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Source: Iranian Geometry Olympiad 2022 P2 Advanced, Free

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309 posts

#1 h Dec 13, 2022, 5:17 PM • 4 Y

Y by itslumi, GeoKing, Rounak_iitr, ItsBesi

We are given an acute triangle $ABC$ with $AB\neq AC$ . Let $D$ be a point of $BC$ such that $DA$ is tangent to the circumcircle of $ABC$ . Let $E$ and $F$ be the circumcenters of triangles $ABD$ and $ACD$ , respectively, and let $M$ be the midpoints $EF$ . Prove that the line tangent to the circumcircle of $AMD$ through $D$ is also tangent to the circumcircle of $ABC$ .

Proposed by Patrik Bak, Slovakia

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#2 h Dec 13, 2022, 6:00 PM

Y by

Solved with @primesarespecial and @theproblemissolved as far as I can remember.
Sketch/ Key claim/ wtv

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#3 h Dec 13, 2022, 6:17 PM • 1 Y

Y by AhmadGgx

Let $DP$ be the tangent from $D$ to $(ABC)$ . Obviously $EF$ is the perpendicular bisector of $AD$ , so $PD$ is tangent to $(AMD)$ iff $MD$ bisects $\angle APD$ . But $\triangle APD$ is isosceles, so we need $MD \perp AP$ , so proving $\angle FMD = \angle PAD$ is sufficient. We have that $\angle AMF= \angle FMD$ , so we need $\angle PAD= \angle AMF$ . Add the midpoint $N$ of $BC$ ; notice that by angle chasing $\triangle AEF \cup \{M\} \sim \triangle ABC \cup \{N\}$ , so $\angle AMF= \angle ANC= \angle ABP =\angle PAD$ , done (we used that $AN$ and $AP$ are isogonal in $\angle BAC$ ).

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#5 h Dec 13, 2022, 6:35 PM • 1 Y

Y by MS_asdfgzxcvb

Let $O$ be the circumcenter of $(ABC)$ and $DG$ be the other tangent line through $D$ to $(ABC)$ . Notice that $\triangle ADE\sim\triangle AOC$ , so $DE\perp AC$ . And $OF\perp AC$ for the obvious reason. Similarly $DF\perp AB\perp OE$ . So $EDFO$ is a parallelogram. So $M$ be the midpoint of $DO$ which bisects $\angle ADG$ . Hence $\angle DAM=\angle ADM=\angle MDG$ . Done!

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#6 h Dec 13, 2022, 7:05 PM

Y by

Let $\omega = \odot(ABC)$ and $O$ be its center; $K$ be the point of $\omega$ distinct from $A$ such that $DK$ is tangent to $\omega$ ; $N$ be midpoint of $BC$ ; $S$ be reflection of $A$ in line $BC$ and $T$ be reflection of $A$ in perpendicular bisector of segment $BC$ (or line $ON$ ).

We want to show $DK$ is also tangent to $\odot(AMD)$ .

Lemma 1: (Known) Let $ABC$ be a triangle and $D$ be any point on side $BC$ . Let $E,F$ be circumcenters of $\triangle ABD, \triangle ACD$ , respectively. Then,
$\triangle AEF \stackrel{+}{\sim} \triangle ABC$ Proof: Note $EA = EB$ , $FA = FC$ . Further,
$\measuredangle BAE = 90^\circ - \measuredangle ADB = 90^\circ - \measuredangle ADC = \measuredangle CAF$ It follows
$\triangle AEB \stackrel{+}{\sim} \triangle AFC$ which implies $\triangle AEF \stackrel{+}{\sim} \triangle ABC$ , as desired. $\square$
$[asy] size(250); pair B=dir(-165),C=dir(-15),A=dir(120),O=(0,0),N=1/2*(B+C),S=2*foot(A,B,C)-A,K=IP(N--S,unitcircle),T=2*foot(A,N,O)-A,D=2*A*K/(A+K); pair E=circumcenter(A,B,D),F=circumcenter(A,D,C),M=1/2*(E+F); fill(A--D--M--A--cycle,cyan+white+white); fill(A--S--N--cycle,green+white+white); draw(A--D--K,brown); draw(E--F,magenta); draw(D--C,red); draw(S--T); draw(A--M--D,purple); draw(S--A--N,purple); draw(A--B--S--C--A,orange+linewidth(0.8)); draw(A--E--D--F--A,fuchsia+linewidth(0.8)); draw(unitcircle,royalblue); draw(circumcircle(A,O,K),grey); dot("$A$",A,dir(90)); dot("$B$",B,dir(-110)); dot("$C$",C,dir(C)); dot("$O$",O,dir(60)); dot("$N$",N,dir(-60)); dot("$S$",S,dir(S)); dot("$T$",T,dir(T)); dot("$K$",K,dir(-70)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$M$",M,dir(-140)); [/asy]$
Let $f$ be the spiral similarity centered at $A$ taking $E \to B$ and $F \to C$ . Note $f(M) = N$ . Further, as $D$ is reflection of $A$ in line $EF$ , so $f(D) = S$ . It follows
$AEFMD \stackrel{+}{\sim} ABCNS$ In particular, this gives
$\angle AMD = \angle ANS \qquad \qquad (1)$ Claim 2: Points $S,K,N,T$ are collinear.

Proof: As $AK$ is symmedian (in $\triangle ABC$ ), so it follows $K \in NT$ . Also, as $N$ is circumcenter of $\triangle TAS$ with $\angle TAS = 90^\circ$ , it follows $N \in TS$ . It follows both $K,S$ lie on $TN$ , as desired. $\square$

Now note the points $A,O,N,K,D$ lie on circle with diameter $DO$ . We obtain
$\angle AMD = \angle ANS = \angle ANK = 180^\circ - \angle ADK$ It follows $DK$ is tangent to $\odot(AMD)$ at $D$ , which completes the proof. $\blacksquare$

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#7 h Dec 13, 2022, 9:54 PM • 1 Y

Y by Vladimir_Djurica

Clearly $EF$ is the perpendicular bisector of $AD$ and as $M$ lies on $EF$ , then $MA = MD$ . Let $O$ be the circumcenter of triangle $ABC$ .

Claim: $\angle EOF = \angle BAC$
Proof: It's easy see that $EO$ is perpendicular bisector of $AB$ and $FO$ is perpendicular bisector of $AC$ . Then if $I = EO \cap AB$ and $J = FO \cap AC$ we have that $AIOJ$ is cyclic, where $\angle BAC = \angle IAJ = 180^{\circ} - \angle IOJ = \angle EOF \square$

Claim: $DEOF$ is a parallelogram
Proof: It's suffices show that $\angle EDF = \angle EOF$ and $\angle DEO = \angle DFO$ . Notice that
$\angle DEO = \angle DEB + \angle BEO = 2 \angle DAB + \frac{\angle BEA}{2} = 2\angle DAB + \angle ADB$ .
Now, $\angle DFO = \angle DFA + \angle AFO = 2\angle DCA + \frac{AFC}{2}$ but $AD$ is tangent to $\odot (ABC)$ at $A$ , then $\angle DAB = \angle DCA$ . Therefore
$\angle DFO = 2\angle DCA + \frac{AFC}{2} = 2\angle DAB + \angle ADC = 2\angle DAB + \angle ADB = \angle DEO$ .
We observe that
$\angle EDF = \angle EDB + \angle BDF = \angle EDB + \angle CDF = (90^{\circ} - \angle DAB) + (90^{\circ} - (180^{\circ}- \angle DAC)) = \angle DAC - \angle DAB = \angle BAC = \angle EOF$ $\square$

Thus $DEOF$ is a parallelogram whose diagonals intersect at their midpoint, which is $M$ . Then $D$ , $M$ and $O$ are collinear.
Let $T$ be the tangency point from $D$ to the $\odot (ABC)$ different of $A$ ; clearly $DO$ is perpendicular bisector of $AT$ , then
$\angle TDO = \angle ODA = \angle MDA = \angle MAD$ but $\angle TDM = \angle TDO = \angle MAD$ thus $TD$ is tangent to the circumcircle of $AMD$ and is also tangent to the circumcircle of $ABC$ , as desired $\blacksquare$

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#8 h Dec 22, 2022, 8:29 PM • 3 Y

Y by Mango247, Mango247, Mango247

Just bash it :yup:

:yup:

Consider $(ABC)$ as the unit circle and let the other tangent from $D$ to $(ABC)$ touch the circle at $X$
$d=\frac{2ax}{a+x} \implies x=\frac{ad}{2a-d}$
$D=AA\cap BC \implies d=\frac{a(ab+ac-2bc)}{a^2-bc}$ Since $E$ is the circumcenter of $(ABD) :$
$e=\begin{vmatrix} a & a\overline{a} & 1\\ b & b\overline{b} & 1\\ d & d\overline{d} & 1\\ \end{vmatrix}\div \begin{vmatrix} a & \overline{a} & 1\\ b & \overline{b} & 1\\ d & \overline{d} & 1\\ \end{vmatrix}=\begin{vmatrix} a-b & 0 & 0\\ b & 1 & 1\\ d & d\overline{d} & 1\\ \end{vmatrix}\div \begin{vmatrix} a-b & \frac{a-b}{-ab} & 0\\ b & \frac{1}{b} & 1\\ d & \overline{d} & 1\\ \end{vmatrix}$ $e=\frac{ab(d\overline{d}-1)}{ab\overline{d}+d-a-b}=\frac{ab(2a^3b+2a^3c-4a^2bc+2abc^2+2ab^2c-a^2b^2-b^2c^2-a^2c^2-a^4)}{(a^2-bc)(2a^2b^2-ab^2-2abc+a^2c+b^2c-a^3)}$ $e=\frac{ab(a-b)^2(a-c)^2}{(a^2-bc)(a-b)^2(a-c)}=\frac{ab(a-c)}{a^2-bc}$ Similarly $: f=\frac{ac(a-b)}{a^2-bc}$
$\implies m=\frac{e+f}{2}=\frac{a(ab+ac-2bc)}{2(a^2-bc)}=\frac{d}{2}$ Now we wanna prove that $DX$ is tangent to $(AMD)$ wich equivalent to :
$\angle ADX +\angle DMA =180^\circ\iff \Delta :=\frac{a-d}{x-d}\times\frac{d-m}{a-m}\in\mathbb{R}$ $\iff \Delta =\frac{(a-d)(d-\frac{d}{2})}{(\frac{ad}{2a-d}-d)(a-\frac{d}{2})}=\frac{d(a-d)}{d(d-a)}=-1\in\mathbb{R}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\text{[math]}$ $\blacksquare$

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#9 h Feb 16, 2023, 5:28 AM • 1 Y

Y by olympiad.geometer

https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhGplOp_28uzqR2IO3TyrMsEaGPcrAM3YT56lsb3CkVOCC9KBxMebdyTPSkk-yGq9LtLAY6se2VVrfKVYElMLxv0N5k-II7tdi7ZfE_hvYs8Ldzlvd50vozQSeWR5UBr0eWUawz_rj1feobbz0h_jjgtJL5wMjcGlnma4r8GF4RhxqpgiNfEmH95LN7/s1240/diagram%20p2.png

https://i.ibb.co/tmv3bRy/IGO2022-P2.png

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#10 h Mar 31, 2023, 8:58 AM • 1 Y

Y by surpidism.

Let $O$ be the circumcenter of $(ABC)$ . Note that $EF$ is the perpendicular bisector of segment $AD$ . Also, lines $EO$ and $FO$ are the perpendicular bisectors of $AB$ and $AC$ respectively. Notice that,

$\angle DEA = 360 - 2 \angle DBA = 2 \angle B$
and,

$\angle DFC = 360 - 2\angle DAC = 360 - 2 \angle DBA = 2\angle B.$
So, $\triangle DEA \sim \triangle DFC$ . Also,

$\angle DEB = 2 \angle DAB = 2 \angle C$
and,

$\angle DFA = 2 \angle DCA = 2 \angle C.$
So, $\triangle DEB \sim \triangle DFA$ too. Then, we make the following key claim.

Claim : $\triangle ABC \sim \triangle EFA \sim \triangle DEF \sim \triangle EOF.$

Proof : Let $K$ be the midpoint of $AD.$ Then,

$\angle FEA = \frac{\angle DEA}{2} = \angle B.$
And,

$\angle EFA = 90 - \angle DAF = 90 - \angle EBD = \frac{\angle DEB}{2} = \angle C.$
So, $\triangle ABC \sim \triangle EFA$ . Note that $\triangle EFA \cong \triangle DEF$ so, $\triangle ABC \sim \triangle EFA \sim \triangle DEF$ .

Now, let $P$ and $Q$ be the midpoints of $AB$ and $AC$ respectively. Then, it is clear that $E, P, O$ and $F, Q, O$ are collinear. Thus,

$\angle EOF = \angle POF = 180 - \angle POQ = \angle A.$
And,

$\angle FEO = \angle FEA - \angle AEO = \frac{\angle AED}{2} - \angle AEP = \angle B - \frac{\angle AEB}{2} = \angle B - \angle ADB = \angle DAB = \angle C.$
So, $\triangle ABC \sim \triangle FOE$ as well. Establishing the claim.

Next, we will show that $DEOF$ is a parallelogram. This follows by similarities i.e. $\angle OEF = \angle EFD$ and $\angle DEF = \angle EFO$ . Which implies $EO \parallel DF$ and $DE \parallel OF$ implying $DEOF$ parallelogram. This means, as $M$ is the midpoint of $EF$ , it must also be the midpoint of $DO$ . So, $MA = MD = MO$ .

Now, let $T$ be the tangency point such that $TD$ is tangent to $(ABC).$ So, $DO$ is the perpendicular bisector of $AT$ which gives $MA = MT$ too. So points $A, O, T, D$ are concylic with $M$ being center of that circle. Now,

$\angle DMT = 2 \angle DAT = 2 \angle DTA = \angle DMA.$
So by $S.A.S,$ $\triangle DMT \cong \triangle DMA$ which implies $\angle TDM = \angle MAD$ which gives our desired tangency.

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#11 h Oct 15, 2023, 6:27 PM • 1 Y

Y by FabrizioFelen

Very nice problem

Claim 1: $\triangle AEF \stackrel{+}{\sim} \triangle ABC$
Proof: By Salmon's Theorem with $\triangle ABC$ and $D$ , $\triangle AEF \stackrel{+}{\sim} \triangle ABC$ $\square$

By spiral similarity with center A.
$AEFMD \stackrel{+}{\sim} ABCZX$ . So
$\angle AMD = \angle AZX$

Let:
$O: \text{circumcenter of } \triangle ABC$
$X: \text{ the reflexion of } A \text{ in } BC$
$Y: Y\in \odot ABC \text{ and } DY \text{ tangent to } \odot ABC$
$Z: \text{ the midpoint of } BC$
$R: \text{ the reflexion of } A \text{ in bisector of } BC$

Claim 2: $X,Y,Z,R$ are collinear.
Proof:
$ABYC$ armonic quadrilateral, so $\angle AZB=\angle BZY$ , but $\angle AZB=\angle BZX$ , so $X,Y,Z$ are collinear.
But $\triangle ABZ \cong \triangle RCZ\Rightarrow \angle RZC=\angle AZB=\angle BZY$ , so $Y,Z,R$ are collinear.
So $X,Y,Z,R$ are collinear. $\square$

Claim 3: $D,A,O,Z,Y$ are concyclic.
Proof: It's easy to see $DA\perp AO,OZ\perp DZ\text{ and } OY\perp DY$ , So $D,A,O,Z,Y$ are concyclic with diameter $DO$ . $\square$

Finishing:
$\angle AMD = \angle AZX =2\alpha$ , but $OA= OY\Rightarrow \angle AOD=\angle DOY=\alpha$ so $\angle ODY= 90^{\circ}-\angle DOY=90^{\circ}-\alpha$
So: $\angle DMA=90^{\circ}-\alpha$ , so $DY$ is tangent to $\odot AMD$ , As desired.

This post has been edited 2 times. Last edited by dkshield, Oct 21, 2023, 5:38 AM
Reason: latex

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cursed_tangent1434

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cursed_tangent1434

#12 h Dec 30, 2023, 9:47 AM • 1 Y

Y by GeoKing

Let $K$ be the second point such that $DK$ is tangent to $(ABC)$ . Now, we have the following key claim.

Claim : Quadrilateral $DEOF$ is a parallelogram.
Proof : The angle chasing is a bit messy but we can grind through. We use the well known fact that the radical axis is perpendicular to the line joining the centers to conclude that $EO\perp AB$ , $FO\perp AC$ and $EF\perp AD$ . This will be used throughout this proof. Now, $\measuredangle FOE = \measuredangle CAB$ . Then,
$2\measuredangle EDB = \measuredangle DEB = 2\measuredangle DAB = 2\measuredangle ACB = 90+ \measuredangle ACB$ Thus, $\measuredangle EDB= 90 + \measuredangle ACB$ . Also,
$2\measuredangle BDF = \measuredangle CFD = 2\measuredangle CAD - 180$ Then, we have that $\measuredangle BDF +90 = \measuredangle CAD$ . Thus,
$\measuredangle EDF = \measuredangle EDB + \measuredangle BDF = 90 + \measuredangle ACB + \measuredangle CAD - 90 = \measuredangle CAB$ Thus,
$\measuredangle FOE = \measuredangle EDF$ Further note that,
$\measuredangle OED = 360 - ( \measuredangle BDE +90 + \measuredangle ABD) = 180 + \measuredangle BCA + CBA = 2\measuredangle CBA + \measuredangle BAC$ with a similar angle chase, we obtain that $\measuredangle DFO = \measuredangle BAC + 2\measuredangle CBA$ . Thus,
$\measuredangle OEF = \measuredangle DFO$ as well. This implies the required result.

Claim : $DKOA$ is a cyclic quadrilateral.
Proof : Simply note that $\measuredangle OAD = 90^\circ = \measuredangle OKD$ . Thus, $DKOA$ is indeed cyclic.

Now, since parallelograms have diagonals which bisect each other, $M$ is also the midpoint of $DO$ . But, the midpoint of $DO$ is the center of $(AOKD)$ ! Thus, $MD =MA=MK$ . So,
$\measuredangle MDK = \measuredangle ADK + \measuredangle KDM = \measuredangle ADM = \measuredangle MAD$ This means that the line tangent to $(ABC)$ through $D$ which does not pass through $A$ , is in fact tangent to $(AMD)$ . Since there exists exactly one tangent to a circle from a point on the circle, we can conclude that indeed, the line tangent to the circumcircle of $AMD$ through $D$ is also tangent to the circumcircle of $ABC$ and we are done.

This post has been edited 1 time. Last edited by cursed_tangent1434, Dec 30, 2023, 9:48 AM
Reason: small adjustments

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#13 h Oct 3, 2024, 4:21 AM • 2 Y

Y by Funcshun840, fearsum_fyz

Solved in 5 minutes (:skull:), nice problem though
Let $N$ the midpoint of $BC$ and $K$ a point in $(ABC)$ such that $AK$ is symedian, then since $K \to N$ in $\sqrt{bc}$ invert we have that from easy angle chase that $ADKN$ is cyclic, redefine $M$ as the center of this cyclic, then we will prove that $M$ is midpoint of $EF$ which clearly finishes as then trivially $\angle MDN=\angle ADM=\angle DAM$ which implies the tangency.
Now draw perpendicular bisectors of $DB,DC,DN$ , notice that the distance from $DB$ to $DN$ is the same as $DN$ to $DC$ by the midpoint, and by radax we have $E,F,M$ colinear so by thales we have $EM=MF$ as desired thus we are done :cool:

:cool:

.

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SomeonesPenguin

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SomeonesPenguin

#14 h Oct 17, 2024, 7:40 AM • 1 Y

Y by radian_51

Why has nobody trigbashed this?

Let $K$ be the intersection with the $A$ -symmedian and the circumcircle of $(ABC)$ . $K$ lies on the $A$ -Apollonian circle, hence $DK$ is tangent to $(ABC)$ . We claim that it is also tangent to $(AMD)$

Claim. $\angle EAM=\angle CBK$ and $\angle FAM=\angle BCK$

Proof: these clearly have the same sums so it suffices to prove that $\frac{\sin(\angle EAM)}{\sin(\angle FAM)}=\frac{\sin(\angle CBK)}{\sin(\angle BCK)}$
From ratio lemma we have $\frac{\sin(\angle EAM)}{\sin(\angle FAM)}=\frac{FA}{EA}=\frac{AD\sin(\angle ACB)}{AD\sin(\angle ABC)}=\frac{AB}{AC}$
The second ratio also follows from ratio lemma $\frac{\sin(\angle CBK)}{\sin(\angle BCK)}=\frac{\sin(\angle CAK)}{\sin(\angle BAK)}=\frac{AB}{AC}$
Hence the claim is proven. Now easy angle chase yields $\angle DAM=\angle ADK/2$ . $\blacksquare$

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SimplisticFormulas

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SimplisticFormulas

#15 h Jan 17, 2025, 9:55 AM • 1 Y

Y by radian_51

soln
remark

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#16 h Jan 28, 2025, 1:02 PM

Y by

Let $O$ be the circumcenter of $\Delta{ABC}$ . Let $K$ be the point of contact of a tangent from $D$ to the circumcircle other than $A$ , i.e., the second intersection of the $A$ -symmedian with the circumcircle. We will show that $KD$ is tangent to $(AMD)$ .

Notice that $EF$ is the perpendicular bisector of $AD$ . Hence $\angle{MAD} = \angle{MDA}$ .

Claim: $O, M, D$ are collinear.
Proof. Angle chasing to show that $\square{OEDF}$ is a parallelogram. This implies that $M$ is the midpoint of $OD$ .

Now $\angle{KDM} = \angle{KDO} = \angle{ODA} = \angle{MDA} = \angle{MAD}$

Hence by alternate segment theorem, $KD$ is tangent to $(AMD)$ .

Attachments:

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#17 h Jan 28, 2025, 1:03 PM • 1 Y

Y by radian_51

SimplisticFormulas wrote:

somehow this is rated d6 in modsmo but i think this has to be atleast a d7. anyways here’s my sol:

Let $T$ be foot of $A$ -perpendicular on $BC$ and $N$ the midpoint of $BC$ . Let the tangent from $D$ to $(ABC)$ touch it in $K$ different from $A$ .
CLAIM 1: $\triangle ABC \sim \triangle AEF$
PROOF: Indeed, observe that $\angle EAF=\angle EAD +\angle DAF=90-\angle B +90-\angle C=\angle A$ . Also observe that $\angle EBA=\angle 90-\angle ADB=\angle AFC$ . Using $EA=EB$ and $FA=FC$ , we get $\triangle EAB \sim \triangle FAC \implies \frac{AE}{AF}=\frac{AB}{AC} \implies \triangle AEF \sim \triangle ABC$

CLAIM 2: $\angle DM =\angle TAN$
PROOF: This is true since under a spiral similarity at $A$ mapping $\triangle AEF$ to $\triangle ABC$ , $M \mapsto N$ and $AD \cap EM \mapsto T$ [using EM $\perp AD$ ]

CLAIM 3: $AK$ is the symmedian of $\triangle ABC$
PROOF: This is true since $(A,K;B,C)=-1$ , a well known fact

Finally, we shall finish by chasing angles:
$\angle MDK=180- \angle MAD -\angle DAK- \angle- \angle DKA$
$=180- \angle TAN-2 \angle DKA$
= $180- \angle TAN - 2( \angle C + \angle BAK)$
$= 180- \angle TAN- 2( \angle C + \angle NAC)$
$=180- \angle TAN- 2( \angle C + 90- \angle C - \angle TAN)$
$180- \angle TAN- 2( 90-\angle TAN)$
$= \angle TAN=\angle MAD$ and we are done. $\blacksquare$

Nah, I think d6 is appropriate. You overcooked it a little

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mcmp

#18 h Mar 30, 2025, 7:19 AM • 1 Y

Y by ohiorizzler1434

Solved with ohiorizzler1434 and Scilyse in 5 minutes flat.

Construct the circumcentre of $\triangle ABC$ , and $S\in(ABC)$ such that $(AS;BC)=-1$ .

Claim 1: $M$ midpoint of $OD$ .

We show that $DEOF$ is a parallelogram. We basically only need to show that opposite angles are equal; we do each separately.
$\begin{align*} \measuredangle DEO&=\measuredangle DEA+\measuredangle AEO\\ &=2\measuredangle DBA+90^\circ-\measuredangle BAE\\ &=2\measuredangle DBA+\measuredangle ADB\\ &=2\measuredangle DBA+\measuredangle ABD+\measuredangle DAB\\ &=\measuredangle DBA+\measuredangle DAB\\ &=\measuredangle CBA+\measuredangle ACB\\ &=\measuredangle CAB=\measuredangle OFD \end{align*}$ where the last equality comes from symmetry.
$\begin{align*} \measuredangle EDF&=\measuredangle EDA+\measuredangle ADF\\ &=\measuredangle 90^\circ-\measuredangle ABD+\measuredangle 90^\circ-\measuredangle DCA\\ &=\measuredangle CBA+\measuredangle ACB\\ &=\measuredangle CAB\\ &=\measuredangle FOE \end{align*}$ so $M$ has to be the midpoint of $OD$ as well.

Finishing touches

Note now that $M$ is the centre of $(ADSO)$ . Furthermore, $\overline{DO}$ bisects $\angle ADS$ so $\measuredangle MAD=\measuredangle ADM=\measuredangle MDS$ as desired.

This post has been edited 1 time. Last edited by mcmp, Mar 30, 2025, 7:28 AM
Reason: darn it clearly my english is getting worse and worse

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#20 h Mar 31, 2025, 12:53 PM

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Lovely IMO style problem :love:

:love:

Solution:
Let $O$ be the e circumcenter of $\triangle ABC$

Claim: Quadrilateral $\square DEFO$ is a parallelogram
Proof:
Note that $E$ is the center of $\odot(ABD)$ and $O$ is the center of $\odot(ABC)$ thus by Salmon Theorem we get that $\triangle AED\stackrel{+}{\sim} \triangle AOC$
Hence $\angle ADE=\angle ACD=90-\angle B$

Also $180-\angle DAC=180-\angle DAB-\angle BAC=180-\angle C-\angle A=\angle B$ hence we get $DE \perp AC$

On the other hand since $FA=FC$ and $OA=OC$ we get that $FO \perp AC$
Since $DE \perp AC$ and $FO \perp AC \implies DE \parallel FO$

Similarly we get $DF \parallel OE$ hence the quadrilateral $DEFO$ is a parallelogram $\square$

Hence from previous claim we have that points $\overline{D-M-O}$ are collinear and $MD=MO$

Let $T$ be the tangent point from $D$ to $\odot (ABC)$

Claim: $MO=MD=MA=MT$
Proof:

Note that since $FD=FA$ and $ED=EA$ we get that $FE \perp AD \implies FE$ lies on the perpendicular bisector of $AD$ but since $M \in FE \implies M$ lies on the perpendicular bisector of $AD \implies MD=MA$ so $MO=MD=MA$

Similarly since $DA=DT$ and $OA=OT \implies DO \perp AT \implies DO$ lies on the perpendicular bisector of $AT$ but since $M \in DO \implies M$ lies on the perpendicular bisector of $AT \implies MA=MT$ hence $MO=MD=MA=MT \square$
Claim: $TD$ is tangent to $\odot (AMD)$
Proof:

Note that since $DM$ is the side bisector of $AT$ and $DT=DA$ (because they are tangent to $\odot (ABC)$ ) we get that $DM$ is the angle bisector of $\angle ADT \implies TDM=\angle MDA$

Finally: $\angle TDM=\angle MDA \stackrel{MD=MA}{=} \angle MAD \implies \angle TDM=\angle MAD \implies$ $TD$ is tangent to $\odot (AMD)$ $\blacksquare$

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This post has been edited 1 time. Last edited by ItsBesi, Mar 31, 2025, 12:54 PM

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