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 $OF=3OI$.
Problem
Source: Iranian Third Round 2020 Geometry exam Problem4
Tags: geometry
i3435
19.11.2020 03:07
I'm going to restate this in terms of the excentral triangle.
Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$. Let $D=\overline{AH}\cap\overline{BC}$, $E=\overline{BH}\cap\overline{AC}$, and $F=\overline{CH}\cap\overline{AB}$. Let $H_A$ be the reflection of $H$ across $D$, and let $K$ be the reflection of $D$ across $H_A$. Let $X=\overline{EF}\cap\overline{BC}$. Let $L$ be the reflection of $H$ across $O$. Prove that $DL$ is a diameter of $(XKA)$.
This problem can be dilated with a factor of $\frac{1}{2}$ around $H$. Let $N$ be the midpoint of $AH$, let $P$ be the midpoint of $XH$, and let $Q$ be the midpoint of $KH$, or equivalently the midpoint of $DH_A$. It suffices to show that $\overline{PN}\perp \overline{NO}$ and $\overline{PQ}\perp \overline{QO}$.
For $\overline{PN}\perp \overline{NO}$, let $M$ be the midpoint of $BC$. $\overline{PN}||\overline{AX}$, and $\overline{NO}||\overline{HM}$ is well known. $\overline{AX}\perp\overline{HM}$ is well known to be true (by Queue Point magic).
For $\overline{PQ}\perp \overline{QO}$, let $A'$ be the reflection of $A$ across $(ABC)$. $\overline{QO}||\overline{DA'}$ by a homothety with factor $2$ around $H_A$. By my post on my blog here, the non $D$ intersection of the circle with diameter $XH$ and the circle with diameter $DH_A$ is the intersection of $\overline{DA'}$ and $(ABC)$, $Y$. Thus their radical axis is $\overline{DA'}$. Since $P$ is the midpoint of the circle with diameter $XH$ and $Q$ is the midpoint of the circle with diameter $DH_A$, $\overline{PQ}\perp\overline{DA'}$ as desired.
Sugiyem
25.11.2020 05:16
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.
KST2003
01.03.2021 21:10
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 $\triangle ADI_A\sim \triangle AIN$.
MP8148
25.04.2021 01:33
uhh... I'm sorry? [asy][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][/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).
Anajar
30.08.2021 22:28
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. $\blacksquare$ [asy][asy] import graph; size(10cm); 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 = -30.50203599661605, xmax = 36.988086509371534, ymin = -20.52975480516736, ymax = 21.937920192072717; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen ccqqqq = rgb(0.8,0,0); draw((-7.918257438016543,4.482080991735521)--(-8.88,-2.5)--(4.7,-2.36)--cycle, linewidth(1) + blue); /* draw figures */ draw((-7.918257438016543,4.482080991735521)--(-8.88,-2.5), linewidth(1) + blue); draw((-8.88,-2.5)--(4.7,-2.36), linewidth(1) + blue); draw((4.7,-2.36)--(-7.918257438016543,4.482080991735521), linewidth(1) + blue); 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SerdarBozdag
31.08.2021 14:03
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.
mathaddiction
30.04.2022 14:41
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label("$I_B$", (-7.66917095813953,1.8185886511627931), NE * labelscalefactor); dot((3.5512009757623715,8.318193358356366),linewidth(4pt) + dotstyle); label("$I_C$", (3.998412186046518,8.50932135813953), NE * labelscalefactor); dot((-3.1626835504002964,2.500093612387794),linewidth(4pt) + dotstyle); label("$O_1$", (-3.021913265116274,3.1369596279069776), NE * labelscalefactor); dot((2.936095864165344,5.312065717232924),linewidth(4pt) + dotstyle); label("$F$", (3.4381045209302386,4.949719720930231), NE * labelscalefactor); dot((-0.04739647231503352,2.920311735195104),linewidth(4pt) + dotstyle); label("$N$", (0.3728920000000057,2.181140669767444), NE * labelscalefactor); dot((-0.866199607607268,0.8205506288647845),linewidth(4pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] 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 $PO:OI=3:1$
Shayan-TayefehIR
23.08.2022 22:10
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.
Pmshw
02.09.2022 13:23
[asy][asy] import graph; size(8cm); real labelscalefactor = 0.7; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = -9.396979330807888, xmax = 8.090060040405483, ymin = -3.278688246116183, ymax = 5.915886026446351; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); pen qqwuqq = rgb(0,0.39215686274509803,0); pen zzttff = rgb(0.6,0.2,1); draw((-3.610743543170552,4.516445428145997)--(-5.1,0.13)--(0.98,0.07)--cycle, linewidth(1)); draw((0.8037699091061296,0.24069075512234328)--(0.6330791539837861,0.06446066422847296)--(0.8093092448776564,-0.10623009089387044)--(0.98,0.07)--cycle, linewidth(1) + ccqqqq); draw((-1.2335836486041638,2.214005388581789)--(-1.4042744037265074,2.037775297687919)--(-1.2280443128326368,1.8670845425655755)--(-1.0573535577102935,2.043314633459446)--cycle, linewidth(1) + ccqqqq); draw((-2.3632703007341065,3.3081832711762913)--(-2.53396105585645,3.131953180282421)--(-2.3577309649625793,2.9612624251600774)--(-2.187040209840236,3.137492516053948)--cycle, linewidth(1) + ccqqqq); 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label("$D$", (-3.5679662070700973,4.632115278956488), NE * labelscalefactor); dot((-5.1,0.13),dotstyle); label("$K$", (-5.048350492463822,0.2487899339235315), NE * labelscalefactor); dot((0.98,0.07),dotstyle); label("$I_A$", (1.023538178096376,0.19096242277533945), NE * labelscalefactor); dot((-3.6541716366363954,0.11573195694049077),linewidth(4pt) + dotstyle); label("$A$", (-3.602662713759013,0.20252792500497785), NE * labelscalefactor); dot((-3.6391466728687423,1.638261618729313),linewidth(4pt) + dotstyle); label("$H$", (-3.5910972115293744,1.7291742193172477), NE * labelscalefactor); dot((-3.6691966004040486,-1.4067977048483324),linewidth(4pt) + dotstyle); label("$E$", (-3.6257937182182896,-1.3125528670776534), NE * labelscalefactor); dot((-3.6616841185202222,-0.6455328739539208),linewidth(4pt) + dotstyle); label("$N$", (-3.614228215988651,-0.5492297199215186), NE * labelscalefactor); dot((-2.931257454954593,0.10859793541073612),linewidth(4pt) + dotstyle); label("$I$", (-2.8856015755214273,0.20252792500497785), NE * labelscalefactor); dot((-0.9756287274772966,0.08929896770536806),linewidth(4pt) + dotstyle); label("$T$", (-0.9310316987125258,0.17939692054570103), NE * labelscalefactor); dot((-2.3186564229987594,-0.27811695312427637),linewidth(4pt) + dotstyle); label("$O$", (-2.2726299573505884,-0.190699150802728), NE * labelscalefactor); dot((-1.7060553910429255,-0.6648318416592889),linewidth(4pt) + dotstyle); label("$I'$", (-1.6596583391797495,-0.5723607243807954), NE * labelscalefactor); dot((-0.4808533271312583,-1.438261618729315),linewidth(4pt) + dotstyle); label("$F$", (-0.4337151028380715,-1.3472493737665687), NE * labelscalefactor); dot((0.24957333643437085,-0.6841308093646574),linewidth(4pt) + dotstyle); label("$M$", (0.2949115376291522,-0.5954917288400723), NE * labelscalefactor); dot((-2.187040209840236,3.137492516053948),linewidth(4pt) + dotstyle); label("$X$", (-2.1454094328245654,3.232689509170241), NE * labelscalefactor); dot((-4.489863638044564,1.9270913224326904),linewidth(4pt) + dotstyle); label("$Y$", (-4.446944376522621,2.018311775058208), NE * labelscalefactor); dot((-1.0573535577102935,2.043314633459446),linewidth(4pt) + dotstyle); label("$G$", (-1.011990214319995,2.1339667973545917), NE * labelscalefactor); dot((-1.933062309015237,-1.616038860067298),linewidth(4pt) + dotstyle); label("$B$", (-1.8909683837725189,-1.520731907211145), NE * labelscalefactor); dot((-2.6678265856212415,1.0697695581781308),linewidth(4pt) + dotstyle); label("$C$", (-2.6195950242397426,1.1624646100649656), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] $\textbf{Claim}$: $N,I,G$ 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 ($OF=OI'+I'F=3OI$).
Infinityfun
07.01.2023 16:55
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 $F=F'$
VicKmath7
23.08.2023 14:49
Add the two other excenters and rephrase as an orthocenter configuration as follows:
Let $ABC$ be a triangle with orthic triangle $DEF$ and orthocenter $H$. Let $EF \cap BC=T$ and $K \in AH$ lying outside the triangle is such that $DK=2DH$. If $P$ is the $T$-antipode in $(ATK)$, show that $NP=3NH$, where $N$ is the center of the nine-point circle.
Let $M$ be the midpoint $BC$; it's well-known that $H$ is the orthocenter of $\triangle ATM$. Reflect $D$ in $M$ to obtain $R$; by PoP $DM \cdot DT=DH \cdot DA \iff DT \cdot DR=DK \cdot DA$, so $R \in (ATK)$.
Redefine $P$ as the intersection of the perpendicular at $R$ to $BC$ with $OH$. If we show that $P \in (ATR)$, the redefinition will be correct, so $OM$ will be a midline in the trapezoid $HDRP$, thus $O$ is midpoint of $HP$, and since $N$ is the midpoint of $OH$, we will obtain $HP=3HN$.
Observe that $\angle AHM=180^{o}-\angle ATR$, so it's sufficient to show that $\angle APR=\angle AHM$. Indeed, if $A'$ is the $A$-antipode in $(ABC)$, then $A' \in PR$, so $PA' \parallel AH$ and $PA'=PR+RA'=PR+HD=2OM-HD+HD=2OM=AH$, so $APA'H$ is a parallelogram, as desired (we used again that $OM$ is midline in the trapezoid $HDRP$).
$\textbf{Remark.}$ Fairly straightforward for this position once you restate it as an orthocenter problem.
IMUKAT
25.11.2023 02:24
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.