AoPS - Problem

AlastorMoody

27.12.2019 18:47

Vietnam MO 2020 P4 Part a) wrote: $H_A$ is the reflection of $H$ in $BC$, $A'$ is the reflection of $A$ over $O$ and $O_A$ is the center of $\odot (BOC)$. Prove that $H_AD$ and $O_AA'$ intersect on $(O)$. Solution: We'll use the well-known facts that $D$ is the center of $\odot (BHC)$ and $O_A$ is the Image of $D$ under Inversion around $(O)$. Let $H_AD$ $\cap$ $\odot (ABC)$ $=$ $K$. Let $O_AO$ $\cap$ $\odot (ABC)$ $=$ $X,Y$ \begin{align*} DX \cdot DY=OD \cdot DO_A=DH_A \cdot DK \end{align*}Hence, $OH_AO_AK$ is cyclic. Moreover, By converse of Reim's Theorem, $ODA'K$ is cyclic. Hence, $\angle H_AOO_A$ $=$ $\angle DOA'$ $=$ $\angle H_AKA'$ $=$ $\angle H_AKO_A$ $\implies$ $K$ $\in$ $O_AA'$ $\qquad \blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14cm); 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 = -12.957859795170341, xmax = 14.82374888938452, ymin = -9.3256294851372, ymax = 6.455235120105774; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen ttffqq = rgb(0.2,1,0); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); draw((-5.780843931650028,5.77633547708488)--(-9.38,-1.87)--(2.13872609253741,-2.1646370790134144)--cycle, linewidth(0.4) + rvwvcq); /* draw figures */ draw((-5.780843931650028,5.77633547708488)--(-9.38,-1.87), linewidth(0.4) + rvwvcq); 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Just a look at part b), makes me want to give up living

AlastorMoody

27.12.2019 21:39

Ok I guess I'm ready to give up living

Vietnam MO 2020 P4 part b) wrote: Let $X$ is a point satisfy $AXDA'$ is the parallelogram. Prove that $\odot (AHX)$, $\odot (ABF)$, $\odot (ACE)$ have a common point different than $A$. Solution: Start off by noticing $XAA'D$, $XAOH$, $HOA'D$ are parallelograms. Hence, $\angle XHA$ $=$ $\angle HAO$ $=$ $\angle DOA'$ $=$ $\angle DKA'$. Let $\odot (ABF)$ $\cap$ $\odot (ACE)$ $=$ $M$. We will show that $M$ is the reflection of $K$ over $N_9$ of $\Delta ABC$ (where $K$ $=$ $H_AD$ $\cap$ $O_AA'$) From Converse of Generalised Reim's Theorem, If $\mathcal{C}$ is the conic passing through $AFBDCE$, then since, $BF||CE$ $\implies$ $M$ $\in$ $\mathcal{C}$. Also, recall $K$ $\in$ $H_AD$, Hence $\angle DKC$ $=$ $90^{\circ}-C$ $=$ $\angle DEC$ $\implies$ $DEKC$ is cyclic. Note: $\angle DEC$ $=$ $90^{\circ}-C$ because $D$ is reflection of $E$ over $CH$. Similarly, we can prove $AFEK$ is cyclic and again by Converse of Generalised Reim's Theorem, we get $K$ $\in$ $\mathcal{C}$. Since, $\angle BMA$ $=$ $\angle EKD$, by reflection argument, we have $M$ as reflection of $K$ over $N_9$. Just note: $\angle XHA$ $=$ $\angle DKA'$ $=$ $\angle XMA$. Hence, $\odot (AHX)$ passes through $M$ $\qquad \blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(17cm); 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 = -19.484037391308085, xmax = 16.54125478112415, ymin = -11.610530895722585, ymax = 8.853016215339267; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen ttffqq = rgb(0.2,1,0); pen ubqqys = rgb(0.29411764705882354,0,0.5098039215686274); pen qqqqcc = rgb(0,0,0.8); draw((-5.780843931650028,5.77633547708488)--(-9.343973485138488,-2.065976330517489)--(2.8962055924152796,-1.9867848124014835)--cycle, linewidth(0.4) + rvwvcq); /* draw figures */ draw((-5.780843931650028,5.77633547708488)--(-9.343973485138488,-2.065976330517489), linewidth(0.4) + rvwvcq); 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HolyMath

27.12.2019 21:48

above how beautiful

mmathss

27.12.2019 21:50

trito11 wrote: Let ABC is a non-isosceles acute triangle with the circumscribed cycle (O) and the orthocenter H. D, E, F are the reflection of O in the lines BC, CA and AB. a) $H_a$ is the reflection of H in BC, A' is the reflection of A at O and $O_a$ is the center of (BOC). Prove that $H_aD$ and OA' intersect on (O). b) Let X is a point satisfy AXDA' is the parallelogram. Prove that (AHX), (ABF), (ACE) have a comom point different A My solution to part (a).Observe that $A$ is the anti Steiner point of $AH$ and $P$ is the anti steiner point of the Euler line where $P$ is intersection of $HD$ and $(ABC)$,So by a well known lemma that angle between two lines passing through orthocenter is equal to to half measure of arc subtended by the 2 anti steiner points,So we get that $\angle AA'P=\angle H_AAO$.So it needs to be proven that $\angle AHO=\angle AA'O_A$.Now it is well known too that $D$ and $O_A$ are inverses w.r.t $(ABC)$,So $\angle AA'O_A=\angle ODA'=\angle AHO$. The last line follows since $D$ and $A'$ are reflections of $A$ and $H$ in the midpoint of $BC$

mmathss

27.12.2019 21:52

Also interesting to note $D$ is the $O_A$ Humpty point in $\triangle AO_AA'$

Pathological

29.12.2019 19:41

Here's a different solution to (b). Firstly we observe that $AXHO$ is also a parallelogram, because $\overrightarrow{AX} = \overrightarrow{A'D} = \overrightarrow{OH}.$ Hence, $(\triangle AHX)$ is the reflection of $(\triangle AOH)$ over $AH.$ We also have that $(\triangle ABF), (\triangle ACE)$ are the reflections of $(\triangle AOB), (\triangle AOC)$ over $AB, AC$, respectively. Let $O_b, O_c, O_h$ denote the circumcenters of $\triangle AOB, \triangle AOC, \triangle AOH$, respectively. Let $O_b', O_c', O_h'$ denote the reflections of $O_b, O_c, O_h$ over $AB, AC, AH$, respectively. To solve the problem, it suffices only to show that $O_b', O_c', O_h'$ are collinear. Notice that $O_b, O_b'$ are points on $OF$ which are symmetric w.r.t. the midpoint of $OF.$ Similarly, $O_c, O_c'$ are points on $OE$ which are symmetric w.r.t. the midpoint of $OE$, and $O_h, O_h'$ are points on $EF$ which are symmetric w.r.t. the midpoint of $EF.$ From here, because $O_b, O_c, O_h$ are collinear (on the perpendicular bisector of $AO$), it's clear by Menelaus that $O_b', O_c', O_h'$ are also collinear. $\square$

jbaca

30.12.2019 00:50

Solution for a). WLOG, assume that $AC>AB$. Let $P=\overline{H_aD}\cap \overline{O_aA'}$. Observe that $D$ and $O_a$ are inverses respect to $(BAC)$, thus $$\frac{OA'}{OD}=\frac{OO_a}{OA'}$$which together with $\angle DOA'=\angle A'OO_a$ implies that $\bigtriangleup DOA'\sim \bigtriangleup A'OO_a$, so $\angle OA'O_a=\angle ODA'$. By symmetry, we have $$\angle ODA'=\angle ODH_a=O_aDP$$hence $\angle ODP=\angle OA'P$, therefore $ODA'P$ is cyclic and the required assertion follows straightforwardly. Solution for b). Let $N$ be the nine-point center of $\bigtriangleup ABC$ (in other words, the midpoint of $\overline{OH}$). Recall that $HA'$ bisects segment $BC$. Moreover, $AH$ is twice the distance from $O$ to $BC$ (1). We conclude that $HOA'D$ is a parallelogram, so $H$ is the midpoint of $XD$ and the point where $XA'$ meets $AD$ coincides with $N$. Let $Q$ the symmetric of $P$ with respect to $N$. We prove that $Q$ is the required second intersection point of circles $(AHX),\ (ABF)$ and $(AEC)$. First, notice that by the analogous versions of (1) for $AB$ and $AC$, we also infer that $FC$ and $BE$ bisect each other at $N$, thus $FQXAE$ and $CPA'DB$ are homothetic with center $N$ and ratio $-1$. Then $$\angle QXA=\angle DPA'=\angle DOA'=\angle H_aOD=\angle H_aHD=\angle AHX$$so $AHQX$ is cyclic. Second, $$\angle FQA=\angle CPD=\angle CPH_a=180^\circ-\angle H_aAC=180^\circ-\angle BAO=180^\circ -\angle FBA$$thus $AQFB$ is inscribed as well, and finally, $$\angle AQE=\angle DPB=\angle BAH_a=\angle OAC=\angle ECA $$i.e. $AQEC$ is cyclic as well. The proof is complete. $\blacksquare$

Physicsknight

09.01.2020 05:17

(a) Using reflection by $BC$'s perpendicular bisector, we can prove that $DA'$, $H_aO_a$ intersect on $(O)$ . Obviously $H_a\in\odot(O)$. Let $A'D\cap(O)=G\neq A'$. As $\overline{A',D,G}$ by applying inversion of center $(O)$ and $\mathrm{pow}(OB^2)$, we can conclude that $O, G, O_a, A'$ are concyclic. Note that, $\widehat{O_aGA'}=\widehat{O_aOA'}=\widehat{H_aAA'}=\widehat{H_aGA'}\implies\overline{G,H_a,O_a}$ (b) Obviously $HOA'D$ is a parallelogram, hence $AXHO$ is also a parallelogram. Therefore, $\odot(AXH)$ passes through the point $O_1$, which is the reflection of $O$ by $AH$. Now, applying inversion of center $A$ with arbitrary $\mathrm{power}$ to convert the original problem into this lemma. Lemma. Let $(A')$ be the circumcircle of $\triangle ABC$, $AA'$ cuts $\odot(BA'C)$ at $H\neq A'$. $O$ is the reflection of $A$ by $BC$. $F, E,O_1$ is reflection of $O$ by $AB,AC,AH$. Then $BF, CE, HO_1$ are concurrent. Proof of the Lemma Note that, $A$ is the $H-\mathrm{excenter}$ of $\triangle HBC$. It follows that $BF,CE, HO_1$ are concurrent at $O'$, which is the isogonal conjugate of $O$ wrt $\odot(HBC)$.

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jayme

10.01.2020 14:29

Dear Mathlinkers, 1. Reim's theorem for G, D, A' collinear; (OA'D) intersect (O) at X ; Ha, D, X collinear 2. Monge's theorem for concurrence of OD, GHa, A'X at point T 3. Pascal's theorem.... 4. with isoceles triangle, T and Oa are identic... Sincerely Jean-Louis

jayme

23.01.2020 17:46

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%202.pdf p. 44... Sincerely Jean-Louis

Physicsknight

29.01.2020 10:25

Another solution a) Let $S=H_aD\cap A'O_a $, $I=OO_a=BC $. $G $ is the symmetrical point of $O $ through $D $, $\widehat {GBO}=\widehat {GCO}=90^{\circ} $. $OA'^2=OC^2=\overline {OI}\cdot\widehat {OG}=\tfrac 12\overline{OD}\cdot\overline {2OO_a}=\overline {OO_a}\cdot\overline {OD}\implies\widehat {OA'S}=\widehat {ODA'}=\widehat {ODH_a}$. Note, $OA'DS $ is cyclic $\implies\widehat {DSA'}=\widehat {DOA'}=\widehat {H_aAA'}\implies S\in\odot (O)$. b) Let $J $ be the midpoint of $OH$, $K=ED\cap HC$.$HOA'D $ is parallelogram $\implies HO=A'D=AX$, and $HO\parallel A'D\parallel AX\implies AXHO\text {is a parallelogram}\implies HX=HD=\tfrac 12 AA'$. $X $ is the symmetrical point of $A'$ through $J $. $D,E,F$ is the symmetrical point of $A, B, C $ through $J $. $\widehat {DKC}=\widehat {DIC}=90^{\circ}\implies IKCD\text {is cyclic} $. $\widehat {ECA}=\widehat {OCA}=\widehat {HCB}=\widehat {IDE}\implies\widehat {ECS}=\widehat {ECA}+\widehat {ACS}=\widehat {IDE}+\widehat {AA'S}=\widehat {IDE}+\widehat {ODS}=\widehat{EDS}\implies S\in\odot (ECD)$. Similarly, $S\in\odot (FBD)\implies (ECD), (FBD), (OA'D) $ are concurrent at $S $. Note, $(ECD)\sim (BFA), (FBD)\sim (CEA), (OA'D)\sim (HXA) $ so $(BFA), (CEA), (HXA) $ are concurrent at $S'$. $S'$ is the symmetrical point of $S $ through $J $.

khanhnx

12.02.2020 18:41

Here is my solution for this problem Solution a) Let $S$ $\equiv$ $O_aA'$ $\cap$ $(O)$ $(S \not \equiv A')$ We have: $(O_aO; O_aB) \equiv 2 (CO; CB) \equiv 2 (BC; BO) \equiv (BD; BO)$ (mod $\pi$) Then: $OB^2 = \overline{OO_a} . \overline{OD}$ So: $(DO; DH_a) \equiv (DA'; DO) \equiv (A'O; A'O_a) \equiv (SA'; SO) \equiv (DO; DS)$ (mod $\pi$) or $H_a$, $S$, $D$ are collinear Hence: $H_aD$ intersects $O_aA'$ at a point $S$ lies on $(O)$ b) Let $S'$ be a point which satisfies $ASDS'$ is parallelogram; $N$ be midpoint of $OH$ Since: $ASDS'$, $AXDA'$ are parallelograms and $N$ is midpoint of $AD$ then: $N$ is also midpoint of $SS'$, $A'X$ So: $\triangle AS'X$ $\stackrel{+}{=}$ $\triangle DSA'$ Hence: $(S'A; S'X) \equiv (SD; SA') \equiv (AH_a; AA') \equiv (DO; DH) \equiv (HH_a; HD) \equiv (HA; HX)$ (mod $\pi$) or $S'$ $\in$ $(AHX)$ We have: $(CE; CS) \equiv (CE; CA) + (CA; CS) \equiv (CA; CO) + (BA; BS) \equiv (AO; AC) + (BA; BS) \equiv (AB; AH)$ $+ (BA; BS) \equiv (SB; SH_a) + (BA; BS) \equiv (H_aS; BA) \equiv (DE; DS)$ (mod $\pi$) Then: $S$, $D$, $C$, $E$ lie on a circle So: $(S'A; S'B) \equiv (SD; SE) \equiv (CD; CE) \equiv (CD; CB) + (CB; CA) + (CA; CE) \equiv (CB; CO) + (CB; CA) +$ $(CO; CA) \equiv 2 (CB; CA) \equiv (OB; OA) \equiv (FA; FB)$ (mod $\pi$) or $S'$ $\in$ $(ABF)$ Similarly: $S'$ $\in$ $(ACE)$ Hence: $(AHX)$, $(ABF)$, $(ACE)$ have common point $S'$ which is different from $A$

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FISHMJ25

24.03.2020 01:21

If we $DH_a \cap (ABC)=P$ then $P$ is Feuerbach point of $\Delta I_aI_bI_c$. Now lets complex bash part b). First center of $\Delta BOC$ is $\frac{bc}{b+c}$ so center of $\Delta BCD$ is $ b+c-\frac{bc}{b+c}$. We can similarly calculate circumcenters of $\Delta ACE$ $\Delta ABF$ to be $ a+c-\frac{ac}{a+c}$ $ b+a-\frac{ba}{b+a}$ respectively. Next $X=2a+b+c$ so we need to find circumcenter of $\Delta AHX$. Shift everything by $a$ so we need circumcenter of $0$; $b+c$; $a+b+c$. But this is easy with formula $\frac{pq(\bar p-\bar q)}{\bar p q-\bar q p}$. So when we plug in, we get that center of $\Delta AHX$ shifted by $a$ is $\frac{(\sum a) bc}{bc-a^2}$. So now we just add $a$ and get formula for center, but we notice that al centers have $a$ so we don't add $a$ but rather substract $a$ from other centers(this is translation so collinearity is preserved). So equivalently we need to prove collinearity of $\frac{(\sum a) bc}{bc-a^2}$, $ \frac{c^2}{a+c}$ and $ \frac{b^2}{a+b}$ . Now just calculate $\frac{(\sum a) bc}{bc-a^2}- \frac{c^2}{a+c}=c\frac{(a+b)(\sum ab)}{(bc-a^2)(a+c)}$ and from this result clearly follows.

Math00954

07.03.2021 15:25

TheHimMan

25.01.2022 15:03

How is the intersection of OA' and HaD inside the circle O?

TheHimMan

25.01.2022 15:04

@trito11 There is a misprint in a)