linewidth(2.) + green); draw((0.,1.)--(-0.49838208602155787,0.5401907493436052), linewidth(1.6) + wrwrwr); draw((-0.49838208602155787,0.5401907493436052)--(0.,0.), linewidth(1.6) + wrwrwr); draw((-0.49838208602155787,0.5401907493436052)--(-0.5419443881458441,0.), linewidth(1.6) + wrwrwr); draw((-0.5419443881458441,0.)--(0.,0.), linewidth(1.6) + wrwrwr); draw((0.,1.)--(0.,0.), linewidth(1.6) + wrwrwr); draw((-0.49838208602155787,0.5401907493436052)--(-0.27097219407292206,0.), linewidth(1.6) + wrwrwr); draw((0.,0.2814052630635615)--(-0.5419443881458441,0.), linewidth(1.6) + wrwrwr); draw((-0.389438385033842,0.2814052630635615)--(0.,0.2814052630635615), linewidth(1.6) + wrwrwr); draw((0.,0.)--(0.5419443881458441,0.), linewidth(1.6) + wrwrwr); draw((-0.49838208602155787,0.5401907493436052)--(0.5419443881458441,0.), linewidth(1.6) + wrwrwr); draw(circle((-0.194719192516921,0.5157026315317808), 0.3046486842341096), linewidth(2.) + green); /* dots and labels */ dot((0.,1.),linewidth(2.pt) + dotstyle); label("$A$", (-0.009143071403737501,1.0174716955928018), NE * labelscalefactor); dot((0.,0.),linewidth(2.pt) + dotstyle); label("$B$", (-0.005040855465478811,-0.04679780053968129), NE * labelscalefactor); dot((-0.49838208602155787,0.5401907493436052),linewidth(2.pt) + dotstyle); label("$C$", (-0.5442267115008499,0.5375124308165351), NE * labelscalefactor); dot((-0.5419443881458441,0.),linewidth(2.pt) + dotstyle); label("$P$", (-0.5931977629143075,-0.04679780053968129), NE * labelscalefactor); dot((-0.27097219407292206,0.),linewidth(2.pt) + dotstyle); label("$M$", (-0.27988932332881106,-0.04679780053968129), NE * labelscalefactor); dot((-0.3195799007083312,0.11546302250205742),linewidth(2.pt) + dotstyle); label("$N$", (-0.29424707911271647, 0.09267754136111412), NE * labelscalefactor); dot((0.,0.2814052630635615),linewidth(2.pt) + dotstyle); label("$E$", (0.023674656102332024,0.2749706107679791), NE * labelscalefactor); dot((-0.389438385033842,0.2814052630635615),linewidth(2.pt) + dotstyle); label("$D$", (-0.380393613816149,0.2893283665518845), NE * labelscalefactor); dot((0.5419443881458441,0.),linewidth(2.pt) + dotstyle); label("$F$", (0.5528605121377032,-0.04679780053968129), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Observe that $MP^2 = MB^2 = MN \cdot MC$ , so $MP$ is tangent to $(PNC)$ , so $\angle PCM = \angle FPE$ .
Observe that since $AB$ is a diameter of $\omega$ and $PB$ is tangent to $\omega$ at $B$ , we have $EB\perp PF$ . So, $\angle FPE = \angle PFE$ .

Claim: $C,E,F$ are collinear.
Proof: since $PC^2=PB^2 = \left(\frac{1}{2}PB \right) (2PB) = PM \cdot PF$ , hence $PC$ is tangent to $(CMF)$ , hence $\angle PFC = \angle PCM = \angle FPE = \angle PFE, \text{ proven } \square$
Finally, observe that $ED\parallel FP$ so $\angle DEC = \angle PFC = \angle PCM$ . By inscribed angle theorem, $PC$ is tangent to $(CDE)$ at $C$ . Since $PC$ is also tangent to $\omega$ at $C$ , we conclude that $(CDE)$ and $\omega$ are tangent at $C$ $\blacksquare$

bro wrote his solution only to flex his asy skill :p

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