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Source: Iranian Third Round 2020 Geometry exam Problem4

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Mr.C

#1 h Nov 18, 2020, 11:18 AM

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Triangle $ABC$ is given. Let $O$ be it's circumcenter. Let $I$ be the center of it's incircle.The external angle bisector of $A$ meet $BC$ at $D$ . And $I_A$ is the $A$ -excenter . The point $K$ is chosen on the line $AI$ such that $AK=2AI$ and $A$ is closer to $K$ than $I$ . If the segment $DF$ is the diameter of the circumcircle of triangle $DKI_A$ , then prove $OF=3OI$ .

This post has been edited 4 times. Last edited by Mr.C, Nov 18, 2020, 11:22 AM

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i3435

#2 h Nov 19, 2020, 12:07 AM • 2 Y

Y by Ali3085, PRMOisTheHardestExam

Woah this is nice

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#3 h Nov 25, 2020, 2:16 AM • 2 Y

Y by RodSalgDomPort, Brian_Xu

Beautiful problem.

Define $E=AI\cap BC$ , $M=AD\cap \odot (ABC)$ with $M\neq A$ , $N=AI\cap \odot (ABC)$ with $N\neq A$ , and $X=I_{A}D\cap \odot (BIC)$ with $X\neq I_{A}$ , $T$ as the reflection of $I$ wrt to $O$ and $F’$ as the reflection of $I$ wrt $T$ .

We will prove that $\angle DI_{A}F’=\angle DKF’=90^{\circ}$ which will solve the problem

$\textbf{Claim 1:}$ $M,I,X$ are collinear.
$\textit{Proof:}$ Notice that
$(B,C;I,X)\stackrel{I_{A}}{=}(B,C;E,D)=-1$ This implies $IX$ , and tangent from $B$ and $C$ to $\odot (BIC)$ is concurrent. It is well known that MB and MC are both tangent to $\odot (BIC)$ . Hence, we have $M,I,X$ are collinear.

$\textbf{Claim 2:}$ $\angle DI_{A}F’=90^{\circ}$ .
$\textit{Proof:}$ Note that $MINT$ is a parallelogram because $IO=OT$ and $MO=ON$ . Moreover, $TN\parallel I_{A}F’$ because $IT=TF’$ and $IN=NI_{A}$ . Thus, we will have $MI\parallel TN\parallel I_{A}F’$ . Therefore,
$\angle DI_{A}F’=180^{\circ}-\angle IXI_{A}=180^{\circ}-90^{\circ}=90^{\circ}$
$\textbf{Claim 3:}$ $\angle DKF’=90^{\circ}$ .
$\textit{Proof:}$ Notice that we have $IM=TN$ (because $MINT$ is a parallelogram), $AI=\frac{AK}{2}$ , $TN=\frac{I_{A}F’}{2}$ , and $\triangle MAI\thicksim \triangle I_{A}AD$ . So,
$\frac{I_{A}F’}{AK}=\frac{TN}{AI}=\frac{IM}{AI}=\frac{I_{A}D}{AD}$ Hence, $\frac{I_{A}F’}{I_{A}D}=\frac{AK}{AD}$ . Because $\angle KAD=\angle F’I_{A}D=90^{\circ}$ , we have that $\triangle AKD\thicksim \triangle I_{A}F’D$ . Therefore,
$\angle I_{A}F’D=\angle AKD=\angle I_{A}KD\implies DI_{A}F’K \text{is cyclic} \implies \angle DKF’=90^{\circ}-\angle DI_{A}F’=180^{\circ}-90^{\circ}=90^{\circ}$
By combining claim 2 and 3, the problem is done.

This post has been edited 1 time. Last edited by Sugiyem, Nov 25, 2020, 2:44 AM

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#4 h Mar 1, 2021, 6:10 PM

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Let $M$ and $N$ be midpoints of arc $BC$ and arc $BAC$ , and let $L$ be the reflection of $A$ over $N$ .
Claim: $L$ lies on $(DKI_A)$ .
Proof: Let $NI_{A}$ cut $(BIC)$ at $T$ . Then inverting with radius $NB$ sends $A$ to $D$ , and $T$ to $I_A$ , so it follows that $A,D,T,I_A$ are concyclic and $\angle DTI_A=\angle DAI_A=90^\circ$ . Meanwhile, $\angle ITI_A=90^\circ$ , so $D,I,T$ are collinear. Hence, $\angle ADI=\angle DI_AN$ , and so $\triangle ADI\sim\triangle AI_AN$ . Finally,
$AD\cdot AN=AI\cdot AI_A\Longrightarrow AD\cdot AL=AK\cdot AI_A$ and so $D,K,I_A,L$ are concyclic.
Now let $F'$ be the point on ray $IO$ such that $IF'=4IO$ , and let $V$ be the midpoint of $IF'$ . Then $V$ is the reflection of $I$ over $O$ , and so $IMVN$ is a parallelogram and $VN\parallel AI$ . Therefore, it follows that $F'L$ is also parallel to these lines, and hence perpendicular to $AD$ . Hence it suffices to show that $\angle F'I_AD=90^\circ$ . Notice that $IN\parallel MV\parallel I_AF'$ . Therefore,
$\angle F'I_AD=\angle F'I_AA+\angle AI_AD=\angle NIA+\angle ANI=90^\circ$ where the second step follows from the fact that $\triangle ADI_A\sim \triangle AIN$ .

This post has been edited 1 time. Last edited by KST2003, Mar 1, 2021, 6:11 PM

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888 posts

MP8148

#5 h Apr 24, 2021, 10:33 PM

Y by

uhh... I'm sorry?

$[asy] size(8cm); defaultpen(fontsize(10pt)); pair D = dir(120), K = dir(200), Ia = dir(340), A = foot(D,K,Ia), I = 2A-(A+K)/2, M = (I+Ia)/2, L = extension(D,A,Ia,foot(Ia,D,I)), O = (M+L)/2, F = 2*origin-D; draw(unitcircle); draw(D--K--Ia--D--L); draw(D--I^^Ia--L, dotted); draw(M--L); draw(I--O--F, blue); draw(circumcircle(A,L,M), dotted); dot("$D$", D, dir(120)); dot("$K$", K, dir(200)); dot("$I_A$", Ia, dir(340)); dot("$A$", A, dir(135)); dot("$I$", I, dir(50)); dot("$M$", M, dir(50)); dot("$L$", L, dir(210)); dot("$O$", O, dir(90)); dot("$F$", F, dir(300)); [/asy]$

Let $M = \overline{AI} \cap (ABC)$ and $L = 2O-M$ . It is well known that $I$ is the orthocenter of $\triangle DLI_A$ .

Use Cartesian Coordinates with $A = (0,0)$ , $D = (0,t)$ , $K = (-2r,0)$ , and $I_A = (r+2s,0)$ . It follows that $I = (r,0)$ and $M=\tfrac 12 (I+I_A) = (r+s,0)$ .

Set $L = (0,l)$ and using the fact that $\overline{I_AL} \perp \overline{DI}$ , we have $\frac{l-0}{0-(r+2s)} = -\left(\frac{t-0}{0-r}\right)^{-1} = \frac{r}{t} \implies l = \frac{-r(r+2s)}{t}.$ Since the projection of $F$ to $\overline{KI_A}$ and $A$ are reflections over the midpoint of $\overline{KI_A}$ , we can set $F = (-r+2s,f)$ for some $f$ . Using $\overline{FK} \perp \overline{DK}$ we have $\frac{f-0}{(-r+2s)-(-2r)} = -\left(\frac{t-0}{0-(-2r)}\right)^{-1} = \frac{-2r}{t} \implies f=\frac{-2r(r+2s)}{t}.$ Finally we compute $O = \frac{M+L}{2} = \left(\frac{r+s}{2}, \frac{-r(r+2s)}{2t}\right)$ and $\frac{3I+F}{4} = \left(\frac{3r+(-r+2s)}{4}, \frac{-2r(r+2s)}{4}\right) = \left(\frac{r+s}{2}, \frac{-r(r+2s)}{2t}\right) = O,$ which implies the conclusion.

(I didn't even realize $O$ , $I$ , $F$ collinear in my initial solution. I guess this makes the finish a bit faster).

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#6 h Aug 18, 2021, 3:49 PM

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An interesting problem.

Attachments:

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130 posts

Anajar

#7 h Aug 30, 2021, 7:28 PM

Y by

Nice Problem! Here's a different solution.
I'll use the following well-known lemma which can be proved using Pythagorean
Lemma: Let $\ell$ be the radical axis of circles $\omega_1$ and $\omega_2$ with centers $O_1$ and $O_2$ , respectively. Let $K$ be an arbitrary point and $H$ be the feet of perpendicular from $K$ to $\ell$ . Then the following relation holds:
$|P_{\omega_1}^K-P_{\omega_2}^K|=2 \cdot KH \cdot O_1O_2$ where $P_{\omega_1}^K$ and $P_{\omega_2}^K$ are the powers of $K$ with respect to $\omega_1$ and $\omega_2$ , respectively.

Back to our problem, Let $M$ and $N$ be the midpoints of arcs $BC$ and $BAC$ in $(ABC)$ . Reflect $A$ about $N$ to get $A'$ and reflect $I$ about $A$ to get $I'$ . Note that $I'$ lies on $(I_AI_BI_C)$ and $(D,A;I_C,I_B)=-1$ hence:
$AA'\cdot AD=2AD\cdot AN=2AI_C\cdot AI_B=2AI' \cdot AI_A=AK\cdot AI_A \implies A'\in (DKI_A)$ From above we deduce that $NA'\cdot ND=NA\cdot ND = NI_C\cdot NI_B$ , i.e. $NI_A$ is the radical axis of $(DKI_A)$ and $(I_AI_BI_C)$ . Let $P$ and $J$ be the centers of $(I_AI_BI_C)$ and $(DKI_A)$ , respectively. $P$ is the reflection of $I$ in $O$ . Let $(NAI)\cap (IBC)= S$ then $\angle NSI= \angle ISI_A = 90^{\circ}$ so $N$ , $S$ , $I_A$ are collinear and $D$ lies on $SI$ which is the radical axis of $(NAI)$ and $(IBC)$ .
Now note that $DS \perp NI_A \perp JP$ $(*)$ ,so using the lemma for $(DKI_A)$ and $(I_AI_BI_C)$ gives us:
$2DS\cdot JP= DI_C\cdot DI_B= DB\cdot DC = DI\cdot DS \implies 2JP=DS$ Now using $(*)$ we get that $P$ is the midpoint of $IF$ which proves the problem. $\blacksquare$
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#8 h Aug 31, 2021, 11:03 AM

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This problem is immediate after proving the following one:

Problem: In triangle $ABC, H$ is the orthocenter, $DEF$ is the orthic triangle, $K=EF\cap BC$ , $AH \cap (ABC)=D'$ , $L$ is the reflection of $D$ across $D'$ , $H'$ is the reflection of $H$ across $O$ . Prove that $KH'$ is the diameter of $(AKLH')$ .

Proof: If $A'$ is the antipode of $A, A'H'AH$ is parallelogram. $H'A' \cap BC=T$ , $A'H \cap (ABC)\cap AK=S$ , $H'T \perp KT$ and $\angle AKT =\angle AHS=180-\angle AH'T \implies AKH'T$ is a cyclic quadrilateral with diameter $KH'$ . $DTA'D'$ is rectangle and $HD=DD'=D'L \implies LT=HA'=AH' \implies AH'TL$ is isoscele trapezoid $\implies LAKH'T$ is a cyclic quadrilateral with diameter $KH'$ . $\square$

Apply this to $I_AI_BI_C$ in the main problem. $H$ goes to incenter, $NPC$ center goes to circumcenter, $H$ ' goes to $F$ . $3HN_9=N_9H'$ finishes the proof.

This post has been edited 2 times. Last edited by SerdarBozdag, Aug 31, 2021, 11:38 AM

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#9 h Apr 30, 2022, 11:41 AM • 1 Y

Y by PRMOisTheHardestExam

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Let $M$ and $P$ be the midpoint of the minor and major arc $BC$ of $(ABC)$ . Let $A'$ be the reflection of $A$ over $P$ .
Claim 1. $D,K,I_A$ and $A'$ are concyclic.
Proof.
Let $I_B$ and $I_C$ be the the $B$ - and $C$ -excenters respectively, then $I$ is the orthocenter of $\triangle I_AI_BI_C$ . Meanwhile $P$ is the midpoint of $I_BI_C$ .

By a well-known fact, $I$ is the orthocentre of $\triangle DPI_A$ . Hence
$AA'\times AD=2AP\times AD=2AI\times AI_A=AK\times AI_A$ as desired. $\blacksquare$

Now let $O_1$ be the center of $(DKI_AA')$ , $l_1$ be the line through $O_1$ parallel to $DI$ while $l_2$ be the line through $P$ perpendicular to $AD$ . Let $N$ be the center of $(I_AI_BI_C)$ and $Q$ be the midpoint of $IF$ .

Claim 2. Both $N$ and $Q$ lie on $l_1$
Proof.
Obviously $Q$ lies on $l_1$ by midpoint theorem.
\newline\newline For $N$ , we notice that
$PA'\times PD=PA\times PD=PI_B\times PI_C$ Hence $P$ lies on the radical axis of $(I_AI_BI_C)$ and $(DKI_AA')$ . Therefore $PI_A$ is the radical axis of these two circles. From above, $DI\perp PI_A$ , hence $O_1N\perp DI$ as desired. $\blacksquare$

Claim 3. Both $Q$ and $N$ lie on $l_2$
Proof.
Obviously $N$ lies $l_2$ as $P$ is the midpoint of $I_BI_C$ . Now, notice that $FA'$ and $AI$ are both perpendicular to $AD$ , hence by intercept theorem, $PN$ is perpendicular to $AD$ as well. $\blacksquare$

Hence $N=Q$ as they are the intersection of $l_1$ and $l_2$ . Notice that $N,O,I$ are collinear with $NO:OI=1:1$ by applying Euler line on $\triangle I_AI_BI_C$ . Hence $P$ lies on $OI$ with $PO:OI=3:1$

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Shayan-TayefehIR

#10 h Aug 23, 2022, 7:10 PM • 1 Y

Y by Mahdi_Mashayekhi

Firstly , we'll redefine points with seeing triangle $\triangle I_aI_bI_c$ ( which $I_b$ and $I_c$ are $B$ and $C$ excenters. ) as our $\triangle ABC$ . So the new problem will describe like this :
In a triangle $\triangle ABC$ , let $D , E , F$ be foot of altitudes from $A , B , C$ to corresponding sides. Also $H$ and $N$ are orthocenter and nine-point circle center of this triangle , $T$ is intersection point of the lines $EF$ and $BC$ and $K$ is a point on the line $AD$ such that $DK=2HD$ . So if the segment $TS$ is the diameter of the circumcircle of triangle $\triangle ATK$ , then $NS=3NH$ .

Now if $O$ is the circumcenter of the triangle $\triangle ABC$ , then since $N$ is the midpoint of $OH$ , we'll show that $OH=OS$ ( another word , set point $S$ as the reflection point of $H$ with respect to $O$ and we'll show that $\angle TAS=\angle TKS=90$ . )
Let $X$ be the second intersection point of the circumcircle of $\triangle ABC$ and $AFHE$ . So by radical axis theorem in circles $(AFHE) , (ABC) , (BFEC)$ , points $A , X , T$ are collinear , then since $O$ lies on the perpendicular bisector of $AX$ , so $AS \parallel XH$ and $\angle SAT=90$ .
Let points $G$ and $M$ be centroid of the triangle and midpoint of segment $BC$ , and the parallel line from $A$ to $BC$ intersect the circumcircle in point $A'$ . Also if $H'$ is the reflection of $H$ with respect to $BC$ , then $\angle H'MT=\angle XMD=\angle XAD$ and the quadrilateral $TAMH'$ is cyclic , so by power of the point $D$ with respect to its circumcircle , we have :
$DA.DK=DT.DM , AA'=2DM , DK=2DH' \implies \triangle DAA' \sim \triangle TDK \implies A'D \perp TK$ Now since $G$ is the nine-point circle and circumcircle's internal homothety center , then $A' , G , D$ are collinear and $GD \perp TK$ . As the result , while $HS=3HG$ and $HK=3HD$ , then $GD \parallel SK$ and $\angle SKT=90$ . So we're done.

This post has been edited 1 time. Last edited by Shayan-TayefehIR, Aug 23, 2022, 7:10 PM

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Pmshw

#11 h Sep 2, 2022, 10:23 AM

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$\textbf{Claim}$ : $N,I,G$ are collinear.
Proof:
$DG.DI_A=DC.DB=DA.DN$ so $\angle NGI_A=90=\angle IGI_A$ hence this proves the claim.

$\textbf{Claim}$ : $AN=EN$
Proof:
We have $KH||NL$ so we have $\triangle KAH \sim \triangle ANL$ so We have $AN=\frac{AH.AL}{KA}=\frac{AH}{2}=\frac{AE}{2}$ .

$\textbf{Claim}$ : $F,I,O$ are collinear and $\frac{IF}{II'}=2$
Proof:
We have: $AN=NE$ so $I_AM=MF=TI'=NI$ so $\frac{I_AF}{I'T}=\frac{2NI}{NI}=2=\frac{II_A}{IT}$ and this proves the claim and subsequently the problem ( $OF=OI'+I'F=3OI$ ).

This post has been edited 4 times. Last edited by Pmshw, Sep 2, 2022, 10:29 AM

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100 posts

#13 h Jan 7, 2023, 1:55 PM

Y by

It seemed straightforward to bash.
Let $F'$ be a point on $IO$ such that $3 IO = OF'$
We want to show that $F=F'$
Let $E$ and $L$ be the midpoint of arcs $BC$ such that $ADE$ are collinear.
Let $AE$ and $BC$ intersect circumcircle of $DI_AK$ at $M$ and $X$ .
By using Power of a point it is easy but time consuming to show that $AE=EM$ hence $F'M\perp AE$
Let $L'$ be a point on $IL$ such that $3IL=LL'$
Again using power of a point we can show that $L'X \perp BC$ hence $F'X\perp BC$
We can finish by saying that circle $(DKMF'XI_A) has diameter DF'$ so $F=F'$

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1386 posts

#14 h Aug 23, 2023, 11:49 AM

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Solution

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IMUKAT

#15 h Nov 24, 2023, 11:24 PM • 1 Y

Y by flippantsummit67

Let $N$ be the midpoint of arc $BAC$ and let $M$ be the midpoint of arc $BC$ . It is easy to see that $D$ , $A$ , and $N$ are colinear.

Let $Q$ be the intersection of $DI_A$ with $(BIC)$ . By power of a point, $DA\cdot DN=DB\cdot DC= DQ\cdot DI_A$ so $ANI_AQ$ is cyclic.

Now, since $II_A$ is a diameter of $(BIC)$ we have $\angle IQI_A=90$ but also $\angle NQI_A=\angle NAI_A=\angle NAM=90$ which implies that $N$ , $Q$ , and $I$ are colinear.

Let $T$ be the intersection of $NQ$ with $(ABC)$ . Since $\angle NTM=\angle NQI_A=90$ we have $TM\parallel QI_A$ .

Now, we take an homothety with center $I$ and factor $\frac{1}{2}$ which sends $K$ to $K'$ , $D$ to $D'$ and $A$ to $L$ . Let $I'$ be the reflection of $I$ over $O$ . I claim $I'$ is the image of $F$ under this homothety which finishes the problem.

Note that $NIMI'$ is a parallelogram. First, we see $\angle I'MD'=\angle I'MN+\angle NMT=\angle TNM+\angle NMT= 90.$

Let $D'L$ meet $NI'$ at $P$ . Since $NI'$ is parallel to $IM$ and $AN$ and $LP$ are both perpendicular to $AI$ , $ANPL$ is a rectangle.

Hence $\angle D'PI'=90$ so $D'PI'M$ is cyclic. It suffices to show that $K'$ lies on this circle or equivalently, $LK'\cdot LM=LD'\cdot LP$ . This is also equivalent to $AI\cdot LM=LD'\cdot AN$ .

Finally, $\angle AIN=\angle TIM=90-\angle IMT=\angle LD'M$ so $D'LM \sim AIN$ and we are done.

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