In triangle $ABC$, $90^{o}> \angle A> \angle B> \angle C$. Let the circumcenter and orthocenter of the triangle be $O$ and $H$. $OH$ intersects $BC$ at $T$ and the circumcenter of $(AHO)$ is $X$. Prove that the reflection of $H$ over $XT$ lies on the circumcircle of triangle $ABC$.
Problem
Source: Turkey National Mathematical Olympiad 2022 P5
Tags: geometry
BarisKoyuncu
23.12.2022 22:49
Let $K$ be the second intersection of $(AHO)$ and $(ABC)$. Notice that proving $|TH|=|TK|$ is sufficient. This can be shown by proving $\triangle KHA\sim\triangle KTO$, which can be proven by simple angle-chasing and the law of sines.
Let $K$ be the second intersection of $(AHO)$ and $(ABC)$. Notice that proving $|TH|=|TK|$ is sufficient. We have $\angle TKH\overset{?}{=}\angle THK=\angle OAK=\angle OKA$ so we need to prove that $\angle AKH=\angle OKT$. See that $\angle HAK=\angle TOK$ so it is sufficient to prove that $\triangle KHA\sim\triangle KTO$ or $\frac{|AH|}{|OT|}=\frac{|AK|}{|OK|}$. By the law of sines, we get $\frac{|AH|}{\cos{A}}=\frac{|AB|}{\sin{C}}=2R$ and $\frac{|OT|}{\cos{A}}=\frac{|BO|}{\sin{BTO}}$. Then, $\frac{|AH|}{|OT|}=2R\cdot\frac{\sin{BTO}}{|BO|}$. Also, $\frac{|AK|}{|OK|}=\frac{2R\cdot\sin{KCA}}{|BO|}$. Hence, we need to prove that $2R\cdot\frac{\sin{BTO}}{|BO|}=\frac{2R\cdot\sin{KCA}}{|BO|}\Leftrightarrow\sin{BTO}=\sin{KCA}$. Let $AH\cap BC=D$. As $\angle BTO=90^\circ-\angle DHT=90^\circ-\angle AKO=\frac 12\cdot\angle KOA=\angle KCA$, we are done.
FabrizioFelen
24.12.2022 18:55
Working in complex numbers, where $\odot (ABC)$ is the unit circle, let $a=1$, $b=b$ and $c=c$, then $h=1+b+c$ and $o=0$. Since $X$ is the circumcenter of triangle $AHO$, we have $$x=\frac{\begin{vmatrix} 1 & 1 & 1 \\ 0 & 0 & 1 \\ h & h \overline{h} & 1 \end{vmatrix}}{\begin{vmatrix} 1 & 1 & 1 \\ 0 & 0 & 1 \\ h & \overline{h} & 1 \end{vmatrix}}=\frac{h-h\overline{h}}{h-\overline{h}}=\frac{b+c+1}{1-bc}.$$Furthermore, since $T\in \overline{OH}$ and $T\in \overline{BC}$, we have $$\frac{t}{\overline{t}}=\frac{h}{\overline{h}}=\frac{bc(b+c+1)}{b+c+bc}\quad \quad ...\quad (\bigstar)$$and $$t+bc\overline{t}=b+c\quad \quad ...\quad (\blacklozenge).$$From $(\bigstar)$ and $(\blacklozenge)$ we get $$t=\frac{(b+c)(b+c+1)}{1+bc+2b+2c}.$$Let $x_1=\dfrac{1}{1-bc}$ and $t_1=\dfrac{b+c}{1+bc+2b+2c}$, note that $x=hx_1$, $t=ht_1$ and $\overline{t_1}=t_1$, $\overline{x_1}=1-x_1$. Let $W$ is the reflection of $H$ across $XT$, then $$w=\frac{(t-x)\overline{h}+\overline{t}x-\overline{x}t}{\overline{t}-\overline{x}}=h\cdot \frac{t_1-x_1 +\overline{t_1}x_1-\overline{x_1}t_1}{\overline{t_1}-\overline{x_1}}$$, replacing $\overline{t_1}=t_1$ and $\overline{x_1}=1-x_1$, we have $$w=h\cdot \frac{t_1-x_1 +t_1(x_1-\overline{x_1})}{{t_1}-\overline{x_1}}=h\cdot \frac{t_1-x_1 +t_1(2x_1-1)}{{t_1}-\overline{x_1}}=\frac{hx_1(2t_1-1)}{t_1-\overline{x_1}}$$$$\Longrightarrow w=\dfrac{(b+c+1)\cdot \dfrac{1}{1-bc}\cdot \dfrac{-(bc+1)}{1+bc+2b+2c}}{\dfrac{(bc+1)(b+c+bc)}{(1-bc)(1+bc+2b+2c)}}$$$$\Longrightarrow w=-\frac{b+c+1}{b+c+bc}.$$Finally, since $\overline{w}=-\dfrac{b+c+bc}{b+c+1}=\dfrac{1}{w}$ we get $W\in \odot(ABC)$.
GuvercinciHoca
25.12.2022 00:13
Let $D,E,F$ be the foot of the perpendiculars from $A,B,C$ respectively and let $H'$ be the reflection of $H$ . $HH'\cap TX=P$ we know that the $P$ is the mid-point of $HH'$ thus $H'\in (ABC)\iff P\in (DEF)$ Let $N,K,M,S$ be the mid-points of $HO,AO,AH,BC$ respectively. It's well-known that $M\in (DEF)$ and $OS=MH$ which implies $N\in MS$ and $NM=NS$ (since $OS\parallel MH$). On the other hand we have $XM\parallel BC$ thus if we let $Q=XN\cap BC$ we get $XN=NQ$. Which implies $\triangle XTQ$ is isosceles hence $\angle PTN=\angle DTN.$ Also we have $\angle NPH=\angle NXH=\angle OAH$ and $\angle NDH=\angle NMH=\angle OAH \implies NP=ND.$ $$ NP=ND \iff P\in (DEF) \iff H'\in (ABC) $$$\square$
TerrificChameleon
25.12.2022 00:19
To prove that the reflection of $H$ over $HT$ lies on the circumcircle of triangle $ABC$, we will use the fact that the circumcenter of a triangle is the reflection of the triangle's orthocenter over its circumcircle.
First, let's consider the triangle $AHO$. Since $H$ is the orthocenter of triangle $ABC$, it is also the orthocenter of triangle $AHO$. Since the circumcenter of triangle $AHO$ is $X$, the reflection of $H$ over the circumcircle of triangle $AHO$ is $X$.
Next, consider the reflection of $H$ over $HT$. Since the reflection of a point over a line is the same as the reflection of that point over the line's perpendicular bisector, the reflection of $H$ over $HT$ is the same as the reflection of $H$ over the perpendicular bisector of $HT$.
Let's call the intersection of the circumcircle of triangle $ABC$ and the perpendicular bisector of $HT$ $Y$. It follows that $X$ and $Y$ are reflections of each other over the perpendicular bisector of $HT$.
Since $X$ is the reflection of $H$ over the circumcircle of triangle $AHO$, it follows that $Y$ is the reflection of $H$ over the circumcircle of triangle $ABC$. Therefore, the reflection of $H$ over $HT$ lies on the circumcircle of triangle $ABC$.
Wikiliks
25.12.2022 01:03
TerrificChameleon wrote:
Since $H$ is the orthocenter of triangle $ABC$, it is also the orthocenter of triangle $AHO$. Since the circumcenter of triangle $AHO$ is $X$, the reflection of $H$ over the circumcircle of triangle $AHO$ is $X$.
Is this a typo, a joke, or am I missing something?
Wikiliks
25.12.2022 01:15
I gave the problem to ChatGPT. Its solution is rather similar to TerrificChameleon's, but I like it better:
Since the circumcenter of a triangle is the intersection of the perpendicular bisectors of the sides of the triangle, it follows that $X$ is the midpoint of $OH$. Furthermore, because $HT$ is an altitude of $\triangle AHT$, it follows that $HT$ is the perpendicular bisector of $AH$.
Since $HT$ is the perpendicular bisector of $AH$, it follows that the reflection of $H$ over $HT$ lies on the perpendicular bisector of $BC$. Thus, it remains to show that the reflection of $H$ over $HT$ lies on the circle with diameter $BC$.
Denote by $H'$ the reflection of $H$ over $HT$. Since $H'HTH$ is a parallelogram, it follows that $H'O = OH$. Furthermore, because $X$ is the midpoint of $OH$, it follows that $XH' = XH$. Therefore, the segment $H'X$ is an isogonal image of the segment $HX$.
Since $H$ is the orthocenter of triangle $ABC$, it follows that $HX$ is an altitude of $\triangle ABC$. Thus, the segment $H'X$ is also an altitude of $\triangle ABC$. Since $\triangle ABC$ is a right triangle with the right angle at $B$, it follows that the circumcircle of $\triangle ABC$ passes through the feet of the altitudes of $\triangle ABC$. Therefore, the reflection of $H$ over $HT$ lies on the circumcircle of triangle $ABC$.
We have thus shown that the reflection of $H$ over $HT$ lies on both the circumcircle of triangle $ABC$ and the perpendicular bisector of $BC$. Since these two curves intersect at exactly one point (which must be the reflection of $H$), it follows that the reflection of $H$ over $HT$ is indeed the unique point at which the circumcircle of triangle $ABC$ and the perpendicular bisector of $BC$ intersect. This completes the proof.
Wikiliks
25.12.2022 01:36
Official solution: Let $AH$ intersect $(ABC)$ at $S \neq A$. Let $(AHO)$ intersect $(ABC)$ at $M \neq A$. Since $OA = OM$, we have $\angle MHO = \angle OHS$. In other words, $M$ and $S$ are the reflections of each other across $OH$. Therefore, the perpendicular bisector of $MH$, the perpendicular bisector of $SH$ (which is $BC$), and the line $OH$ coincide. Thus, the perpendicular bisector of $MH$ is the line $TX$.
laikhanhhoang_3011
25.12.2022 06:58
$AH \cap (O)=H_1, \, (T;TH_1) \cap (O)=\left \{ H_1;H \right \}.$ Angel chase we get: $A,O,H,H'$ is cyclic and note that $TX$ is perpendicular bisector of $HH'.$ Q.E.D