Let $I, O, \omega, \Omega$ be the incenter, circumcenter, the incircle, and the circumcircle, respectively, of a scalene triangle $ABC$. The incircle $\omega$ is tangent to side $BC$ at point $D$. Let $S$ be the point on the circumcircle $\Omega$ such that $AS, OI, BC$ are concurrent. Let $H$ be the orthocenter of triangle $BIC$. Point $T$ lies on $\Omega$ such that $\angle ATI$ is a right angle. Prove that the points $D, T, H, S$ are concyclic. Proposed by ltf0501
Problem
Source: 2020 Taiwan TST Round 2 Mock Exam P6
Tags: geometry, geometry proposed
Li4
24.04.2020 11:02
An easy corollary of the properties of Feuerbach point.
zuss77
24.04.2020 12:42
Statement without words: https://www.youtube.com/watch?v=loW190Nt99I
TelvCohl
24.04.2020 14:06
Let $ \triangle XYZ $ be the circumcircle mid-arc triangle of $ \triangle ABC $ and $ J $ be the reflection of $ I $ in $ BC, $ then $ D, T, X $ are collinear and from $ \triangle AIO \stackrel{+}{\sim} \triangle IJX $ we get $ JX $ passes through the reflection $ U $ of $ A $ in $ OI. $ Let $ V $ be the reflection of $ X $ in $ OI, $ then $$ (A,U;B,C) \stackrel{I}{=} (X, V; Y, Z) = X (X, V; Y, Z) = I (D, O;C,B )\ , $$so $ U\in DS. $ Perform the inversion with center $ D $ that fixed $ \Omega $ we conclude that $ D, H, S, T $ are concyclic. $ \qquad \blacksquare $
Attachments:
ltf0501
24.04.2020 14:44
TelvCohl wrote: Let $ \triangle XYZ $ be the circumcircle mid-arc triangle of $ \triangle ABC $ and $ J $ be the reflection of $ I $ in $ BC, $ then $ D, T, X $ are collinear and from $ \triangle AIO \stackrel{+}{\sim} \triangle IJX $ we get $ JX $ passes through the reflection $ U $ of $ A $ in $ OI. $ Let $ V $ be the reflection of $ X $ in $ OI, $ then $$ (A,U;B,C) \stackrel{I}{=} (X, V; Y, Z) = X (X, V; Y, Z) = I (D, O;C,B )\ , $$so $ U\in DS. $ Perform the inversion with center $ D $ that fixed $ \Omega $ we conclude that $ D, H, S, T $ are concyclic. $ \qquad \blacksquare $ Wow!!! What a fantastic solution.
ltf0501
29.04.2020 06:33
Here is the official solution provided by me. Let $N$ be the second intersection of $AI$ and $\Omega$, then $N, D, T$ are collinear. Now I give two solutions. Solution 1. (with lots of properties.) Let $D'$ be the reflection of $D$ w.r.t $EF$, where $E, F$ are the tangents of $\omega$ and $CA, AB$, respectively. It is known that $AD', OI, BC$ are collinear. (Maybe you can see here) Let $K$ be the foot from $D$ to $EF$, then $A', I, K, T$ are collinear, where $A'$ is the antipode of $A$ w.r.t $\Omega$. Also note that both $BI \cap CH$ and $CI \cap BH$ lie on $EF$, so \[ K(D, E; T, H) \stackrel{K}{=} (D, DI \cap EF; I, H) = -1 \]Since $DK \perp EF$, we have $\measuredangle TKD = \measuredangle DKH$. Let $N'$ be the antipode of $N$ w.r.t $\Omega$, then \[ \measuredangle DTK = \measuredangle NTA' = \measuredangle NN'A' = \measuredangle HDK \]Hence $\triangle KDT \stackrel{+}{\sim} \triangle KHD$. By $D'$ is the symmetric point of $D$ w.r.t $EF$, it follows that $\triangle K'DT \stackrel{+}{\sim} \triangle KHD'$, so \[ \measuredangle D'TD = \measuredangle D'TK + \measuredangle KTD = \measuredangle HD'K + \measuredangle KDH = \measuredangle D'HD, \]that is, $D, H, D', T$ are concyclic. On the other hand, since \[ \measuredangle STD = \measuredangle STN = \measuredangle SAN = \measuredangle SD'D, \]$D, S, D', T$ are concyclic. It follows that $D, T, H, S$ are concyclic, as desired. Solution 2. Notice that \[ \measuredangle STD = \measuredangle STN = \measuredangle SAN = \measuredangle SAI, \ \measuredangle SHD = \measuredangle SHI, \]so $D, T, H, S$ are concyclic is equivalent to $\measuredangle SAI = \measuredangle SHI$, that is, $A, I, S, H$ are concyclic. Let $Q$ be the second intersection of $OI$ and $\odot(BIC)$. By radical axis theorem we have $Q \in \odot(AIS)$. Hence we just need to prove that $A, I, Q, H$ are concyclic. Consider the inversion with center $I$, then the statement becomes the following: Let $I$ be the orthocenter of $\triangle ABC$. $H$ is on the plane such that $I$ is the incenter(or excenter, the proof is similar) of $\triangle HBC$. Prove that $A, H, Q$ are collinear. Now we just look at the inverting statement. Let $O_1$ be the circumcenter of $\triangle ABC$, $D', M$ be the foot from $I, O_1$ to $BC$, respectively. Notice that the reflection $O_1'$ of $O_1$ w.r.t $BC$ is the circumcenter $\triangle IBC$, which is the second intersection of $HI$ and $\odot(HBC)$, so $AO_1 \| IO_1' \| HI$. Notice that the reflection $I_h$ of $A$ w.r.t $M$ is the excenter of $\triangle HBC$, satisfying that $H, I, I_h$ are conllinear. Also known that $MI_h \| HD'$, so $AM \| HD'$. Combined with $O_1M \| ID'$, it follows that $\triangle AO_1M$ is homothetic to $\triangle HID$. Thus $AH, O_1I, MD$ are concurrent, that is, $A, H, Q$ are collinear, as desired.
Abhaysingh2003
07.09.2020 01:43
TelvCohl wrote: $$ X (X, V; Y, Z) = I (D, O;C,B )\ $$ Can anyone please explain this step?
Abhaysingh2003
07.09.2020 04:50
Bump. Please anyone?
Ninjasolver0201
07.09.2020 05:54
Abhaysingh2003 wrote: TelvCohl wrote: $$ X (X, V; Y, Z) = I (D, O;C,B )\ $$ Can anyone please explain this step? Just draw the diagram and see the projections. It is done using projective geometry.
Abhaysingh2003
07.09.2020 05:58
@above I know Projective Geometry. But that step seems fishy and is not related to perspectivities. I will be glad if someone explains it. I just want to know why $X(X,V;Y,Z)=I(D,O;C,B)$.
dolly33
07.09.2020 06:07
Abhaysingh2003 wrote: TelvCohl wrote: $$ X (X, V; Y, Z) = I (D, O;C,B )\ $$ Can anyone please explain this step? just note that $XX\perp ID, XV\perp IO, XY\perp IC, XZ\perp IB$. (as $X,Y,Z$ are the midpoints of the arcs) Therefore the ratios are preserved.
KevinCheng
08.01.2021 12:56
Let $A'$ be the reflection of $A$ in $OI$, we can prove that: (1) $A',D,S$ are collinear. (2) $H,T,X_{104}$ are collinear , where $X_{104}$ is the isogonal conjugate of the point at infinity of $OI$ WRT $\triangle ABC$. (3) $X_{104}A'\bot BC$. Then the problem can be easily proved.
SerdarBozdag
25.08.2021 19:05
Some extensions: $1)$ Reflection of $OI$ across $EF$ is parallel to $D'H$ where $D'$ is the reflection of $D$ across $EF$. $2)$ Intersection of $(THDS)$ and $(I)$ other than $D$ is on $FeH$.
Mahdi_Mashayekhi
07.03.2023 22:10
Let $\omega$ meet $AB,AC$ at $F,E$. Let $D'$ be reflection of $D$ across $EF$. Let $K,K'$ be midpoints of arcs $BC$ and $BAC$. Let $DD'$ and $AH$ meet $EF$ at $X$ and $Y$. Let $AI$ meet $BC$ at $Z$. It's well known that $T,D,K$ and $T,X,I$ are collinear. Lemma $: AD',IO,BC$ are concurrent.
Let $H'$ be orthocenter of $DEF$. it's well known that $H'$ lies on $IO$. Note that $AZ || DD'$ so we want to prove $\frac{AI}{IZ} = \frac{D'H'}{H'D}$. Let $DH'$ meet $\omega$ at $R$. we have $\frac{D'H'}{H'D} = \frac{DR}{DH} = \frac{\sin{\frac{A}{2}+B}}{\sin{\frac{A}{2}}} = \frac{CA}{CZ} = \frac{b}{\frac{ab}{b+c}} = \frac{b+c}{a} = \frac{AI}{IZ}$
Claim $: TXD$ and $DXH$ are similar.
Note that $(D,Y ; I,H) = -1$ and $\angle DXY = \angle 90$ so $\angle IXY = \angle HXY \implies \angle HXD = \angle TXD$. Also Note that $IX.IT = IF^2 = ID^2$ so $\angle XTD = \angle XDH$
Claim $: D'XT$ and $HXD'$ are similar.
$\frac{TX}{XD'} = \frac{TX}{XD} = \frac{XD}{XH} = \frac{XD'}{XH}$ and also $\angle TXD' = \angle D'XH$.
Claim $: TD'H$ and $TXD$ are similar.
$\frac{TD'}{D'H} = \frac{TX}{D'X} = \frac{TX}{DX}$ and $\angle TD'H = \angle TD'X + \angle XD'H = \angle 180 - \angle D'XH = \angle DXH = \angle TXD$.
Claim $: TAK'$ and $TXD$ are similar.
$\angle TAK' = \angle 180 - \angle TKK' = \angle 180 - \angle TDY = \angle 180 - \angle XDY - \angle XTD = \angle TXD$ and $\angle XTD = \angle YDX = \angle K'KA = \angle K'TA$
Now we have that $TD'H$ and $TAK'$ are similar so $\angle TAD' = \angle TK'H$ which implies that $H,K',S$ are collinear. Claim $: HAIS$ is cyclic.
$\angle HSA = \angle K'SA = \frac{B-C}{2} = \angle DIK = \angle AIH$
Now we have $\angle DHS = \angle IHS = \angle IAS = \angle KAS = \angle KTS = \angle DTS$ so $DTHS$ is cyclic as wanted. we're Done.
teomihai
07.03.2023 22:23
Mahdi_Mashayekhi wrote: Let $\omega$ meet $AB,AC$ at $F,E$. Let $D'$ be reflection of $D$ across $EF$. Let $K,K'$ be midpoints of arcs $BC$ and $BAC$. Let $DD'$ and $AH$ meet $EF$ at $X$ and $Y$. Let $AI$ meet $BC$ at $Z$. It's well known that $T,D,K$ and $T,X,I$ are collinear. Lemma $: AD',IO,BC$ are concurrent.
Let $H'$ be orthocenter of $DEF$. it's well known that $H'$ lies on $IO$. Note that $AZ || DD'$ so we want to prove $\frac{AI}{IZ} = \frac{D'H'}{H'D}$. Let $DH'$ meet $\omega$ at $R$. we have $\frac{D'H'}{H'D} = \frac{DR}{DH} = \frac{\sin{\frac{A}{2}+B}}{\sin{\frac{A}{2}}} = \frac{CA}{CZ} = \frac{b}{\frac{ab}{b+c}} = \frac{b+c}{a} = \frac{AI}{IZ}$
Claim $: TXD$ and $DXH$ are similar.
Note that $(D,Y ; I,H) = -1$ and $\angle DXY = \angle 90$ so $\angle IXY = \angle HXY \implies \angle HXD = \angle TXD$. Also Note that $IX.IT = IF^2 = ID^2$ so $\angle XTD = \angle XDH$
Claim $: D'XT$ and $HXD'$ are similar.
$\frac{TX}{XD'} = \frac{TX}{XD} = \frac{XD}{XH} = \frac{XD'}{XH}$ and also $\angle TXD' = \angle D'XH$.
Claim $: TD'H$ and $TXD$ are similar.
$\frac{TD'}{D'H} = \frac{TX}{D'X} = \frac{TX}{DX}$ and $\angle TD'H = \angle TD'X + \angle XD'H = \angle 180 - \angle D'XH = \angle DXH = \angle TXD$.
Claim $: TAK'$ and $TXD$ are similar.
$\angle TAK' = \angle 180 - \angle TKK' = \angle 180 - \angle TDY = \angle 180 - \angle XDY - \angle XTD = \angle TXD$ and $\angle XTD = \angle YDX = \angle K'KA = \angle K'TA$
Now we have that $TD'H$ and $TAK'$ are similar so $\angle TAD' = \angle TK'H$ which implies that $H,K',S$ are collinear. Claim $: HAIS$ is cyclic.
$\angle HSA = \angle K'SA = \frac{B-C}{2} = \angle DIK = \angle AIH$
Now we have $\angle DHS = \angle IHS = \angle IAS = \angle KAS = \angle KTS = \angle DTS$ so $DTHS$ is cyclic as wanted. we're Done. very nice!