A new geometric model in "Quan Hinh" Topic August 2019

by andrenguyen, Mar 28, 2021, 2:52 PM

Problem: Let $ABC$ be an acute and scalene triangle. Denote $D$ be a point moving on $BC$ such that $AD$ is not perpendicular to $BC$ . Assume that there are points $E, F$ on segment $AC, AB$ such that $\angle ADB = \angle BEC = \angle CFA$ . 3 lines $AD, BE, CF$ meet each other creating triangle $XYZ$ . Prove that the orthocenter of $XYZ$ moves on a fixed circle while $D$ moves on $BC$ .

("Quan Hinh" Topic August 2019 - Proposed by Nguyen Duc Toan)

Solution:

https://scontent-sin6-1.xx.fbcdn.net/v/t1.15752-9/166278288_1037123566811121_5309973010513133108_n.png?_nc_cat=107&ccb=1-3&_nc_sid=ae9488&_nc_ohc=BkOgEMGnub8AX-kJEdD&_nc_ht=scontent-sin6-1.xx&oh=1a0c7cd5b3fa51c2f4360ed809065238&oe=60862AA9

Let $O$ and $K$ be the circumcenter and the orthocenter of triangle $ABC$ , respectively. Let $H$ be the orthocenter of triangle $XYZ$ . We can easily see that $\angle KAD = \angle KBE = \angle KCF$ . Hence, $A, Z, H, B$ are concyclic; $B, C, X, H$ are concyclic; and $A, C, Y, H$ are concyclic. Hence $\angle KXZ = \angle KCB = \angle KAB = \angle KZX$ . Hence $KX=KZ$ . Similarly, we can lead to $K$ is the circumcenter of triangle $XYZ$ .

Let $\Gamma$ be the rotation with $O$ and rotational angle equal $\angle (DC,DA)$ We have $\Gamma : A \mapsto A', B \mapsto B', C \mapsto C', \triangle ABC \mapsto \triangle A'B'C'$ Hence, $\triangle ABC = \triangle A'B'C'$ and $\angle (BC,B'C') = \angle (BC,DA)$ . Hence, $AD \parallel B'C'$ , or $YZ \parallel B'C'$ . Similarly, $XY \parallel A'B', XZ \parallel A'C'$ Hence, triangles $A'B'C'$ and $XYZ$ are homothetic with each other.

Let $\Omega$ be the homothety mapping triangle $XYZ$ to triangle $A'B'C'$ . Let $N$ be the orthocenter of triangle $A'B'C'$ . We have $\Omega : X \mapsto A', Y \mapsto B', Z \mapsto C', K \mapsto O, H \mapsto N$ Then, $KH \parallel ON$ , and since $\Gamma: K \mapsto N$ , $\angle KON = \angle ADC$ . Hence, $\angle OKH = \angle ADB$ . On ray $KH$ , denote point $H'$ such that $OK=OH'$ . Let $CP$ be the diameter of $\odot(O)$ . We have $AK = BP$ and $\angle PCB = \angle ACK = \angle AYK$ . According to Sine Theorem, we have $\dfrac{KY}{KA}=\dfrac{\sin \angle YAK}{\sin \angle AYK } \wedge \dfrac{PB}{2OC}=\sin \angle PCB \Rightarrow \dfrac{KY}{OC}=2 \sin \angle YAK$ In addition, $\Omega: H \mapsto N, K \mapsto O$ . Therefore, $\dfrac{HK}{KO}=\dfrac{HK}{ON}=\dfrac{KY}{OB}=2 \sin \angle YAK$ We have $H'KO$ is an isosceles triangle at $O$ . According to Sine Theorem, we have
$\dfrac{KH'}{KO}=\dfrac{\sin \angle H'OK }{\sin(90^o-\frac{\angle H'OK}{2})}=\dfrac{\sin \angle H'OK}{\cos(\frac{\angle H'OK}{2})}=2 \sin(\frac{\angle H'OK}{2})=2 \sin \angle YAK = \dfrac{HK}{KO}$ Hence, $H \equiv H'$ , so $OH = OK = const$ . Therefore, $H$ moves on the circle with center $O$ and radius $OK$ which is fixed.

fixed points

trigonometry

rotation

Publication in The Mathematical Reflections - ANTI-STEINER POINT REVISITED

by andrenguyen, Nov 16, 2020, 2:02 PM

My article - Anti-Steiner Point Revisited - is published in The Mathematical Reflections Issue 6, 2020.

You can access this link below for the article.

https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2020-06/mr_6_2020_anti-steiner_point_revisited.pdf

I hope that you can enjoy the beauty of the problems and properties from my article.

publication

Anti-steiner point

A geometric problem using symmetry

by andrenguyen, Oct 18, 2020, 9:31 AM

Problem: Let $ABC$ be the triangle ( $AB>AC$ ) with bisector $AD$ of angle $\angle BAC$ and median $AM$ ( $M$ is the midpoint of segment $BC$ ). Let $E, F$ be the points on segments $AB, AC$ respectively such that $DE=DB$ and $DF=DC$ . On the external bisector of $\angle BAC$ , denote point $T$ such that $\angle ATD = \angle AMD$ . Prove that $A,E,F,T$ are concyclic.

("Quan Hinh" Topic September 2019 - proposed by Nguyen Duc Toan)

Solution:

https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvZS83L2UwNmM0MGJmOGRlMDA4YjFjZjM2MTNlOWZiMzcxOTlmMjM0NjFlLnBuZw==&rn=U29sLVFILXRoYW5nOS0yMDE5LU5EVC5wbmc=

Let $I,J$ be the midpoint of minor arc $BC$ and arc $BAC$ of the circumcircle of triangle $ABC$ . Hence, $J, A, T$ are collinear, and $A, D, I$ are collinear. Besides, $\angle DMJ = \angle DAJ = 90^o$ . Hence, $\angle DJA = \angle DMA = \angle DTA$ . Hence, $J$ is the reflection of $T$ in $AD$ . Let $B', C'$ be the reflections of $B, C$ in $AD$ , respectively. Hence, $B', C', D$ are collinear, and $A, T, B', C', I$ are concyclic. We have $AD$ is the bisector of $\angle C'AF$ and $DC'=DF$ , so $A, C', D, F$ are concyclic. Similarly, we have $A, E, D, B'$ are concyclic. Let $K$ be the intersection of $B'E$ and $C'F$ . Hence, $D$ be the Miquel point of the complete quadrilateral $AK.EF.B'C'$ . Moreover, $C', D, B'$ are collinear. Hence, $A, E, K, F$ are concyclic. In addition, $\angle KB'C' = \angle DAE = \angle DAF = \angle DC'K$ . Hence, $K$ lies on the perpendicular bisector of segment $B'C'$ , so $I, K, T$ are collinear. Hence,
$\angle EAT = \angle DAC' + 90^o = \angle KB'C' +90^o=180^o- \angle EKT$ Hence, $A, E, K, T$ are concyclic, or $A, E, F, T$ are concylic.

You can also find more about this problem through the initial topic at: https://artofproblemsolving.com/community/c6h2309042_concyclic_problem_from_quotquan_hinhquot_topic

This post has been edited 1 time. Last edited by andrenguyen, Oct 19, 2020, 2:56 PM

Symmetric

Concyclic

A beautiful geometric problem from "Quan HInh" Topic November 2019

by andrenguyen, Oct 18, 2020, 4:32 AM

Problem: Let $\odot(I)$ be the excircle with respect to vertex $A$ of triangle $ABC$ . $\odot(I)$ tangents to $BC$ at $D$ . A circle passing through $A$ and $I$ meets the opposite rays of rays $BA. CA$ at $E$ and $F$ , respectively. Let $M$ be the midpoint of segment $EF$ . $DM$ meets $\odot(I)$ at the second point $H$ . The line passing through $F$ and parallel to $BC$ meets the line passing through $E$ and perpendicular to $IH$ at point $K$ . Prove that: $KE+KF = AE+AF$
("Quan Hinh" Topic November 2019 - proposed by Nguyen Duc Toan)

Solution: (Nguyen Duc Toan)

https://scontent.fhan5-7.fna.fbcdn.net/v/t1.15752-9/121571993_351491026061853_2180977616051625270_n.png?_nc_cat=100&_nc_sid=ae9488&_nc_ohc=wANAoV-BweoAX9CuPxT&_nc_oc=AQm_88LXePe2RFQnpD8x9_mNWTAm7xuSDAkU7f1ChqksqrLokYgem0cS0ns5lhBMp0g&_nc_ht=scontent.fhan5-7.fna&oh=fe270e5a7f12745e16bed576457f8298&oe=5FB05C32

The tangent at $H$ of $\odot(I)$ meets three lines $AB, AC$ , and $BC$ at $G, L,$ and $T$ , respectively. $EF$ meets $BC$ and $GL$ at $S, R$ respectively. Let $P, Q$ be the projections of $I$ to $AB, AC$ , respectively. According to Simpson line Theorem, we have $P, M, Q$ are collinear. Moreover, since $\angle AEI = \angle IFQ$ , $\triangle IPE = \triangle IQF$ . Hence, $PE=QF$ , so $AE+AF = AP+AQ=2AP$ We have $M,H,D$ are the projections of $I$ to $RS, TR, TS$ , and $D, M, H$ are collinear. According to the inverse Simpson line Theorem, $T, R, I, S$ are concyclic. Similarly, we have $\triangle IRH = \triangle ISD$ and $TS+TR=TH+TD=2TH$ We have $AP = PI \tan \angle PIA = IH \tan \angle TIH$ as well as $\triangle KEF \sim \triangle TRS$ . Hence, $\dfrac{AE+AF}{TR+TS}=\dfrac{2AP}{2TH}=\dfrac{PI\tan\angle PIA}{IH\tan\angle TIH}=\dfrac{IM\tan\angle EIM}{IM\tan\angle RIM}=\dfrac{EM}{RM}=\dfrac{KE+KF}{TR+TS}$ Therefore, $AE+AF=TR+TS$

This post has been edited 1 time. Last edited by andrenguyen, Oct 18, 2020, 5:48 AM

plane-geometry

trigonometry

Simpson line

A tricky Concurrent lines related problems

by andrenguyen, Oct 16, 2020, 12:11 PM

Problem: (Tran Quang Hung) Let $ABC$ be a triangle inscribed in circle $(O)$ , altitudes $AD$ , $BF$ meet at orthocenter $H$ . $J$ is midpoint of $AH$ . $OJ$ meets $BC$ at $P$ . $Q$ is the point such that $AQ\parallel BC$ , $JQ\perp AB$ . $R$ is a point on ray $DF$ such that $AR$ is tangent to $(ADF)$ . $M$ , $N$ are reflections of $A$ in the lines $QR$ , $QF$ respectively. Prove that the lines $PQ$ , $MN$ and $AD$ are concurrent.

https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYi8wLzFjNmEzYjVkOWRhZDBhNjUzNDczNzE4MGExZTUwMjFhZDg4ZTRmLnBuZw==&rn=RmlndXJlNzQ3OC5wbmc=

Solution: (Nguyen Duc Toan)

First, we have a lemma:

Lemma: Triangle $ABC$ with circumcenter $O$ , orthocenter $H$ . Let $AD, BE, CF$ are the altitudes of the triangle. $EF$ meets $BC$ at $V$ . Let the perpendicular bisectors of two segments $AB$ and $HB$ meet the line through $B$ perpendicular to $BC$ at $M$ and $N$ , respectively. Similarly, Let the perpendicular bisectors of two segments $AC$ and $HC$ meet the line through $C$ perpendicular to $BC$ at $P$ and $Q$ , respectively. Let $J, T, K'$ are the midpoint of segments $AH, MN, PQ$ . Then, $V, T, J, K'$ are in the same line.

https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvMC9iL2EwMWZjZWY5NzZlMjY1NzZiMDkxYTZiZTNmMGRlZjIzZjI3OWMxLnBuZw==&rn=bGVtbWEucG5n

Proof
Define circle $(X,XY)$ is the circle with center $X$ and radius $XY$ , circle $(O)$ is the circle with center $O$ .
Let $G$ is the midpoint of $BC$ . $VH$ meets $AG$ at $K$ . It is well-known that $(I,IA)$ , $(O)$ , $AV, GH$ are concurrent at point $I$ . Hence, $H$ is the orthocenter of triangle $AVG$ . Hence $VK \perp AG$ and $K \in (I)$ . According to Newton's theorem, we have $(VDBC)=-1$ , hence $GB^2=GC^2=\overline{GD}.\overline{GV}=\overline{GH}.\overline{GI}=\overline{GK}.\overline{GA}$ . Hence $(M,MA), (P,PA), (I)$ are coaxial (line $AK$ ), and $(N,NH), (Q,QH), (I)$ are coaxial (line $HI$ ). Hence, $M, J, P$ are collinear, and $N, J, Q$ are collinear. Moreover, we have $\dfrac{NB}{CP}=\dfrac{\frac{BH}{2 \cos C}}{\frac{AC}{2 \sin C}}=\dfrac{BH}{AC}.\dfrac{\sin C}{\cos C}=\dfrac{BD}{AD}.\dfrac{AD}{DC}=\dfrac{BD}{DC}=\dfrac{VB}{VC}$ . According the Thales converse theorem, we have $V, N, P$ are collinear. Similarly, we have $V, M Q$ are collinear. Therefore, we have $VJ$ passes through the midpoints of two segments $MN$ and $PQ$ .

Back to the main problem

https://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvNi9jLzUxMWIzYWQ1NThkMjM1MTExOWYxNzdjZDIxZTFjMDcxMTlhYzdkLnBuZw==&rn=dGFwaHVhbmRhbmFuZy1UUUgucG5n

Let $AA'$ is the diameter of the circumcircle of triangle $ABC$ , $BE$ is the altitude. $A'H$ meet $(O)$ at the second point $L$ . It is easy to prove that $AL, EF, BC$ are concurrent at $V$ . Also, $A, L, E, F, H$ lie in the circle with center $J$ and diameter $AH$ . Hence, $(O)$ meets $(I)$ at two points $A$ and $L$ . Hence, $OI \perp AV$ at $W$ . Hence, $J$ is the orthocenter of triangle $AVP$ and $VJ \perp AP$ at $U$ .

Let $\Phi$ is the inversion with center $A$ and power $AH.AD$ . We have:
$\Phi: H \leftrightarrow D, F \leftrightarrow B E \leftrightarrow C, J \leftrightarrow J'$ ( $J'$ is symmetric with $A$ by $BC$ ) $; P \leftrightarrow P'$ ( $P'$ is symmetric with $A$ by $VJ$ ) $;$ circle $(Q,QA) \leftrightarrow$ the line through $B$ perpendicular with $BC$ , $Q \leftrightarrow Q'$ . Moreover, since $A, J, F, Q$ are concyclic, $Q' \in J'B$ . Let $VJ$ meets the line through $B$ perpendicular to $BC$ at $T$ . Then, $T$ is the circumcenter of triangle $Q'P'A$ .

Furthermore, we have $\Phi:$ line $BT \leftrightarrow$ circle $(Q,QA)$ , circle $(F,FA) \leftrightarrow$ the perpendicular bisector of segment $AB$ . Therefore, if $\Phi: N \leftrightarrow N'$ , $N'$ is the intersection of line $BT$ and the perpendicular bisector of segment $AB$ . Besides, we can easily see that circle $(R,RA)$ is the A-Apollonius circle of triangle $AFD$ . Hence, circle $(R,RA)$ is orthogonal with $(AFD)$ . Hence $\Phi: (R,RA) \leftrightarrow$ the perpendicular bisector of segment $BH$ . Therefore, if $M \leftrightarrow M'$ , $M'$ is the intersection of $BT$ and the perpendicular bisector of segment $BH$ .

Therefore, to prove $PQ, MN, AD$ are concurrent, we just need to prove the line passing two circumcenters of triangle $AM'N'$ and $AQ'P'$ is perpendicular with $AD$ . Indeed, according to the lemma, we have $T$ is midpoint of $M'N'$ . $T$ is also the circumcenter of triangle $AP'Q'$ . Therefore, we finish the proof.

This post has been edited 1 time. Last edited by andrenguyen, Oct 16, 2020, 12:12 PM

Inversion

concurrency

orthocenter model

A tangent circles related problem from "Quan Hinh" Topic - February 2019

by andrenguyen, Sep 12, 2020, 9:23 AM

Problem: Let $ABC$ be the triangle with circumcenter $O$ . Let $J$ be the center of the $A-mixtilinear$ circle of the triangle ( $\odot(J)$ is tangents to $AB, AC$ and internally tangent to $\odot(O)$ ). The tangents of $\odot(J)$ from $B$ and $C$ meet $\odot(O)$ at the second points $M$ and $N$ , respectively. $AM, AN$ meet $BC$ at $X, Y$ , respectively. Prove that the excircle of triangle $AXY$ with respect to vertex $A$ is tangent to $\odot(O)$ .

("Quan Hinh" Topic - February 2019 - proposed by Nguyen Duc Toan)

My official solution:
In this problem, we just prove this lemma:

Lemma: Given two circles $\odot(J)$ and $\odot(O)$ internally tangent to each other ( $J$ lies on the internal of $\odot(O)$ ). A point $A$ moves on $\odot(O)$ . $B, C$ are points on $\odot(O)$ such that $AB, AC$ are tangent to $\odot(J)$ . Then, $BC$ is tangent to a fixed circle $\gamma$ when $A$ is moving, and $\gamma$ is also tangent to $\odot(O)$ .

From this lemma, we can get that $AM, AN$ and $BC$ are tangent to $\gamma$ or $\gamma$ is the excircle of triangle $AXY$ with respect to vertex $A$ . Then, we have $\gamma$ is tangent to $\odot(O)$ , and we finish our proof.

Now, we will prove the lemma by two claims:

https://scontent.fdad3-3.fna.fbcdn.net/v/t1.15752-9/118980585_645869456358264_8282208668614775530_n.png?_nc_cat=108&_nc_sid=b96e70&_nc_ohc=430pOwznKwEAX88kZE7&_nc_ht=scontent.fdad3-3.fna&oh=cdc37741375774a3649eab1e9b922dfe&oe=5F82E3E7

Claim 1: When $A$ is moving, the incenter $I$ of triangle $ABC$ is also moving on a fixed circle.
Indeed, Let $D, E$ be the midpoints of the arc $BC$ and arc $BAC$ of $\odot(O)$ . Let $F, G$ be the tangent points of $\odot(J)$ with $AB, AC$ , respectively. $BC$ meet $TE$ at $M$ .
According to a popular property from mixtilinear circle model, we have $T, E, I$ are collinear and $F, I, G$ are collinear. The line passing through $I$ and perpendicular to $BC$ meets $OT$ at point $L$ . According to the Menelaus's theorem, we have $\dfrac{DO}{DE} \cdot \dfrac{IE}{IT} \cdot \dfrac{JT}{JO} = 1 \Rightarrow \dfrac{IE}{IT} = 2 \dfrac{JO}{JT} = const$ Then, according to Thales's theorem, we have $\dfrac{LT}{LO}=\dfrac{IE}{IT}=const$ Hence, $L$ is a fixed point. Hence, the circle with center $L$ and radius $LT$ is fixed. Therefore, we have $I$ lies on the fixed circle $\odot(L)$ .

Claim 2: The circle with center $K$ tangent to $\odot(O)$ at $T$ and also tangent $BC$ is a fixed circle.
Indeed, according to a popular property of mixtilinear circle model, we have $FG, BC, TD$ are concurrent at $N$ . Also, since $\angle NID = 90^o$ , we have $TI^2=TN.TD$ . Moreover, since $M$ is the orthocenter of triangle $NED$ , $TN.TD=TM.TE$ . Hence, $TI^2=TM.TE$ . Hence, $\dfrac{TM}{TI}=\dfrac{TI}{TE}$ . Let $\Phi$ be the homothety with center $T$ and ratio $\dfrac{\overline{TM}}{\overline{TI}}$ . We have, $\Phi: I \mapsto M, E \mapsto I, L \mapsto K, O \mapsto L, \odot(L) \mapsto \odot(K), \odot(O) \mapsto \odot(L)$ Since $\odot(L)$ and $\odot(O)$ are fixed circles, we have $\dfrac{\overline{TM}}{\overline{TI}}=const$ . Hence, $\odot(K)$ is a fixed circle.

After this proof, we can see that $\odot(K)$ is equivalent to $\gamma$ .

This post has been edited 3 times. Last edited by andrenguyen, Sep 12, 2020, 12:34 PM

An Anti-steiner point of Euler line related problem-"Quan Hinh" Topic April 2019

by andrenguyen, Sep 8, 2020, 2:27 PM

Problem: Triangle $ABC$ is scalene. Let two point $D$ and $E$ lie on rays $BC$ and $CB$ respectively such that $BD=BA$ and $CE=CA$ . The circumcircle of triangle $ABC$ meets the circumcircle of triangle $ADE$ at $A$ and $L$ . Let $F$ be the midpoint of arc $AB$ which not contain $C$ of the circumcircle of $ABC$ . Let $K$ lie on $CF$ such that $\angle FDA = \angle KDE$ . $KL \cap AE = J$ .
Prove that $J$ lies on the Euler line of triangle $ADE$ .

("Quan Hinh" Topic - April 2019 - proposed by Nguyen Duc Toan)

Original Topic: https://artofproblemsolving.com/community/c6h1823305

My official solution:

Lemma: Let $ABC$ be a triangle with circumcenter $O$ . $O'$ is the reflection of $O$ in $BC$ . Let $J$ be the circumcircle of triangle $BOC$ . Then, $AJ, AO'$ are isogonal conjugate lines in angle $\angle BAC$ .

https://scontent-hkt1-1.xx.fbcdn.net/v/t1.15752-9/118886740_626630898014568_8383638100042678367_n.png?_nc_cat=103&_nc_sid=b96e70&_nc_ohc=9NQEN_sckeMAX9Nm-fj&_nc_ht=scontent-hkt1-1.xx&oh=40bc9a7d719bdd1a4a95336a91755d0a&oe=5F7C8673

Proof:
Let $D$ be the intersection of two tangents from $B$ and $C$ of $\odot(O)$ . Denote that $M$ is the midpoint of segment $BC$ . Since, $AD$ is the symmedian, $AD$ and $AM$ are isogonal conjugate lines in angle $\angle BAC$ . Moreover, we have $OM.OD = OB^2 = OA^2 = OJ.OO'$ Hence, $\triangle OAM \sim \triangle ODA$ and $\triangle OJA \sim \triangle OAO'$ . Hence, $\angle OAM = \angle ODA$ and $\angle OAJ = \angle OO'A$ . Then, $\angle DAO' = \angle JAM$ . Hence, $AJ, AO'$ are two isogonal conjugate lines in angle $\angle BAC$ .

Back to our main problem,

https://scontent-hkt1-1.xx.fbcdn.net/v/t1.15752-9/118799487_322305488844777_8852750022538852114_n.png?_nc_cat=109&_nc_sid=b96e70&_nc_ohc=3TzOjJ9WXisAX_uujYj&_nc_ht=scontent-hkt1-1.xx&oh=3755af48b91a734aa6479921b325c441&oe=5F7EABAA

Denote that $O$ and $H$ are the circumcenter and orthocenter of triangle $AED$ , respectively. $OC$ meets $AD$ at $S$ ; $OB$ meets $AE$ at $I$ . We can easily get that $C, O, L$ lie on the perpendicular bisector of segment $AE$ , and $O$ is the incenter of triangle $ABC$ . Then, $F$ is the center of $\odot(AOB)$ . Besides, we have $\angle AEB = 90^o - \dfrac{\angle ACB}{2} = 180^o - (90^o - \dfrac{\angle ACB}{2}) = \angle 180^o - \angle AOB$ Hence, $A, E, O, B$ are concyclic and $F$ is the center of $\odot(AOE)$ . Similarly, we have $O, E, R, S, D$ are concyclic. According to the Lemma, we have that $K$ is the reflection of $O$ in $AE$ .
Now, we need to prove that $L$ is the Anti-Steiner point of $OH$ with respect to triangle $AED$ .
Indeed, let $L'$ be the Anti-Steiner point of $OH$ with respect to triangle $AED$ . We can easily see that $O$ is the orthcenter of triangle $AIS$ . Hence, $L'$ is the Anti-Steiner point of line $OH$ with respect to triangle $AIS$ . Hence, $L' \in \odot (AIS)$ . $IS$ meets $ED$ at $R$ . Hence, $L'$ is the Miquel point of the complete quadrilateral $AR.ES.ID$ . Let $T$ and $U$ be the midpoints of segments $AE$ and $AD$ , respectively. Hence, $(L'I,L'A) \equiv (SI,SA) \equiv (SI,SU) \equiv (TI,TU) \equiv (EI,ER) \equiv (L'I,L'R)$ Hence, $L', R, A$ are collinear. We can easily prove that $OB.OI = OE^2 = OS.OC$ . Hence, $I, B, S, C$ are concyclic. Then, $RB.RC=RS.RI=RL'.RA$ . Hence, $A, B, C, L'$ are concylcic. Therefore, $L' \equiv L$ or $L$ is the Anti-Steiner point of $OH$ with respect to triangle $ADE$ .
Hence, line $LK$ is the reflection of line $OH$ in $AE$ . Hence, $J$ lies on the Euler line of triangle $ADE$ .

Comment: The Lemma in the problem is popular and useful in many other problems' solutions.

You can also find the old "Quan Hinh" Topic beautiful problems in our Facebook group.
https://www.facebook.com/groups/hinhhocphang.geometry

This post has been edited 3 times. Last edited by andrenguyen, Sep 12, 2020, 9:45 AM

A beautiful problem related to Anti-Steiner point and Feuerbach point

by andrenguyen, Sep 4, 2020, 2:31 PM

Problem: Let $(I)$ be the incircle of triangle $ABC$ and $D, E, F$ be the contacts triangle. Let $F_e$ be the Feuerbach point of triangle $ABC$ . Let $K$ be the orthocenter of triangle $DEF$ . Let $S$ be the Anti Steiner point of $I$ WRT triangle $ABC$ . Prove that $KF_e || IS$ .

Original post: https://artofproblemsolving.com/community/c6h1824619p12198966

Solution:

Label that $\odot (O)$ is the circle with center $O$ , $\odot(XYZ)$ is the circumcircle of triangle $XYZ$ . Label that $(d,d')$ is the directed angle of lines $d$ and $d'$ modulo $\pi$ .

Lemma 1: (Ha Huy Khoi) Let $ABC$ be the triangle with circumcenter $O$ , incenter $I$ , and orthocenter $H$ . $K$ is the Anti-Steiner point of $HI$ with respect to $\triangle ABC$ . $KI$ meets $\odot(O)$ at the second point $L$ . $AL$ meets $OI$ at $J$ . Then, $AL$ is symmetric to $OI$ with respect to the perpendicular bisector of segment $AI$ .

https://scontent-hkt1-1.xx.fbcdn.net/v/t1.15752-9/118769645_635303847360805_4860266796520711783_n.png?_nc_cat=107&_nc_sid=b96e70&_nc_ohc=NoNztaTS6DcAX9it21o&_nc_ht=scontent-hkt1-1.xx&oh=d4ac211789833f410cdf543cd429e1f4&oe=5F77DE95

Proof: (Ha Huy Khoi) Indeed, $AH$ meets $\odot(O)$ at the second point $E$ . $I'$ is the reflection of $I$ in $BC$ . Then, $EI'$ passes through $K$ . $AI$ meets $\odot(O)$ at the second point $D$ . Then,
$\rightarrow (II',ID) \equiv (AE,AD) \equiv (KI',KD) \rightarrow$ $I, I', D, K$ are concyclic.
Let $D'$ be the reflection of $D$ in $BC$ , $M$ be the midpoint of segment $BC$ , $DN$ be the diameter of $\odot(O)$ .
We have: $DI^2=DB^2=DM.DN=DD'.DO \rightarrow \triangle DIO \sim \triangle DD'I$
$\rightarrow (IA,IO) \equiv (IO,ID) \equiv (D'D,D'I) \equiv (KD,KI) \equiv (AI,AL)$
$\rightarrow (AI,AL) \equiv (IA,IO) \equiv (IJ,IA) \rightarrow JA=JI$
Hence, $AL$ is symmetric to $OI$ with respect to the perpendicular bisector of segment $AI$ .

Lemma 2: Let $ABC$ be the triangle with circumcenter $O$ and orthocenter $H$ . $T$ is the Anti-Steiner point of $OH$ with respect to triangle $ABC$ . Three altitudes of triangle $ABC$ meet $\odot(O)$ at the second points $D, E, F$ , respectively. $TH$ meets $\odot(O)$ at the second point $K$ . Then, $K$ is the Anti-Steiner of $H$ with respect to triangle $DEF$

Proof: (Ha Huy Khoi)
Let $K'$ be the Anti-Steiner point of $H$ with respect to triangle $DEF$ . $K'H$ meets $(O)$ at the second point $T'$ .
Applying Lemma 1 into the $\triangle DEF$ with its incenter $H$ , we have $DT'$ is symmetric to $OH$ with respect to the perpendicular bisector of segment $DH$ . Then, $DT'$ is the reflection of $OH$ in $BC$ . Similarly, we have $ET', FT'$ are the reflections of $OH$ in $CA, AB$ , respectively.
Hence, $T'$ is the Anti-Steiner point of $OH$ with respect to $\triangle ABC$ .
Therefore, $T' \equiv T, K' \equiv K$ . Hence, $K$ is the Anti-Steiner point of $H$ with respect to triangle $DEF$ .

Back to our main problem,

https://scontent-hkt1-1.xx.fbcdn.net/v/t1.15752-9/118729561_646244236097057_3423383911389957874_n.png?_nc_cat=110&_nc_sid=b96e70&_nc_ohc=OpxwZyoygl8AX8wWo50&_nc_ht=scontent-hkt1-1.xx&oh=0a1ac3d32d9012fdccccdd0aaa3bcca2&oe=5F780964

Let $O, H$ be the circumcenter and the orthocenter of triangle $ABC$ , $X,Y,Z$ be the second intersections of $AI, BI, CI$ and $\odot (O)$ , respectively.
Let $M, N, P$ be the midpoint of segments $EF, FD, DE$ respectively. Let $\Theta$ be the inversion with center $I$ and power $IE^2$ . We have $\Theta: A \leftrightarrow M, B \leftrightarrow N, C \leftrightarrow P, \odot(ABC) \leftrightarrow \odot(MNP)$ Hence we have, $I, O$ and the center of $\odot(MNP)$ are collinear. Hence, $O$ lies on the Euler line of trianlge $DEF$ . Hence, $O, I, K$ are collinear.
Moreover, according to Fontene's theorem, we have $F_e$ is the Anti-Steiner point of $OK$ .
Let $T$ be the Anti-Steiner point of $H$ with respect to triangle $XYZ$ . According to Lemma 2, we have $T, I , S$ are collinear.
Besides, we can easily get $EF \parallel YZ, FD \parallel XZ, DE \parallel XY$ . Then, there is a homothety $\Phi$ that $\Phi: D \mapsto X, E \mapsto Y, F \mapsto Z, \triangle DEF \mapsto \triangle XYZ, \odot(DEF) \mapsto \odot(XYZ), K \mapsto I, F_e \mapsto T$ Therefore, $KF_e \parallel IT$ or $KF_e \parallel IS$ .

Anti-steiner point

Feuerbach Point

Inversion

My solution for IMOC 2019 G5

by andrenguyen, Aug 23, 2020, 5:14 PM

Problem: (IMOC 2019 G5) Given a scalene triangle $\vartriangle ABC$ with orthocenter $H$ and circumcenter $O$ . The exterior angle bisector of $\angle BAC$ intersects circumcircle of $\vartriangle ABC$ at $N \ne A$ . Let $D$ be another intersection of $HN$ and the circumcircle of $\vartriangle ABC$ . The line passing through $O$ , which is parallel to $AN$ , intersects $AB,AC$ at $E, F$ , respectively. Prove that $DH$ bisects the angle $\angle EDF$ .

Solution:

https://scontent.fdad3-1.fna.fbcdn.net/v/t1.15752-9/117770228_633255090650839_4614696949313122246_n.png?_nc_cat=110&_nc_sid=b96e70&_nc_ohc=N0CN3RGUOeoAX-TkhEU&_nc_ht=scontent.fdad3-1.fna&oh=ed7c421b8a6f768f22c3e68a5fb7528b&oe=5F69C468

Let $K$ be the midpoint of minor arc $BC$ , $A'$ be the antipode of $A$ , $CL$ be the altitude of triangle $ABC$ . The line passing through $H$ and parallel $AN$ meets $AB, AC$ at $I$ and $J$ respectively. We can easily get $E, F, I, J$ are concyclic. Since, $EF \parallel AN \parallel AK$ , $K$ is the reflection of $A$ in $EF$ . Then, $\angle EAF = \angle EKF$ . We have $\triangle AHL \sim \triangle AA'C$ . Then, $\dfrac{AH}{AO} = 2 \cdot \dfrac{AH}{AA'} = 2 \cdot \dfrac{AL}{AC} = 2 \cdot \cos \angle A$ Also, we have $\triangle AIH \sim \triangle AOF$ . Then, $\dfrac{AI}{AF} = \dfrac{AH}{AO} = 2 \cos \angle A$ Then, $FIA$ is an isosceles triangle. Then, $\angle AIF = \angle IAF = \angle EKF$ . Therefore, $E, F, K, I$ are concyclic. According to the Reim's Theory, since $IJ \parallel AN$ , we have $I, H, B, D$ are concyclic and $J, H, D, C$ are concyclic. Then, $\angle IDJ = \angle IDH + \angle JDH = \angle HBA + \angle HCA = 2(90^o-\angle BAC) = \angle IFJ$ Hence, $I, F, J, D$ are concyclic. Since $EI = FJ$ and $DH$ bisects angle $\angle IDJ$ , $DH$ bisects angle $\angle EDF$ . $\qquad \blacksquare$

This post has been edited 1 time. Last edited by andrenguyen, Aug 23, 2020, 5:15 PM

Reim s theorem

My solution for the APMO 2019 P3

by andrenguyen, Aug 23, 2020, 3:56 AM

Problem: (APMO 2019 P3) Let $ABC$ be a scalene triangle with circumcircle $\Gamma$ . Let $M$ be the midpoint of $BC$ . A variable point $P$ is selected in the line segment $AM$ . The circumcircles of triangles $BPM$ and $CPM$ intersect $\Gamma$ again at points $D$ and $E$ , respectively. The lines $DP$ and $EP$ intersect (a second time) the circumcircles to triangles $CPM$ and $BPM$ at $X$ and $Y$ , respectively. Prove that as $P$ varies, the circumcircle of $\triangle AXY$ passes through a fixed point $T$ distinct from $A$ .

Solution: First, we have a popular lemma.

Lemma: (Reim's theorem) Given two circles $(O)$ and $(O')$ meet each other at two points $A$ and $B$ respectively. Line $d$ passing through $A$ meets $(O)$ and $(O')$ at second points $M$ and $N$ respectively. Line $d'$ passing through $B$ meets $(O)$ and $(O')$ at second points $P$ and $Q$ respectively. Then, $MP \parallel NQ$ .

Back to the problem,

https://scontent.fhan5-4.fna.fbcdn.net/v/t1.15752-9/118051099_321381685731058_5359914565316307920_n.jpg?_nc_cat=110&_nc_sid=b96e70&_nc_ohc=6v_GHr9fDTcAX9n1vbl&_nc_ht=scontent.fhan5-4.fna&oh=cf68f965bd4fb10db0d1e75ee251f35e&oe=5F65F7CE

$EP, DP$ meet the circle $\gamma$ at second points $K$ and $L$ respectively. According to the Lemma, we have $KB \parallel AM \parallel LC$ . Then $K$ and $L$ are fixed points. Similarly, according to the Lemma, we have $EC \parallel BY, XC \parallel BD$ . $BD \cap EC = V, BY \cap CX = U$ . Then, $V$ is the radical center of three circles $\odot(BDM), \odot(EMC)$ and $\gamma$ . Hence, $V, P, M$ are collinear. Moreover, since $BUCV$ is a parallelogram, $U, P, M, V$ are collinear. Hence, $UX.UC=UP.UM=UY.UB$ , then $Y, X, C, B$ are concyclic and $XY \parallel DE$ . According to the Lemma, then $K, X, Y, L$ are concyclic. Therefore, $KL, YX, BC$ are concurrent at the radical center $I$ of three circles $\odot(KYX), \odot(BYX)$ and $\gamma$ . $IA$ meets $\gamma$ at the second point $T$ . Then $T,Y,X,A$ are concyclic. Since $KL$ and $BC$ are fixed, $I$ is a fixed point. Hence, $T$ is a fixed point. Therefore, $\odot(AXY)$ passes through fixed point $T$ . $\qquad \blacksquare$

This post has been edited 1 time. Last edited by andrenguyen, Aug 23, 2020, 3:56 AM

radical-axis

Fixed point

Reim s theorem