Let $ \Omega_A, \Omega_B, \Omega_C $ be the circle with center $ A, B, C $ passing through $ H, $ respectively, then there is a circle $ \Gamma $ with center $ K $ that is tangent to $ \Omega_A, \Omega_B, \Omega_C $ internally. Notice that $ \odot (I) $ is the image of $ \Gamma $ under the inversion with center $ H $ that fixed $ \odot (ABC) $ we conclude that $ K $ lies on $ IH. $ $ \qquad \blacksquare $

We r e c i p r o c a t e around a circle with center $H$ that sends the ellipse $\mathcal E$ through $A,B,C$ with $H$ as one of the foci to a circle. Then $A,B,C$ get mapped to lines tangent to the circle (whose center lies on the major axis of $\mathcal E$) and these three lines form a triangle that is homothetic to $\triangle ABC$ with center $H$. Taking homothety around $H$ sending this triangle and its incircle back to $\triangle ABC$ and we are done.

Sharygin 2014 Quote: Let $H$ be the orthocenter of a triangle $ABC$. Given that $H$ lies on the incircle of $ABC$ , prove that three circles with centers $A, B, C$ and radii $AH, BH, CH$ have a common tangent.

Let $\triangle DEF$ be the intouch triangle of $\triangle ABC$. Let $P_A,P_B,P_C$ be points on $AH,BH,CH$, respectively, such that $P_AH=P_BH=P_CH=AH+AK$ and let $Q_A,Q_B,Q_C$ be points such that $\triangle P_AP_BP_C$ is the intouch triangle of $\triangle Q_AQ_BQ_C$. Note that $AH=AP_A,BH=BP_B,CH=CP_C,$ there is a homothety that sends $\triangle P_AP_BP_C\cup \{H\}$ to $\triangle DEF\cup \{I\}$ and $\triangle Q_AQ_BQ_C$ to $\triangle ABC$. We will prove that $K$ is the orthocenter of $\triangle Q_AQ_BQ_C$. Consider that $BK^2+CQ_A^2=BK^2+CP_C^2+P_CQ_A^2=CK^2+BP_B^2+P_BQ_A^2=CK^2+BQ_A^2$. Hence, line $Q_AK$ is perpendicular to line $BC$ and hence $K$ is the orthocenter of $\triangle Q_AQ_BQ_C$. By homothety from $\triangle P_AP_BP_C$ to $\triangle DEF$, $H,I,K$ are collinear as desired.

what's inversion Lemma: The locus of points $P$ satisfying $uAP^2+vBP^2+wCP^2=k$ is a line if $A$, $B$, $C$, $u$, $v$, $w$, and $k$ are fixed, $u+v+w=0$, and $u$, $v$, and $w$ are not all zero. Proof: We will prove this using Cartesian coordinates. Let $A=(a_1,a_2)$, $B=(b_1,b_2)$, and $C=(c_1,c_2)$. Then, if $P=(p_1,p_2)$, then $u(a_1-p_1)^2+u(a_2-p_2)^2+v(b_1-p_1)^2+v(b_2-p_2)^2+w(c_1-p_1)^2+w(c_2-p_2)^2=k$. When this is expanded, the coefficient of $p_1^2$ and $p_2^2$ is $u+v+w=0$, which means that this is a linear equation in $p_1$ and $p_2$. Therefore, it suffices to show that the left hand side is not constant. If it is constant, then setting $P$ to be the circumcenter of $ABC$, we get $k=0$, so $P=A$ implies $vc^2+wb^2=0$, which means that one of $v$ and $w$ is negative. Assume without loss of generality $v$ is negative and $w$ is positive. Then, similarly, one of $w$ and $u$ is negative, so $u$ is negative. However, this is a contradiction since $u$ and $v$ are both negative. It suffices to show that there exists $u$, $v$, and $w$ such that \begin{align*} u+v+w&=0\\ (AH^2-AI^2)u+(BH^2-BI^2)v+(CH^2-CI^2)w&=0\\ (AH^2-AK^2)u+(BH^2-BK^2)v+(CH^2-CK^2)w&=0, \end{align*} which is equivalent to showing $$\begin{vmatrix} 1&1&1\\ AH^2-AI^2&BH^2-BI^2&CH^2-CI^2\\ AH^2-AK^2&BH^2-BK^2&CH^2-CK^2\\ \end{vmatrix}=0.$$ Suppose that $AH+AK=BH+BK=CH+CK=x$. Then, $AH^2-AK^2=(AH-AK)x$, so this is equivalent to proving $$\begin{vmatrix} 1&1&1\\ AH^2-AI^2&BH^2-BI^2&CH^2-CI^2\\ AH-AK&BH-BK&CH-CK\\ \end{vmatrix}=\begin{vmatrix} 1&1&1\\ AH^2-AI^2&BH^2-BI^2&CH^2-CI^2\\ 2AH-x&2BH-x&2CH-x\\ \end{vmatrix}=0,$$or $$\begin{vmatrix} 1&1&1\\ AH^2-AI^2&BH^2-BI^2&CH^2-CI^2\\ AH&BH&CH\\ \end{vmatrix}=0.$$ Since $AH^2=4R^2-a^2$ and $AI^2=\frac{bc(b+c-a)}{a+b+c}$, this means that we have \begin{align*} &\begin{vmatrix} 1&1&1\\ AH^2-AI^2&BH^2-BI^2&CH^2-CI^2\\ AH&BH&CH\\ \end{vmatrix}\\ =&\begin{vmatrix} 1&1&1\\ 4R^2-a^2-AI^2&4R^2-b^2-BI^2&4R^2-c^2-CI^2\\ AH&BH&CH\\ \end{vmatrix}\\ =&\begin{vmatrix} 1&1&1\\ -a^2-\frac{bc(b+c-a)}{a+b+c}&-b^2-\frac{ca(c+a-b)}{a+b+c}&-c^2-\frac{ab(a+b-c)}{a+b+c}\\ AH&BH&CH\\ \end{vmatrix}\\ =&\frac1{a+b+c}\begin{vmatrix} 1&1&1\\ -a^2(a+b+c)-bc(b+c-a)&-b^2(a+b+c)-ca(c+a-b)&-c^2(a+b+c)-ab(a+b-c)\\ AH&BH&CH\\ \end{vmatrix}\\ =&-\frac1{a+b+c}\begin{vmatrix} 1&1&1\\ a^3+a^2b+a^2c+b^2c+bc^2-abc&b^2a+b^3+b^2c+c^2a+ca^2-abc&c^2a+c^2b+c^3+a^2b+ab^2-abc\\ AH&BH&CH\\ \end{vmatrix}\\ =&-\frac1{a+b+c}\begin{vmatrix} 1&1&1\\ a^3-ab^2-ac^2&b^3-ba^2-bc^2&c^3-ca^2-cb^2\\ AH&BH&CH\\ \end{vmatrix}\\ \end{align*}We have \begin{align*} AH&=\sqrt{4R^2-a^2}\\ &=\sqrt{\frac{a^2b^2c^2}{4[ABC]^2}-a^2}\\ &=\sqrt{\frac{a^2b^2c^2}{4s(s-a)(s-b)(s-c)}-a^2}\\ &=\sqrt{\frac{a^2b^2c^2}{-\frac14(a^4+b^4+c^4)+\frac12(a^2b^2+b^2c^2+c^2a^2)}-a^2}\\ &=\sqrt{\frac{a^6+a^2b^4+a^2c^4-2a^4b^2-2a^4c^2+2a^2b^2c^2}{16s(s-a)(s-b)(s-c)}}\\ &=\frac{\sqrt{a^2(a^4+b^4+c^4-2a^2b^2-2a^2c^2+2b^2c^2)}}{[ABC]}\\ &=\frac{a(b^2+c^2-a^2)}{[ABC]} \end{align*} if $ABC$ is acute. Therefore, it suffices to show that $$\begin{vmatrix} 1&1&1\\ a^3-ab^2-ac^2&b^3-ba^2-bc^2&c^3-ca^2-cb^2\\ ab^2+ac^2-a^3&ba^2+bc^2-b^3&ca^2+cb^2-c^3\\ \end{vmatrix}=0,$$which is true since adding the second and third rows gives $0$. If $ABC$ is obtuse, assume without loss of generality $\angle BAC>90^{\circ}$. Then, we claim that it is impossible for $H$ to be the focus of an ellipse. Since an ellipse is convex, this means that if $H$ is inside the ellipse and $B$ and $C$ are on the ellipse, then the only points inside $HBC$ that are on the ellipse are $B$ and $C$. However, this is a contradiction since $A$ is inside $HBC$. Therefore, this means that $H$, $I$, and $K$ are collinear.

stroller wrote: We r e c i p r o c a t e around a circle with center $H$ that sends the ellipse $\mathcal E$ through $A,B,C$ with $H$ as one of the foci to a circle. Then $A,B,C$ get mapped to lines tangent to the circle (whose center lies on the major axis of $\mathcal E$) and these three lines form a triangle that is homothetic to $\triangle ABC$ with center $H$. Taking homothety around $H$ sending this triangle and its incircle back to $\triangle ABC$ and we are done. I have a doubt in this solution. Can someone please clarify them. 1. I can't get why the center of $\odot(ABC)$ after duality lies on major axes of $\mathcal E$. 2. I might have misunderstood, but all Ellipses don't go to a circle after inversion through a focus. This can be verified on geogebra.

Amazing problem, probably my all time favorite geo Here is a solution requiring only "elementary" methods, that is things from chapters 1-3 of EGMO, which I think is different from the other ones posted above.

I was trying to design a problem using $\omega_A,\omega_B,\omega_C$ mentioned below and unexpectedly solved this Sol:-Let $I_AI_BI_C$ be the excentral triangle and $D,E,F$ be the feet of $I_A,I_B,I_C$ on $BC,CA,AB$ respectively.Let $I_AD=r_a;I_BE=r_b;I_CF=r_c$ and $\omega_A,\omega_B,\omega_C$ be the circles centered at $A,B,C$ having radius $r_a,r_b,r_c$ respectively.It is well known that $|\overrightarrow{HA}+\overrightarrow{I_AD}|=|\overrightarrow{HB}+\overrightarrow{I_BE}|=|\overrightarrow{HC}+\overrightarrow{I_CF}|$.Let So it isn't hard to prove that there exists a circle centered at $H$ internally tangent to $\omega_A,\omega_B,\omega_C$.Lets say this circle $(H)$ has radius $r$.Now we use well known facts from appolonius problem. Let $R$ be the radical center of $\omega_A,\omega_B,\omega_C$. The inversion centered at $R$ which fixes the $3$ circles maps $(H)$ to $(K')$.Let the radius of $(K')$ be $s$.Note that $AH+AK'=r+s$ and similarly $r+s=BH+BK'=CH+CK'$ so $K=K'$. Now note that $D$ is internal homothetic center of $\omega_B, \omega_C$ so the external homothetic center of $\omega_B, \omega_C$ is $A'=EF \cap BC$. Define $B',C'$ similarly. By appolonius problem $A'-B'-C'$ is radax of $(H) ,(K)$. So $HK \perp A'-B'-C'$.Also $A'-B'-C'$ is perspectrix of $DEF, ABC$.It is well known that it is $\perp HI$. Hence $H-I-K$ are collinear.

that is correct see #2 congratulations

asdf334 wrote: that is correct see #2 congratulations thanks asdf i worked hard on this ... 3 hours lol

Let $\omega_A$, $\omega_B$, and $\omega_C$ be the circles centered at $A$,$B$, and $C$, respectively, passing through point $H$. Let $\Omega_A$ be the circle centered at point $K$ internally tangent to $\omega_A$ at point $T_A$ and define $\Omega_B$ and $\Omega_C$ similarly. Denote $r_A$, $r_B$, and $r_C$ as the radii of $\Omega_A$, $\Omega_B$, and $\Omega_C$, respectively. We have \[r_A=KT_A=AK+AT_A=AK+AH,\]\[r_B=KT_B=BK+BT_B=BK+BH,\]\[r_C=KT_C=CK+CT_C=CK+CH.\] Thus, $r_A=r_B=r_C$, meaning that $\Omega_A$, $\Omega_B$, and $\Omega_C$ are the same circle; denote it by $\omega$ from this point on. Consider the inversion centered at $H$ that fixes $(ABC)$. It sends $\omega_A$, $\omega_B$, and $\omega_C$, to $\overline{AB}$, $\overline{BC}$, and $\overline{CA}$. Hence, $\omega$ is mapped to the incircle, so their centers map. We are done. $\square$

korncrazy wrote: Let $\triangle DEF$ be the intouch triangle of $\triangle ABC$. Let $P_A,P_B,P_C$ be points on $AH,BH,CH$, respectively, such that $P_AH=P_BH=P_CH=AH+AK$ and let $Q_A,Q_B,Q_C$ be points such that $\triangle P_AP_BP_C$ is the intouch triangle of $\triangle Q_AQ_BQ_C$. Note that $AH=AP_A,BH=BP_B,CH=CP_C,$ there is a homothety that sends $\triangle P_AP_BP_C\cup \{H\}$ to $\triangle DEF\cup \{I\}$ and $\triangle Q_AQ_BQ_C$ to $\triangle ABC$. We will prove that $K$ is the orthocenter of $\triangle Q_AQ_BQ_C$. Consider that $BK^2+CQ_A^2=BK^2+CP_C^2+P_CQ_A^2=CK^2+BP_B^2+P_BQ_A^2=CK^2+BQ_A^2$. Hence, line $Q_AK$ is perpendicular to line $BC$ and hence $K$ is the orthocenter of $\triangle Q_AQ_BQ_C$. By homothety from $\triangle P_AP_BP_C$ to $\triangle DEF$, $H,I,K$ are collinear as desired. As written, the part with $AH = AP_A$ etc. is wrong since $AH \neq AP_A$. When the solution defined $P_A$, it’s slightly ambiguous where or not $P_A$ lies closer to $A$ or $H$ on line $AH$, but, in both cases, $AH \neq AP_A$. Also, as a note, for most inversion solutions here (#2, #10, #13), a direct inversion at $H$ doesn’t fix $(ABC)$. You would need an inversion + reflection of all points about $H$ in order to actually fix $(ABC)$. With this minor fix, the points where $\Omega_A$ (using notation in post 2) and $(ABC)$ intersect map to $B$ and $C$ under this inversion+reflection.

Consider the circles centered at $A$, $B$, and $C$ respectively that pass thru $H$. Observe that the circle with center $K$ and radius $AH+AK$ is internally tangent to the three original circles. Invert at $H$ such that $A$ is mapped to the reflection of $H$ over $BC$. Now the three orginal circles are mapped to the sides of $\triangle ABC$. So The circle with center $K$ is mapped to the incircle. The collinearity follows.

This problem slaps. Draw circles $(A)$, $(B)$, $(C)$ centered at $A$, $B$, $C$ through $K$. Note the circle $\Gamma$ with center $H$ and radius $AH + AK$ is tangent to all three by design; let the points of tangency be $D$, $E$, $F$ respectively. Now draw the tangents from $D$, $E$, $F$ to $\Gamma_H$ and note we have a triangle homothetic to $ABC$ by design, say $A'B'C'$. By the Radical Axis theorem, the radical axis of $(B)$ and $(C)$ passes through $K$ and $A'$. It follows that $A'K \perp BC$ and since $BC \parallel B'C'$, $A'K \perp B'C'$. Therefore a homothety sending $A'B'C'$ to $ABC$ has center on line $HK$. But from here, note that the incenter of $\triangle A'B'C'$ is $H$, implying that $I$ lies on $HK$, as desired.