Solution: Let $E$ be the intersection of $I_1I_3$ and $AM, E'$ be the intersection of the second tangent of $(I_1)$ through $C$ with $AM$. Since $BCE'A$ is a circumscribed quadrilateral then $CE'+AB=AE'+BC$. This means $AE'-CE'=AB-BC=AD-CD$. So $CE'AD$ is a circumscribed quadrilateral, or $CE'$ is tangent to $(I_3)$. Hence $E'$ is the intersection of $I_1I_3$ and $AM$, or $E'\equiv E.$ $\Rightarrow \angle I_1CI_3=\frac{1}{2}\angle BCD=\angle 180^o-\angle MI_2N=\angle I_1I_2I_3.$ Therefore $I_1,I_2,I_3,C$ are concyclic. But the refections of $C$ with respect to $I_2I_3, I_1I_2$ lie on g then applying Steiner's line we obtain the orthocenter of triangle $I_1I_2I_3$ lies on $g$.

At first look, I remembered Steiner line and tdl's post http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=247411&hilit=tangential+quadrangle. It is the case that C is a degenerate circle. Nice proof, livetolove212. Please point out if there is a mistake.

I am not familiar with Steiner's line, so I concluded in a different way; after showing that $C,I_1,I_2,I_3$ are concyclic, consider the second point $P_3$ where $BC$ intersects the circumcircle of $I_1I_2I_3$; then $\angle I_3P_3C+\angle I_1MP_3=\angle I_3I_2C+\angle I_2MC=\angle I_2NM+\angle I_2CM+\angle I_2MN=90^{\circ}$ because each of the three last angles is half one of the angles of $MCN$. Or $I_3P_3\perp I_1I_2$. But it is well known that the symmetric of the orthocenter $H$ with respect to one side $BC$ of a triangle $ABC$ is the second point where $AH$ intersects the circumcircle of $ABC$, or $P_3$ is the symmetric of $H$ with respect to $I_1I_2$, and $H$ is on $g$ which is the symmetric of $BC$ with respect to $I_1I_2$. The same reasoning may be performed using points $P_1,P_2$, the second points where $CD$, $CE$ meet the circumcircle of $I_1I_2I_3$.

Let $\alpha,\beta,\gamma,\delta$ denote the half-angles of $A,B,C,D$ so that $\alpha+\beta+\gamma+\delta=180^\circ$. Also let $a,b,c,d$ denote the tangents from $A,B,C,D$ to the incircle $(I)$ with inradius $r$. Now set $\theta=\angle{DNA}/2$. By a simple angle chase, $\angle{I_1I_2I_3}=\gamma$. Define $H_1=I_1H\cap AN$ and $H_3=I_3H\cap AN$, where $H$ is the orthocenter of $\triangle{I_1I_2I_3}$. Then $\angle{MI_1H_1}=\angle{NI_3H_3}=90^\circ-\gamma$, and we need to show that $H_1=H_3$, i.e. $NH_3-MH_1=NM$. By the law of sines on $\triangle{MI_1H_1}$, $\triangle{MI_1B}$, and $\triangle{MAB}$ (in this order), \begin{align*} MH_1 = \frac{MI_1\cos\gamma}{\cos\theta} = \frac{BM\sin\beta\cos\gamma}{\sin(\beta+\gamma-\theta)\cos\theta} &= \frac{AM\sin(2\beta+2\gamma-2\theta)\sin\beta\cos\gamma}{\sin(2\beta)\sin(\beta+\gamma-\theta)\cos\theta} \\ &= \frac{AM\cos(\beta+\gamma-\theta)\cos\gamma}{\cos\beta\cos\theta}. \end{align*}and similarly by the law of sines on $\triangle{NI_3H_3}$, $\triangle{NI_3D}$, and $\triangle{NAD}$, \begin{align*} NH_3 = \frac{NI_3\cos\gamma}{\cos(\theta-\gamma)} = \frac{DN\sin\delta\cos\gamma}{\sin(\theta+\delta)\cos(\theta-\gamma)} &= \frac{AN\sin(2\theta+2\delta)\sin\delta\cos\gamma}{\sin(2\delta)\sin(\theta+\delta)\cos(\theta-\gamma)} \\ &= \frac{AN\cos(\theta+\delta)\cos\gamma}{\cos\delta\cos(\theta-\gamma)}. \end{align*}Since $NM=AN-AM$, it remains to show that $NH_3-AN=MH_1-AM$, i.e. \[AN\left(\frac{\cos(\theta+\delta)\cos\gamma}{\cos\delta\cos(\theta-\gamma)}-1\right) = AM\left(\frac{\cos(\beta+\gamma-\theta)\cos\gamma}{\cos\beta\cos\theta}-1\right).\]But by the law of sines on $\triangle{AND}$ and $\triangle{AMB}$, \begin{align*} \frac{AN}{AM} = \frac{\frac{AD\sin(2\delta)}{\sin(2\theta)}}{\frac{AB\sin(2\beta)}{\sin(2\gamma-2\theta)}} = \frac{(a+d)\sin(2\delta)\sin(2\gamma-2\theta)}{(a+b)\sin(2\beta)\sin(2\theta)} &= \frac{r(\cot\alpha+\cot\delta)\sin(2\delta)\sin(2\gamma-2\theta)}{r(\cot\alpha+\cot\beta)\sin(2\beta)\sin(2\theta)} \\ &= \frac{\sin\beta\sin(\alpha+\delta)\sin(2\delta)\sin(2\gamma-2\theta)}{\sin\delta\sin(\alpha+\beta)\sin(2\beta)\sin(2\theta)} \\ & =\frac{\sin(\beta+\gamma)\cos\delta\sin(2\gamma-2\theta)}{\sin(\gamma+\delta)\cos\beta\sin(2\theta)}. \end{align*}Furthermore, by the product-to-sum and sum-to-product identities, \begin{align*} \frac{\left(\frac{\cos(\beta+\gamma-\theta)\cos\gamma}{\cos\beta\cos\theta}-1\right)}{\left(\frac{\cos(\theta+\delta)\cos\gamma}{\cos\delta\cos(\theta-\gamma)}-1\right)} &= \frac{\cos\delta\cos(\theta-\gamma)}{\cos\beta\cos\theta} \frac{\cos(\beta+\gamma-\theta)\cos\gamma-\cos\beta\cos\theta}{\cos(\theta+\delta)\cos\gamma-\cos\delta\cos(\theta-\gamma)} \\ &= \frac{\cos\delta\cos(\theta-\gamma)}{\cos\beta\cos\theta} \frac{\cos(\beta-\theta)+\cos(\beta+2\gamma-\theta)-\cos(\beta-\theta)-\cos(\beta+\theta)}{\cos(\theta-\gamma+\delta)+\cos(\theta+\gamma+\delta)-\cos(\theta-\gamma-\delta)-\cos(\theta-\gamma+\delta)} \\ &= \frac{\cos\delta\cos(\theta-\gamma)}{\cos\beta\cos\theta} \frac{\cos(\beta+2\gamma-\theta)-\cos(\beta+\theta)}{\cos(\theta+\gamma+\delta)-\cos(\theta-\gamma-\delta)} \\ &= \frac{\cos\delta\cos(\theta-\gamma)}{\cos\beta\cos\theta} \frac{\sin(\beta+\gamma)\sin(\gamma-\theta)}{\sin(\theta)\sin(\gamma+\delta)} \\ &= \frac{\sin(\beta+\gamma)\cos\delta\sin(2\gamma-2\theta)}{\sin(\gamma+\delta)\cos\beta\sin(2\theta)} \\ &= \frac{AN}{AM}, \end{align*}as desired.

yes,the key step is to prove $C,I_1,I_2,I_3$ are concyclic.

Other way, Let $I$ is the center of incircle of $ABCD$ then we have $\angle DCI=\angle ICB$ we must prove that: $\angle I_{1}CB=\angle ICI_{3}$(+) Then we must: $\frac{sin(ICI_{3})}{sin(I_{3}CD)}=\frac{sin(I_{1}CB)}{sin(I_{1}CI)}$ That is similar with: $\frac{II_{1}.CB}{I_{1}B.CI}=\frac{I_{3}D.CI}{I_{3}I.CD}$(*) On the other hand we have that: $\angle DAI_{3}=\angle IAI_{1}, \angle IAI_{3}=\angle I_{1}AB$ Hence (*) is similar with : $(\frac{CI}{AI})^{2}=\frac{CB.CD}{AB.AD}$ (**) And (**) is true. we have (+) then $\angle I_{1}II_{3}=\frac{\angle BCD}{2}$ we have QED.

Sorry for my not relevant question. Which program did you draw these figures? I really need one in my PC. Thanks in advance.

Geometry sketchpad

newsun wrote: Sorry for my not relevant question. Which program did you draw these figures? I really need one in my PC. Thanks in advance. you can also use Geogebra for drawing geometric figures.

goodar2006 wrote: newsun wrote: Sorry for my not relevant question. Which program did you draw these figures? I really need one in my PC. Thanks in advance. you can also use Geogebra for drawing geometric figures. Could you give me the link of this, goodar2006..

just push the rightmost green button on the top of the page you want to reply. (I tried to attach a picture but I couldn't, I don't know whats wrong with it....)

Let $\angle DAI_3 =\angle I_3AN = \alpha$, and $\angle NAI_1 = \angle I_1AB = \theta$. If $I$ is the incenter of the circle inscribed in $ABCD$, then $\angle IAI_3 = \theta$ as well. Let $T$ be the point on the line $AI_1$ such that $\triangle ABT \sim \triangle AII_3$, and $Q = I_3T \cap AN$. We claim that Q is the orthocenter of $\triangle I_1I_2I_3$. Since $\triangle ABT \sim \triangle AII_3$, $\triangle ABI \sim \triangle ATI_3$, so $BI$ rotated clockwise by $\theta$ is parallel to $I_3Q$. But $\angle BI_1I_2 = 90+\theta$, so $I_3Q \perp I_1I_2$. Also, since $\triangle ABT \sim \triangle AII_3$, $\angle ATI_3 =\angle ABI$, so $\triangle ATQ \sim \triangle ABI_1$, which in turn means that $\triangle AI_1Q \sim \triangle ABT \sim \triangle AII_3$. Thus $I_1Q$ rotated clockwise by $\alpha$ is parallel to $I_3I$. But $\angle DI_3I_2 = 90+\alpha$, so $I_1Q \perp I_3I_2$.

.

I'd just like to underline that the crux of this problem is the following lemma: if in a tangential quadrilateral $ABCD$, a circle $(I_1)$ touching angle $ABC$ and a circle $(I_3)$ touching angle $ADC$ are such that one of their inner common tangents passes through $A$, then the other inner common tangent passes through $C$.

One common tangent passes through $A$ if and only if $\angle BAI_1=\angle IAI_3$ and $\angle I_1AI=\angle I_3AD$. The condition for $C$ is similar. Thus, a straightforward proof of this fact using sine ratios is possible.

From the lemma, it becomes clear that $\angle I_1CI_3=\frac12 \angle BCD=\angle I_1I_2I_3$, whence $CI_3I_1I_2$ is cyclic, and thus the reflections of $g$ wrt the sides of $I_1I_2I_3$ -- namely $CD$, $BC$, and the second inner common tangent of $(I_1),(I_3)$ -- pass through a point on the circumcircle of $I_1I_2I_3$. By the Steiner line lemma, we are now done.

[asy][asy] unitsize(100); pen n_purple = rgb(0.7,0.4,1), n_blue = rgb(0,0.6,1), n_green = rgb(0,0.4,0), n_orange = rgb(1,0.4,0.1), n_red = rgb(1,0.2,0.4); pair pole(pair a, pair b) { return extension(a, rotate(90, a) * (0, 0), b, rotate(90, b) * (0, 0)); } pair S1, S2, S3, S4, A, B, C, D, M, N, I1, I2, I3, I; real theta = 60; S1 = dir(80); S2 = dir(170); S3 = dir(270); S4 = dir(20); A = pole(S2, S3); B = pole(S1, S2); C = pole(S4, S1); D = pole(S3, S4); M = extension(A, A + dir(theta), B, C); N = extension(A, A + dir(theta), C, D); I1 = incenter(A, B, M); I2 = incenter(M, N, C); I3 = incenter(N, D, A); I = origin; draw(C--N, gray(0.6)); draw(I1--I2, gray(0.8)); draw(N--I3, gray(0.8)); draw(B--I^^D--I, gray(0.8)); draw(A--B--C--D--cycle); draw(A--N); draw(incircle(A, B, M), n_green); draw(incircle(M, N, C), n_green); draw(incircle(N, D, A), n_green); draw(circumcircle(I1, I2, I3), gray(0.6) + dashed); draw(A--I1^^A--I3, n_red); draw(A--I, n_red); draw(C--I1^^C--I3, n_purple); draw(C--I, n_purple); draw(unitcircle, n_blue); dot(A^^B^^C^^D^^M^^N^^I1^^I2^^I3^^I); label("$A$", A, dir(dir(A - B) + dir(A - D))); label("$B$", B, dir(dir(B - C) + dir(B - A))); label("$C$", C, rotate(-90) * dir(N - C)); label("$D$", D, dir(dir(D - A) + dir(D - C))); label("$M$", M, dir(110)); label("$N$", N, dir(90)); label("$I_1$", I1, dir(190)); label("$I_2$", I2, dir(30)); label("$I_3$", I3, dir(270)); label("$I$", I, dir(90)); [/asy][/asy] Let $I$ be the incenter of quadrilateral $ABCD$. Lemma: If $M$ and $N$ are points on segments $BI$ and $DI$, then $\angle MAN = \tfrac{1}{2} \angle BAD$ iff $\angle MCN = \tfrac{1}{2} \angle BCD$. Proof: First, we must prove the identity $\tfrac{AB \cdot AD}{CB \cdot CD} = \tfrac{AI^2}{CI^2}$, which may be proven easily by complex bash. Now, we proceed to the proof of the lemma. Note that $\angle MCN = \tfrac{1}{2} \angle BCD$ iff $\angle BCM = \angle ICN$ iff $\angle MCI = \angle NCD$. Since $\angle BCD < 180^{\circ}$ we obtain $\angle BCI = \angle ICD < 90^{\circ}$ and thus, letting $\theta = \tfrac{1}{2} \angle BCD$, we see that $f: (0, \theta) \to \mathbb{R}$ defined by $f(x) = \tfrac{\sin (\theta - x)}{\sin x}$ is increasing and thus injective. This all means that $\angle MCN = \tfrac{1}{2} \angle BCD$ iff $\tfrac{\sin \angle BCM}{\sin \angle MCI} = \tfrac{\sin \angle ICN}{\sin \angle NCD}$, which happens iff, by Ratio Lemma, \[\frac{BM}{IM} \cdot \frac{IC}{BC} = \frac{IN}{DN} \cdot \frac{DC}{IC} \iff \frac{BM \cdot DN}{IM \cdot IN} = \frac{BC \cdot DC}{IC^2}.\]Similarly, $\angle MAN = \tfrac{1}{2} \angle BAD$ iff \[\frac{BM \cdot DN}{IM \cdot IN} = \frac{BA \cdot DA}{IA^2}.\]By our identity $\tfrac{AB \cdot AD}{CB \cdot CD} = \tfrac{IA^2}{IC^2}$, these are equivalent, proving the lemma. $\square$ Now, we pass to the main proof. First, since $\overline{AI_1}$ and $\overline{AI_3}$ are bisectors of $\angle (\overline{AB}, g)$ and $\angle (g, \overline{AD})$ respectively, we obtain $\angle I_1AI_3 = \tfrac{1}{2} \angle BAD$. Thus the lemma applies with $M = I_1$ and $N = I_3$, informing us that $\angle I_1CI_3 = \tfrac{1}{2} \angle BCD$. By an easy angle chase, $\angle I_1I_2I_3 = \tfrac{1}{2} \angle BCD$. Thus we conclude that $C$ belongs to $(I_1I_2I_3)$. Finally, consider the reflection of $C$ over $\overline{I_1I_2}$. It must lie on the reflection of line $AM$ over $\overline{I_1I_2}$, which is $g$. Thus, this reflection lies on $g$; similarly, the reflection of $C$ over $\overline{I_2I_3}$ is on $g$ too. However, since $C$ is on $(I_1I_2I_3)$, the two aforementioned reflections must be collinear with the orthocenter of $\triangle I_1I_2I_3$. Thus it lies on $g$, as desired.

To easy for g8 i think: Let $H$ be the point on ray $AN$ such that $AH \cdot AI=AI_1 \cdot AI_3$.Obviously there is spiral similarity between triangles $AI_1I$ , $AI_3H$.A simple angle chasing gives then $I_3H$ is perpendicular to $I_1I_2$.Similarly $I_1H$ is perpendicular to $I_2I_3$ so $H$ is the orthocenter of $I_1I_2I_3$.

April wrote: Let $ABCD$ be a circumscribed quadrilateral. Let $g$ be a line through $A$ which meets the segment $BC$ in $M$ and the line $CD$ in $N$. Denote by $I_1$, $I_2$ and $I_3$ the incenters of $\triangle ABM$, $\triangle MNC$ and $\triangle NDA$, respectively. Prove that the orthocenter of $\triangle I_1I_2I_3$ lies on $g$. Proposed by Nikolay Beluhov, Bulgaria Let $I$ be the incenter of $ABCD$. Claim. If $P$ and $Q$ are points on $\overline{BI}$ and $\overline{DI}$ then $\angle PAQ=\tfrac{1}{2}\angle DAB \iff \angle PCQ=\tfrac{1}{2}\angle DCB$. (Proof) For all $P$ define $S,T$ phantom points with $\angle PAS=\tfrac{1}{2}\angle DAB$ and $\angle PCT=\tfrac{1}{2}\angle DCB$ on line $\overline{DI}$. Note that $P \mapsto S$ and $P \mapsto T$ are projective maps. Now we check the claim for three choices of $P$. Cases $P=B$ and $P=I$ are obvious. If $R=\overline{BI} \cap \overline{AC}$ coincides with $I$ then the claim is trivial as both maps are reflections in line $\overline{AC}$. Denote $\angle BAC=x, \angle BCA=y, \angle DAB=2\alpha, \angle DCB=2\beta$; let $K$ be the point with $\angle KAB=\alpha+x$ and $\angle KCB=\beta+y$. Then we show $K$ lies on $\overline{DI}$. Observe that $K$ is equidistant to $\overline{DA}, \overline{DC}$ if $\sin \beta \sin(\alpha-x)=\sin \alpha \sin (\beta-y)$. However, $I$ is equidistant to $\overline{BA}, \overline{BC}$ yields $\sin \beta \sin(\alpha-x)=\sin \alpha \sin (\beta-y)$ and we're done. $\blacksquare$ Now $\angle I_1I_2I_3=180^{\circ}-\angle NI_2M=90^{\circ}-\tfrac{1}{2}\angle NCM=\tfrac{1}{2}\angle DCB$ so $\angle I_1CI_3=\tfrac{1}{2}\angle DCB=\angle I_1I_2I_3 \implies C \in \odot(I_1I_2I_3)$. Consequently, $g$ coincides with line joining reflections of $C$ in the sides of $\triangle I_1I_2I_3$. Thus, $g$ passes through the orthocenter of $\triangle I_1I_2I_3$, as desired. $\blacksquare$

I am splicing together ideas from all the solutions above and the official shortlist to get what I think is a short solution to this nice problem. Lemma: We have $\angle I_1 C I_3 = \frac{1}{2} \angle BCD$. Proof. The idea is to show the second common internal tangent passes through $C$. To that end, let $T$ be a point on $\overline{AM}$ with $\overline{CT}$ tangent to $(I_1)$. Then $AT - CT = AB - BC = AD - CD $ since (possibly concave) $ATCB$ and $ABCD$ are circumscribed. Thus quadrilateral $ADCT$ is circumscribed too, and this incircle is $(I_3)$. [asy][asy] size(5cm); pair W = dir(110); pair X = dir(170); pair Y = dir(270); pair Z = dir(10); pair A = 2*W*X/(W+X); pair D = 2*X*Y/(X+Y); pair C = 2*Y*Z/(Y+Z); pair B = 2*Z*W/(Z+W); draw(A--B--C--D--cycle, blue); draw(unitcircle, lightblue+dotted); pair M = 0.55*C+0.45*B; pair N = extension(A, M, C, D); pair I = origin; pair I_1 = incenter(A, B, M); pair I_3 = incenter(A, D, N); draw(A--M, blue); draw(I_1--C--I_3, red); pair T = extension(I_1, I_3, A, M); draw(C--(1.7*T-0.7*C), orange); draw(incircle(A, B, M), deepgreen+dashed); draw(incircle(A, N, D), deepgreen+dashed); dot("$A$", A, dir(100)); dot("$D$", D, dir(D)); dot("$C$", C, dir(310)); dot("$B$", B, dir(B)); dot(M, dir(45)); dot("$I$", I, dir(0)); dot("$I_1$", I_1, dir(135)); dot("$I_3$", I_3, dir(150)); dot("$T$", T, dir(T)); /* TSQ Source: W := dir 110 X := dir 170 Y := dir 270 Z := dir 10 A = 2*W*X/(W+X) R100 D = 2*X*Y/(X+Y) C = 2*Y*Z/(Y+Z) R310 B = 2*Z*W/(Z+W) A--B--C--D--cycle 0.1 lightcyan / blue unitcircle 0.1 heavycyan / lightblue dotted M .= 0.55*C+0.45*B R45 N := extension A M C D I = origin R0 I_1 = incenter A B M R135 I_3 = incenter A D N R150 A--M blue I_1--C--I_3 red T = extension I_1 I_3 A M C--(1.7*T-0.7*C) orange incircle A B M 0.1 lightgreen / deepgreen dashed incircle A N D 0.1 lightgreen / deepgreen dashed */ [/asy][/asy] Now the lemma is proven since $\overline{CI_3}$ bisects $\angle TCD$ and $\overline{CI_1}$ bisects $\angle BCT$. $\blacksquare$ Back to the problem. Note $M \in \overline{I_1 I_2}$ and $N \in \overline{I_1 I_3}$. Claim: Quadrilateral $C$, $I_1$, $I_2$, $I_3$ is cyclic. Proof. By angle chasing, $\angle I_1 I_2 I_3 = 180^{\circ} - \angle M I_2 N = 90^{\circ} - \frac{1}{2} \angle MCN = \frac{1}{2} \angle BCD$. Together with the lemma this proves the result. $\blacksquare$ [asy][asy] size(14cm); pair W = dir(110); pair X = dir(170); pair Y = dir(270); pair Z = dir(10); pair A = 2*W*X/(W+X); pair D = 2*X*Y/(X+Y); pair C = 2*Y*Z/(Y+Z); pair B = 2*Z*W/(Z+W); draw(A--B--C--D--cycle, blue); draw(unitcircle, lightblue+dotted); pair M = 0.65*C+0.35*B; pair N = extension(A, M, C, D); draw(incircle(A, B, M), deepgreen+dashed); draw(incircle(C, N, M), deepgreen+dashed); draw(incircle(A, N, D), deepgreen+dashed); draw(A--N--C, blue); pair I = origin; pair I_1 = incenter(A, B, M); pair I_2 = incenter(C, M, N); pair I_3 = incenter(A, D, N); draw(I_1--I_3--I_2--cycle, deepgreen); draw(I_2--N, deepgreen); draw(circumcircle(I_1, I_3, I_2), red); pair H = orthocenter(I_1, I_2, I_3); pair S_1 = foot(C, I_2, I_3); pair S_2 = foot(C, I_3, I_1); pair S_3 = foot(C, I_1, I_2); pair T_1 = 2*S_1-C; pair T_2 = 2*S_2-C; pair T_3 = 2*S_3-C; pair S_H = midpoint(C--H); draw(S_3--S_2, orange); draw(T_3--C--T_1, orange+dashed); draw(H--C--T_2, orange+dashed); draw(T_2--A, blue); draw(I_2--S_3, deepgreen); draw(I_1--C--I_3, blue); dot("$A$", A, dir(100)); dot("$D$", D, dir(D)); dot("$C$", C, dir(310)); dot("$B$", B, dir(B)); dot("$M$", M, dir(45)); dot("$N$", N, dir(N)); dot("$I$", I, dir(0)); dot("$I_1$", I_1, dir(135)); dot("$I_2$", I_2, dir(100)); dot("$I_3$", I_3, dir(150)); dot("$H$", H, dir(H)); dot(S_1); dot(S_2); dot(S_3); dot(T_1); dot(T_2); dot(T_3); dot(S_H); /* TSQ Source: !size(14cm); W := dir 110 X := dir 170 Y := dir 270 Z := dir 10 A = 2*W*X/(W+X) R100 D = 2*X*Y/(X+Y) C = 2*Y*Z/(Y+Z) R310 B = 2*Z*W/(Z+W) A--B--C--D--cycle 0.1 lightcyan / blue unitcircle 0.1 heavycyan / lightblue dotted M = 0.65*C+0.35*B R45 N = extension A M C D incircle A B M 0.1 lightgreen / deepgreen dashed incircle C N M 0.1 lightgreen / deepgreen dashed incircle A N D 0.1 lightgreen / deepgreen dashed A--N--C blue I = origin R0 I_1 = incenter A B M R135 I_2 = incenter C M N R100 I_3 = incenter A D N R150 I_1--I_3--I_2--cycle 0.1 yellow / deepgreen I_2--N deepgreen circumcircle I_1 I_3 I_2 0.1 orange / red H = orthocenter I_1 I_2 I_3 S_1 .= foot C I_2 I_3 S_2 .= foot C I_3 I_1 S_3 .= foot C I_1 I_2 T_1 .= 2*S_1-C T_2 .= 2*S_2-C T_3 .= 2*S_3-C S_H .= midpoint C--H S_3--S_2 orange T_3--C--T_1 orange dashed H--C--T_2 orange dashed T_2--A blue I_2--S_3 deepgreen I_1--C--I_3 blue */ [/asy][/asy] To finish, consider the Simson line $\ell$ of $C$ with respect to $\triangle I_1 I_2 I_3$. It should bisect $\overline{CH}$, where $H$ is the desired orthocenter. The reflection of $C$ across $\overline{I_2 I_3}$ and $\overline{I_1 I_2}$ is known to lie on $\overline{AMN}$, hence $\overline{AMN}$ is the image of $\ell$ under a homothety with ratio $2$. Thus $\overline{AMN}$ passes through $H$.

Essentially identical to what you can find above but wtv. With eisirrational (in the background trying to length bash) and Th3Numb3rThr33. [asy][asy] size(6cm); defaultpen(fontsize(10pt)); pair I,WW,X,Y,Z,A,B,C,D,M,NN,I1,I2,I3,H,EE; I=(0,0); WW=dir(100); X=dir(170); Y=dir(300); Z=dir(30); A=2*WW*X/(WW+X); B=2*X*Y/(X+Y); C=2*Y*Z/(Y+Z); D=2*Z*WW/(Z+WW); M=(2*B+7*C)/9; NN=extension(A,M,C,D); I1=incenter(A,B,M); I2=incenter(M,NN,C); I3=incenter(NN,D,A); H=orthocenter(I1,I2,I3); EE=extension(I1,I3,A,M); draw(incircle(A,B,M),gray); draw(incircle(M,NN,C),gray); draw(incircle(NN,D,A),gray); draw(C--reflect(I1,I3)*foot(I1,A,M)); draw(circumcircle(I1,I2,I3)); draw(unitcircle); draw(A--B--C--D--A--NN--C); dot("$A$",A,NW); dot("$B$",B,S); dot("$C$",C,dir(30)); dot("$D$",D,NE); dot("$M$",M,S); dot("$N$",NN,SE); dot("$I_1$",I1,W); dot("$I_2$",I2,SE); dot("$I_3$",I3,N); dot("$E$",EE,dir(240)); [/asy][/asy] Let $\omega$, $\omega_1$, $\omega_2$, $\omega_3$ be the incircles of $ABCD$, $\triangle ABM$, $\triangle MNC$, $\triangle NDA$ respectively. Claim 1. The other common internal tangent of $\omega_1$ and $\omega_3$ passes through $C$. Proof. Let the second tangent from $C$ to $\omega_1$ intersect $g$ at $E$. Then $ABCD$ and $ABCE$ are tangential, so by Pitot's theorem, \begin{align*} AB+CD&=BC+DA,\\ AB+CE&=BC+EA. \end{align*}Subtracting and rearranging, \[CD+EA=DA+CE,\]so $ADCE$ is tangential, and $E$ is the insimilicenter of $\omega_1$, and $\omega_3$, as desired. $\blacksquare$ Claim 2. $C$ lies on $(I_1I_2I_3)$. Proof. Note that \[\angle I_1I_2I_3=\angle I_2NM+\measuredangle NMI_2=\frac{\angle CNM+\angle NMC}2=\frac{\angle BCD}2,\]and furthermore \[\angle I_1CI_2=\angle I_1CE+\angle ECI_3=\frac{\angle BCE+\angle ECD}2=\frac{\angle BCD}2,\]as desired. $\blacksquare$ Finally $g$ is the Steiner line of $C$ with respect to $\triangle I_1I_2I_3$, so the orthocenter of $\triangle I_1I_2I_3$ lies on $g$, and we are done.

Beautiful problem. Here goes my solution. [asy][asy] size(13cm); pair X,Y,Z,W,A,B,C,D; X=dir(110); Y=dir(170); Z=dir(270); W=dir(10); pair I=(0,0); A=2*X*Y/(X+Y); D=2*Y*Z/(Y+Z); C=2*Z*W/(Z+W); B=2*W*X/(W+X); draw(unitcircle,red+dotted); draw(A--B--C--D--A,green); pair M=(3*B+6*C)/9; pair N=extension(A,M,C,D); draw(A--M--N,blue); draw(C--D--N,green); pair I1=incenter(A,B,M); pair I3=incenter(A,D,N); pair I2=incenter(M,C,N); pair K=foot(C,I1,I3); pair C4=2*K-C; draw(incircle(A,B,M),red); draw(incircle(A,D,N),red); draw(incircle(M,C,N),red); draw(I1--I2--I3--I1,purple); draw(I2--I3--N,purple); draw(C--I,cyan); draw(I--D,orange); draw(I--B,orange); draw(circumcircle(I1,I2,I3),dashed+purple+white); draw(A--C4,blue); draw(C4--I1,blue+white+white); draw(C4--I3,blue+white+white); draw(A--I1,blue+white+white); draw(A--I3,blue+white+white); draw(C--I1,blue+white+white); draw(C--I2,blue+white+white); draw(C--I3,blue+white+white); dot("$A$",A,NW); dot("$B$",B,NE); dot("$C$",C,S); dot("$D$",D,SW); dot("$M$",M,NE); dot("$N$",N,S); dot("$I$",I,NW); dot("$I_1$",I1,NW); dot("$I_2$",I2,NW); dot("$I_3$",I3,NW); dot("$C'$",C4,NW); [/asy][/asy] Proof: We present the following claim which is the key to this problem. Claim : $C \in \odot(I_1I_2I_3)$. Proof: We present a proof using mOvInG pOiNtS. We firstly claim that $\angle I_1CI_3=\frac{\angle ACD}{2}$. Let $f$ be the map which sends every point $P\in\overline{ID}$ to a point $Q\in\overline{IB}$ such that $\angle PAQ=\frac{\angle BAD}{2}$ and let $g$ be a map that sends every point $P \in \overline{BI}$ to $Q \in \overline{DI}$ such that $\angle PCQ=\frac{\angle BCD}{2}$. Notice that $f: P \mapsto BI \mapsto DI \mapsto Q$ is projective since $BI \mapsto DI$ is just a rotation by a fixed angle $\frac{\angle BAD}{2}$. By a similiar reasoning we have that $g$ is projective. So to prove that $f \equiv g$ we need to check for $3$ cases. $P=B$. Proof: Notice that in this case $Q=I$ and now this is trivial $\square$. $P=I$ Proof: Notice that in this case $Q=D$ and now this is trivial $\square$ $P$ is the incentre of $\triangle{ABC}$ Proof: Notice that in this case $Q$ becomes the incentre of $\triangle{ACD}$ and the problem becomes trivial again $\square$ Now notice that $\angle I_3AI_1=\frac{\angle BAD}{2}$ so we have that $\angle I_3CI_1=\frac{\angle BCD}{2}$. But now notice that $$\angle I_3I_2I_1=180^\circ-\angle MI_2N=180^\circ-(90^\circ+\frac{\angle MCN}{2})=90^\circ-\frac{\angle MCN}{2}=\frac{\angle BCD}{2}$$which gives the desired result $\blacksquare$ Now back to the main problem. Let $C'$ be the replection of $C$ over $\overline{I_1I_3}$. Then notice that $$\angle I_1C'C=\angle I_1CC'=90^\circ-\angle I_3I_1C=90^\circ-\angle I_3I_2C=\angle CMI_2$$$\implies$ $(C'I_1MC)$ is cyclic. So we have that $$\angle I_1MC'=\angle I_1CC'=\angle I_1C'C=\angle CMI_2=\angle NMI_2$$$\implies$ $C' \in \overline{MN}$. But it's clearly visible that the reflection of $C$ in lines $\overline{I_1I_2},\overline{I_2I_3}$ are also in $\overline{MN}$ $\implies$ $\overline{MN}$ is the stiener line of $C$ w.r.t $\triangle{I_1I_2I_3}$ but it's well known that the stiener line passes through the orthocentre with which we are done $\blacksquare$

Nice problem ! (Though on the easier side for a G8) .

Here's a length-based solution with Daniel Hong and Ryan Li. Let $I$ denote the incenter of $ABCD$, and let $X = \overline{CI}\cap \overline{I_1I_2}$ and $Y = \overline{CI}\cap \overline{I_2I_3}$. Claim: $X$ and $Y$ are the $M$- and $N$-excenters in $\triangle CMN$. Proof. This is clear since $\overline{CI}$ is the external angle bisector of $\angle MCN$, while $\overline{I_1I_2}$ and $\overline{I_2I_3}$ are the $M$- and $N$-internal angle bisectors. $\blacksquare$ [asy][asy] size(250); defaultpen(fontsize(10pt)); pair A, B, C, D, I, M, N, I1, I2, I3, X, Y, H, P, Q, R, S, U, V, W; I = (0,0); P = dir(110); Q = dir(20); R = dir(270); S = dir(170); A = extension(P, rotate(90, P)*I, S, rotate(90, S)*I); B = extension(Q, rotate(90, Q)*I, P, rotate(90, P)*I); C = extension(R, rotate(90, R)*I, Q, rotate(90, Q)*I); D = extension(S, rotate(90, S)*I, R, rotate(90, R)*I); M = (6*C+4*B)/10; N = extension(A, M, C, D); I1 = incenter(A, B, M); I2 = incenter(C, M, N); I3 = incenter(A, N, D); H = orthocenter(I1, I2, I3); X = extension(I1, I2, C, I); Y = extension(I2, I3, C, I); U = foot(I1, B, M); V = foot(I3, C, D); W = foot(Y, C, D); draw(A--B--C--D--cycle^^A--N--C, orange); draw(Circle(I, 1), red); draw(B--I^^D--I^^X--I^^N--I3^^I1--X, dashed+lightblue); draw(incircle(A, B, M), heavygreen+dotted); draw(incircle(C, M, N), heavygreen+dotted); draw(incircle(A, D, N), heavygreen+dotted); draw(I1--H--I3, heavycyan); draw(X--N^^Y--M, heavycyan); draw(I1--U^^I3--V^^Y--W, purple+dotted); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(250)); dot("$D$", D, dir(D)); dot("$M$", M, dir(30)); dot("$N$", N, dir(N)); dot("$I$", I, dir(270)); dot("$I_1$", I1, dir(130)); dot("$I_2$", I2, dir(40)); dot("$I_3$", I3, dir(100)); dot("$X$", X, dir(300)); dot("$Y$", Y, dir(90)); dot("$H$", H, dir(270)); dot("$U$", U, dir(20)); dot("$V$", V, dir(270)); dot("$W$", W, dir(270)); [/asy][/asy] Let the altitude from $I_1$ to $\overline{I_2I_3}$ hit $\overline{AMN}$ at $H_1$ and the altitude from $I_3$ to $\overline{I_1I_2}$ hit $\overline{AMN}$ at $H_3$. We will show $MH_1 = MH_3$. The key is that $\overline{NX}\parallel \overline{H_1I_1}$ and $\overline{MY}\parallel \overline{H_3I_3}$, and so by similar triangles we have $$MH_1 = \frac{MN\cdot MI_1}{MX} \quad \text{and} \quad MH_3 = \frac{MN\cdot YI_3}{YN}.$$Hence it suffices to show that $$\frac{MI_1}{YI_3} = \frac{MX}{YN} = \frac{XY\cos \angle N/2}{XY\cos \angle M/2} \iff MI_1\cos \angle M/2 = YI_3\cos \angle N/2.$$Let $U$, $V$, $W$ be the feet from $I_1$ to $\overline{BC}$, $I_3$ to $\overline{CD}$, and $Y$ to $\overline{CD}$. Then we have \begin{align*} YI_3\cos \angle N/2 &= VW = NV - NW \\ &= \tfrac{1}{2}(NA+ND-AD) - \tfrac{1}{2}(NM+MC+CN) \\ &= \tfrac{1}{2}(AM+CD-AD-MC) \end{align*}and $$MI_1\cos M/2 = UM = \tfrac{1}{2}(MA+MB-AB).$$Thus we just need to show \begin{align*} MI_1\cos \angle M/2 = YI_3\cos \angle N/2 &\iff AM+CD-AD-MC = MA+MB-AB \\ &\iff AB+CD = AD+BC, \end{align*}which is just Pitot's theorem! This completes the proof. Remark: The real work is in constructing $X$ and $Y$, since they allow you to access $I_1H_1$ and $I_3H_3$ with similar triangles. The length calculation that follows is fairly straightforward.

[asy][asy] size(10cm); defaultpen(fontsize(10pt)); defaultpen(linewidth(0.4)); dotfactor *= 1.5; pair P = dir(110), Q = dir(170), R = dir(270), S = dir(25), O = origin, A = extension(P,P+dir(Q--foot(Q,P,O)),Q,Q+dir(R--foot(R,Q,O))), D = extension(Q,Q+dir(R--foot(R,Q,O)),R,R+dir(S--foot(S,R,O))), C = extension(R,R+dir(S--foot(S,R,O)),S,S+dir(P--foot(P,S,O))), B = extension(S,S+dir(P--foot(P,S,O)),P,P+dir(Q--foot(Q,P,O))), N = C+dir(D--C)*abs(C-D)*0.4, M = extension(A,N,B,C), I1 = incenter(A,B,M), I2 = incenter(M,N,C), I3 = incenter(A,N,D), H = orthocenter(I1,I2,I3), P1 = extension(I1,H,D,C), P3 = extension(I3,H,B,C), S = extension(I1,I3,A,M); draw(A--B--C--D--A^^C--N, orange); draw(incircle(A,B,M)^^incircle(A,D,N)^^unitcircle, cyan); draw(A--N, red); draw(I1--I2--I3--I1^^I2--N, blue); draw(circumcircle(I1,I2,I3), purple); draw(I1--P1^^I3--P3, heavygreen); draw(I1--C--I3, magenta); draw(C--extension(C,S,A,B), red+dashed); dot("$A$", A, dir(135)); dot("$B$", B, dir(60)); dot("$C$", C, dir(300)); dot("$D$", D, dir(225)); dot("$M$", M, dir(45)); dot("$N$", N, dir(345)); dot("$H$", H, dir(0)); dot("$I_1$", I1, dir(135)); dot("$I_2$", I2, dir(30)); dot("$I_3$", I3, dir(150)); dot("$P_1$", P1, dir(240)); dot("$P_3$", P3, dir(60)); dot("$S$", S, dir(90)); [/asy][/asy] Denote $\omega_1$ to be the incircle of $\triangle ABM$ and $\omega_3$ to be the incircle of $\triangle ADN$. Key Claim: $CI_1I_2I_3$ is cyclic. Proof. Note that $$\angle I_1I_2I_3 = \angle I_2MN + \angle I_2NM = \tfrac{1}{2} \angle BCD,$$so it suffices to show that $\angle I_1CI_3 = \tfrac{1}{2} \angle BCD$. Suppose that the second tangent from $C$ to $\omega_3$ intersects $l$ at $S$. By definition $ASCD$ is circumscribed, so $$AS-CS = AD-CD = AB - CB,$$implying $ASCB$ is circumscribed as well, so $\overline{CS}$ is a common tangent to $\omega_1$ and $\omega_3$. This implies $$\angle I_1CI_3 = \angle I_1CS + \angle I_3CS = \tfrac{1}{2}(\angle BCS + \angle DCS) = \tfrac 12 \angle BCD,$$which is what we wanted. $\square$ Suppose that $(CI_1I_2I_3)$ intersects $\overline{CD}$ again at $P_1$ and $\overline{CB}$ again at $P_3$. Note that $$\angle I_2I_1P_1 = 180^\circ - \angle I_2CP_1 = \tfrac 12 \angle MCN = 90^\circ - \tfrac 12 (\angle CMN + \angle MNC) = 90^\circ - \angle I_1I_2I_3,$$so $\overline{I_1P_1} \perp \overline{I_2I_3}$. Similarly $\overline{I_3P_3} \perp \overline{I_2I_1}$, so $H = \overline{I_1P_1} \cap \overline{I_3P_3}$ is the orthocenter of $\triangle I_1I_2I_3$. Finally, Pascal's on $I_1P_1CP_3I_3I_2$ gives $H \in l$ as desired. $\blacksquare$

April wrote: Let $ABCD$ be a circumscribed quadrilateral. Let $g$ be a line through $A$ which meets the segment $BC$ in $M$ and the line $CD$ in $N$. Denote by $I_1$, $I_2$ and $I_3$ the incenters of $\triangle ABM$, $\triangle MNC$ and $\triangle NDA$, respectively. Prove that the orthocenter of $\triangle I_1I_2I_3$ lies on $g$. Proposed by Nikolay Beluhov, Bulgaria

Note that $M\in\overline{I_1 I_2}$ and $N\in\overline{I_2 I_3}$. Let $E$ be the $N$-Excenter of $\triangle{MNC}$. Let the perpendicular from $I_1$ to $I_2I_3$ meet $g,\overline{I_2I_3}$ at $H,F$ respectively. Let the projections from $I_3,I_1,E$ to $g$ be $P,Q,R$ respectively. Finally let $S$ be the point on the extension of $\overline{ER}$ in the side of $R$ such that $QI_1=RS$. It is well known that \begin{align*} 2AP&=AD+AN-DN\\ 2AQ&=AB+AM-BM\\ 2RM&=MC+CN-MN\\ \end{align*}Hence \begin{align*} &2\left(AQ-AP-RM\right)\\ &=({\color{red} AB}+{\color{blue} AM}-{\color{green} BM})-({\color{red} AD}+{\color{blue} AN}-{\color{magenta} DN})-({\color{green} MC}+{\color{magenta} CN}-{\color{blue} MN})\\ &=(AB-AD)+(AM+MN-AN)-(BM+MC)+(DN-CN)\\ &=AB-AD-BC+DC=0 \end{align*}where in the last equality we use the fact that $ABCD$ is circumscribed. Hence $A,P,Q$ lie on $g$ in this order and $PQ=RM$. It follows that $\triangle I_1QP,\triangle SRM$ are reflections of each other through the perpendicular bisector of $PM$. Hence $I_1SMP$ is an isosceles trapezoid, therefore a cyclic quadrilateral. We have \[\angle I_1 MQ=90^{\circ}-\angle RME=\angle MER\]Hence $\triangle I_1MQ\sim\triangle MER$. Therefore \[\frac{QI_1}{QM}=\frac{RM}{RE}\implies ER\cdot QI_1=QM\cdot RM\implies ER\cdot RS=PR\cdot RM\]so $P,S,M,E$ are concyclic. In total $P,I_1,S,M,E$ are concyclic. We have $I_1M\perp ME$ and $I_1F\perp FE$, hence $I_1,M,E,F$ are concyclic. Thus $P,I_1,S,M,E,F$ are concyclic. We have $HF\perp I_3F$ and $I_3P\perp PH$, so $P,H,F,I_3$ are concyclic. Note that \[\angle I_2I_3H=\angle FPM=\angle FI_1I_2=90^{\circ}-\angle I_1I_2F\]Hence $I_3H\perp I_1I_2$, so $H$ is the orthocenter of $I_1I_2I_3$. $\blacksquare$

The key of the proof is just an angle condition and a similarity. Let $X, Y$ be the $N$-excenter and $M$-excenter of $\triangle CNM$. It's easy to verify that $X = NI_2 \cap CI$ and $Y = MI_2 \cap CI$. Let $T$ be the orthocenter of $\triangle I_1I_2I_3$, then $TI_a \parallel NY, TI_3 \parallel MX$. Hence to show $T$ lies on $MN$, it suffices to show $MI_1 / I_1Y = XI_3 / I_3N$. Lemma.Let $I$ be the incenter of quadrilateral $ABCD$, then $\angle BAI_1 = \angle IAI_3$, and also $\angle BCI_1 = \angle ICI_3$. proof Of course it can be done by trigonometric functions bashing, thus here provides a proof using isogonal conjugate. First we have $\angle BAI_1 = \tfrac{1}{2}\angle BAM = \tfrac{1}{2}(\angle BAC - \angle MAC) = \angle IAD - \angle I_3AD = \angle IAI_3$. Note that $\angle BIA + \angle CII_3 = \angle BIA + \angle CID = 180^\circ$ since $I$ is the incenter of $ABCD$. Hence $I$ has an isogonal conjugate WRT $ABCI_3$. Combines that $AI_1, AI$ are isogonal WRT $\angle BAI_3$ and $BI, BI_1$ are isogonal WRT $\angle ABC$, we conclude that $I_1$ is the isogonal conjugate of $I$ WRT quadrilateral $ABCI_3$, so $\angle BCI_1 = \angle ICI_3$. By the Lemma above, we conclude that $\triangle CMY \cup \{I_1\} \sim \triangle CXN \cup \{I_3\}$, hence $MI_1 / I_1Y = XI_3 / I_3N$, as desired.

mela_20-15 wrote: To easy for g8 i think: Let $H$ be the point on ray $AN$ such that $AH \cdot AI=AI_1 \cdot AI_3$.Obviously there is spiral similarity between triangles $AI_1I$ , $AI_3H$.A simple angle chasing gives then $I_3H$ is perpendicular to $I_1I_2$.Similarly $I_1H$ is perpendicular to $I_2I_3$ so $H$ is the orthocenter of $I_1I_2I_3$. Dark magic.

Let $I$ be incenter of $ABCD.$ Obviously $I_1\in BI,I_3\in DI,I_2=MI_1\cap NI_3.$ Lemma: if $X\in BI,Y\in CI$ satisfy $\angle BAD=2\angle XAY,$ then also $\angle BCD=2\angle XCY.$ Proof: fix $ABCD$ and move $X$ along $BI.$ Since $\angle XAY$ is fixed, mapping $X\mapsto Y$ of $BI$ onto $CI$ is projective. Thus it's suffice to prove lemma for $3$ positions of $X;$ trivial enough when $X$ coincide with $B,I$ and incenter of $\triangle ABC \text{ } \Box$ Clearly $\angle I_1AI_3=\frac{1}{2}\angle BAD,$ so by lemma $$\angle I_1CI_3=\frac{1}{2}\angle BCD=\angle I_2MN+\angle MNI_2=\angle I_1I_2I_3\implies C\in \odot (I_1I_2I_3).$$But $g$ passes through reflections of $C$ over $I_1I_2$ and $I_2I_3,$ i.e. it's the Steiner line of $C$ wrt $\triangle I_1I_2I_3.$ The conclusion follows.

Claim: The external tangent $\ell\neq\overline{AM}$ of $(I_1)$ and $(I_3)$ passes through $C.$ Proof. Let $X$ be the intersection between $\overline{AM}$ and the tangent to $(I_1)$ from $C.$ Since $ABCD$ and $ADCX$ are circumscribed, $$XC+AB-BC-AX=XC-CD+AD-AX=0$$and $ABCX$ is circumscribed. Hence, $\overline{CX}=\ell.$ $\blacksquare$ Claim: $CI_2I_1I_3$ is cyclic. Proof. Since $\overline{CI_3}$ and $\overline{CI_1}$ bisect $\angle DCX$ and $\angle XCB,$ respectively, we see $$\angle I_3CI_1=\tfrac{1}{2}\angle DCB=90-\tfrac{1}{2}\angle BCN=180-\angle MI_2N=\angle I_3I_2I_1.$$$\blacksquare$ Notice the reflection of $C$ in $\overline{I_1I_2}$ lies on $g$ as $\overline{I_1I_2}$ bisects $\angle NMC.$ Similarly, the reflections of $C$ in $\overline{I_2I_3}$ and $\overline{I_3I_1}$ also lie on $g,$ so $g$ is the Steiner line of $C$ with respect to $\triangle I_1I_2I_3.$ $\square$

New clean solution (perhaps?) Let a tangent to $(I_1)$ parallel to $ND$ intersect side $BC$ at $P$ and $g$ at $H'.$ Let the line through $H'$ parallel to $BC$ intersect $CD$ at $Q.$ Then \begin{align*} H'Q + AD &= PC + AD \\ &= BC -BP + AD \\ &= CD + AB - BP \\ &= CD + AH' - H'P \\ &= AH' + CD - CQ \\ &= QD + H'A \\ \end{align*}so $H'Q$ tangent to $(I_3).$ Now see $I_1H'$ parallel to the external angle bisector of $\angle AND$ since $PH' \parallel ND,$ which is perpendicular to $I_2I_3.$ Similarly, $I_1I_2$ parallel to the external angle bisector of $\angle AH'Q$ since $QH' \parallel BC,$ which is perpendicular to $H'I_3.$ So $H'=H,$ done! $\blacksquare$

Let $\omega_1$, $\omega_2$, $\omega_3$ be incircles of $\triangle ABM$, $\triangle CMN$, and $\triangle AND$, respectively. Let $g$ be tangent to $\omega_1$ and $\omega_3$ at $X_1$ and $X_3$, respectively. Let the tangents from $C$ to $\omega_1$ meet $\omega_1$ at $Z_1$ on $BC$ and $Y_1\neq Z_1$. Define $Y_3$ and $Z_3$ similarly. We claim that $C,Y_1,Y_3$ are collinear. $~$ Note that \[2X_1X_2=2AX_1-2AX_2=AB+AM-BM-AN-AD-DN\]and \[2CY_1-2CY_3=2CZ_1-2CZ_3=2(CM+CN)+AM+BM-AB-AN-ND+AD\]The difference between the two is \[2(CM+CN)+2BM-2AB-2ND+2AD=2(CM+CN+BM-ND+CD-CB)=0\]so $CY_1-CY_3=X_1X_3$, so $Y_1Y_3\le X_1X_3\le Y_1Y_3\implies X_1X_3=Y_1Y_3$ which implies the claim. Thus, \[\angle I_1CI_3=\angle Y_1CI_1+\angle Y_3CI_3=\tfrac12 \angle C=\angle I_1I_2I_3\]so $C$ lies on $(I_1I_2I_3)$ $~$ Let $H$ be the orthocenter of $\triangle I_1I_2I_3$. $I_1I_2$ bisects $\angle ANC$ so $C_1$, reflection of $C$ over $I_2I_3$, is on $g$. Similarly, $C_2$, reflection of $C$ over $C_2C_3$ is on $g$. Let $CC_1\perp I_1I_2$ at $D_1$ and $CC_2\perp I_2I_3$ at $D_2$ then $D_1D_2$ is the Simson line and bisects $CH$ so $H$ lies on $g$.

after getting a hint on steiner line, i solved a g8!! also no latex every problem my writeup is spending 30 minutes which could be used on actually doing math wtfwtfwtfwtwtfwtwfwtwtfazsdfjasdfsdfjkahfd;gldisfkghsjkdgvxonoxoooonkncskscksckskfnss We claim that CI_1I_2I_3 is cyclic; let $T_1$ be the tangent point on AMN from C to the incircle of ABM; we prove that CT_1 is also tangent to the incircle of ADN. Indeed, by basic tangent lengths we have AT_1-CT_1=AB-BC=AD-CD, so ADCT is circumscribed as well. It follows that I_1CI_3=I_1CT+TCI_3=BCT/2+TCD/2=BCD/2, from which it's evident that I_1I_2I_3=180-MI_2N=90-MCN/2=BCD/2=I_1CI_3. Next, by Steiner's line the reflection of C over each of the side of I_1I_2I_3 with H forms a line; in particular, by properties of reflecting C over I_1I_2 and I_2I_3, they lie on AMN, implying AMNH is a line. $\blacksquare$ Remark: This is well motivated for the tangent thing because we first see with a good diagram to try to prove CI_1I_2I_3; by that backwards thought process we'd want BCD=2I_1I_2I_3.

How so difficult. Solved with hints. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); 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 = -13.554554304713665, xmax = 24.56474803959741, ymin = -19.66603602614126, ymax = 21.709407003858495; /* image dimensions */ pen ffvvqq = rgb(1,0.3333333333333333,0); pen qqwwzz = rgb(0,0.4,0.6); pen ffwwqq = rgb(1,0.4,0); pen cczzqq = rgb(0.8,0.6,0); /* draw figures */ draw(circle((-0.03829551184039537,-3.0091969558477922), 8.58333894791307), linewidth(0.7) + dotted); draw((-9.15710010967521,1.9400427161824167)--(17.102746555317268,15.511009422411483), linewidth(0.7) + ffvvqq); draw((17.102746555317268,15.511009422411483)--(4.906139316119445,-13.362122095551303), linewidth(0.7) + ffvvqq); draw((4.906139316119445,-13.362122095551303)--(-8.002606748178124,-9.650204852671033), linewidth(0.7) + ffvvqq); draw((-8.002606748178124,-9.650204852671033)--(-9.15710010967521,1.9400427161824167), linewidth(0.7) + ffvvqq); draw((-9.15710010967521,1.9400427161824167)--(12.40070558068635,-15.51718887018527), linewidth(0.7) + qqwwzz); draw((4.906139316119445,-13.362122095551303)--(12.40070558068635,-15.51718887018527), linewidth(0.7) + ffwwqq); draw(circle((6.658450958946224,-12.525238741396153), 1.2885474749018233), linewidth(0.7) + linetype("2 2") + qqwwzz); draw(circle((-4.486461954315393,-6.718282097583698), 3.7894378634983794), linewidth(0.7) + linetype("2 2") + qqwwzz); draw(circle((3.358569639628092,0.660979186176298), 6.8823662275163295), linewidth(0.7) + linetype("2 2") + qqwwzz); draw(circle((2.708547255445126,-6.507700811334957), 7.198090163832131), linewidth(0.7) + qqwwzz); draw((-4.486461954315393,-6.718282097583698)--(3.358569639628092,0.660979186176298), linewidth(0.7) + blue); draw((-4.486461954315393,-6.718282097583698)--(6.658450958946224,-12.525238741396153), linewidth(0.7) + blue); draw((6.658450958946224,-12.525238741396153)--(3.358569639628092,0.660979186176298), linewidth(0.7) + blue); draw((6.658450958946224,-12.525238741396153)--(12.40070558068635,-15.51718887018527), linewidth(0.7) + blue); draw((4.906139316119445,-13.362122095551303)--(-7.969552291275887,4.692000415352528), linewidth(0.7) + dotted + cczzqq); /* dots and labels */ dot((-9.15710010967521,1.9400427161824167),linewidth(4pt) + dotstyle); label("$A$", (-10.186132889899726,2.5094049465180377), NE * labelscalefactor); dot((-8.002606748178124,-9.650204852671033),linewidth(4pt) + dotstyle); label("$B$", (-9.007185394714847,-10.795859637077541), NE * labelscalefactor); dot((4.906139316119445,-13.362122095551303),linewidth(4pt) + dotstyle); label("$C$", (4.747202049108737,-14.557263548895701), NE * labelscalefactor); dot((17.102746555317268,15.511009422411483),linewidth(4pt) + dotstyle); label("$D$", (17.32264199774744,15.983090600792043), NE * labelscalefactor); dot((6.138061942938785,-10.445781237267795),dotstyle); label("$M$", (6.824395254910666,-10.290596425042267), NE * labelscalefactor); dot((12.40070558068635,-15.51718887018527),linewidth(4pt) + dotstyle); label("$N$", (12.999834515402888,-16.073053185001527), NE * labelscalefactor); dot((6.658450958946224,-12.525238741396153),linewidth(4pt) + dotstyle); label("$I_{2}$", (6.487553113429272,-13.60287748171796), NE * labelscalefactor); dot((3.358569639628092,0.660979186176298),linewidth(4pt) + dotstyle); label("$I_{3}$", (3.1191316986153335,1.3234399500187207), NE * labelscalefactor); dot((-4.486461954315393,-6.718282097583698),linewidth(4pt) + dotstyle); label("$I_{1}$", (-5.694904336814473,-6.585332870116915), NE * labelscalefactor); dot((0.11346413336866767,-5.567140030133641),linewidth(4pt) + dotstyle); label("$H$", (-0.1370090023714744,-5.032701092654239), NE * labelscalefactor); dot((-1.4012449617969458,-4.517992246134442),linewidth(4pt) + dotstyle); label("$X$", (-2.214202208173403,-5.855508230510407), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $\omega_i$ denote the incircle with center $I_i$. Claim: $\angle I_1CI_3 = \frac{1}{2} \angle BCD$ Proof. Let $g'$ denote the second internal tangent to $\omega_1$ and $\omega_3$, noting that the other internal tangent is given by $g$. Then it suffices to show that $C \in g'$. Thus we let $X \in \overline{AM}$ be such that $\overline{CX}$ tangent to $\omega_1$. Then we claim that $CXAD$ is circumscribed. To see this note that we have the following relations from $ABCX$ and $ABCD$, \begin{align*} CX + AD &= CD + AX\\ AB + CD &= AD + BC \end{align*}Adding the two equations we find $CX + AB = AX + BC$. As a result $CXAB$ is circumscribed. This implies the claim as then $\angle XCI_1 + \angle XCI_3 = \frac{1}{2}(\angle XCB + \angle XCD)$. $\blacksquare$ Claim: $CI_1I_3I_2$ is cyclic. Proof. It suffices to show that $\angle I_1I_2I_3 = \frac{1}{2} \angle BCD$. Deleting $I_1$ and $I_3$ we wish to show that $180 - \angle MI_2N = \frac{1}{2} \angle BCD$. Note that, \begin{align*} 180 - \angle MI_2N &= 180 - \left(90 + \frac 12 \angle MCN\right)\\ &= 90 - \frac{1}{2} \angle MCN\\ &= 90 - \frac 12 \left(180 - \angle BCD \right)\\ &= \frac{1}{2} \angle BCD \end{align*}as desired. $\blacksquare$ Claim: Line $g$ is the Steiner line of $C$ with respect to $\triangle I_1I_2I_3$. Proof. This follows as $I_1I_2$ bisects $\angle CMN$ and $I_3I_2$ bisects $\angle MNB$. $\blacksquare$ Then from properties of the Steiner line our conclusion follows.

Cute! Claim 1: $CI_1I_2I_3$ is cyclic

Next, clearly $AMN$ is the steiner line of $C$ WRT $I_1I_2I_3$. Done.

What a joke! By simple angle chasing, $AI$ and $g$ are isogonal lines wrt. $\angle I_3AI_1$. $(\spadesuit)$ Furthermore, note that $\angle AI_3I=180^\circ-\angle AI_3D=180^\circ-(90^\circ+\frac{\angle AND}{2})=90^\circ-\angle ANI_3$ Similarly, $\angle AI_1I=90^\circ-\angle AMI_1$. Now, we just focus on points $A,I,I_1,I_3,M$ and $N$. We can easily finish by spiral similarity Let $K$ be a point such that $\triangle AII_1 \overset{+}{\sim} \triangle AI_3K$ From $(\spadesuit)$, we can conclude that $K$ lies on $g$. We have $\triangle AII_1 \overset{+}{\sim} \triangle AI_3K$; $\angle AKI_3=\angle AI_1I=90^\circ-\angle AMI_1$ $\rightarrow I_3K \perp I_1M \equiv I_1I_2$ We have $\triangle AI_3I \overset{+}{\sim} \triangle AKI_1$;$\angle AKI_1=\angle AI_3I=90^\circ-\angle ANI_3$ $\rightarrow I_1K \perp I_3N \equiv I_1I_2$ So, $K$ is the orthocenter of $\triangle I_1I_2I_3$ which lies on line $g$. Remark: At first glance, the problem looks so fancy since there are too many circles. The first idea that pops into my mind is to apply Monge’s theorem, but it didn’t help much so I ended up with angle-chasing. The key to this problem is to reduce the figure, forget about all circles, and change all conditions to angle conditions.

hehe Let $I$ be the incenter of $ABCD$. As \[\angle AII_1+\angle DIC = \angle AIB + \angle DIC = 180 ^{\circ},\]$I$ has an isogonal conjugate in $AI_1CD$. As $\angle I_1AI_2=\angle BAD/2=\angle DAI$, this isogonal conjugate is $I$. Thus \[\measuredangle I_1I_3C=\measuredangle AI_3D=\measuredangle MI_2C=\measuredangle I_1I_2C\]so $I_1I_2I_3C$ is cyclic. To finish, by Iran the simson line of $I_1I_2I_3$ with respect to $C$ is the miline of $CMN$, so by simson line orthocenter bisection we conclude.

2009 G8 [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10.33849693688569cm); 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 = -6.3006272917184285, xmax = 5.037869645167262, ymin = -4.6747582366769285, ymax = 1.9141009458983518; /* image dimensions */ pen ffefdv = rgb(1.,0.9372549019607843,0.8352941176470589); pen ffwwqq = rgb(1.,0.4,0.); pen ffcctt = rgb(1.,0.8,0.2); pen qqffff = rgb(0.,1.,1.); pen zzffff = rgb(0.6,1.,1.); pen ffdxqq = rgb(1.,0.8431372549019608,0.); pen eqeqeq = rgb(0.8784313725490196,0.8784313725490196,0.8784313725490196); draw((-1.7693050841942273,-0.8248099356998294)--(-1.1147577444415655,-3.5258920276887893)--(0.5862308824283616,-0.76526659691777)--cycle, linewidth(1.) + zzffff); /* draw figures */ draw(circle((-0.08,-0.8351237657055474), 2.004195599236761), linewidth(1.) + ffefdv); draw((-1.6781553930806994,-4.440260208581232)--(-0.7345095425602122,1.637179242402999), linewidth(1.) + ffwwqq); draw(circle((-1.1147577444415655,-3.5258920276887893), 0.416433453627681), linewidth(1.) + ffcctt); draw(circle((-1.7693050841942273,-0.8248099356998294), 0.6447953899830091), linewidth(1.) + ffcctt); draw(circle((0.5862308824283616,-0.76526659691777), 1.6737134053275853), linewidth(1.) + ffcctt); draw((-1.6781553930806994,-4.440260208581232)--(3.9603296372618035,-0.4114778542466089), linewidth(1.) + qqffff); draw((3.9603296372618035,-0.4114778542466089)--(-0.7345095425602122,1.637179242402999), linewidth(1.) + qqffff); draw((-0.7345095425602122,1.637179242402999)--(-2.5705820908832644,-0.8199178443234377), linewidth(1.) + qqffff); draw((-2.5705820908832644,-0.8199178443234377)--(-0.5307229291187352,-3.6204024562374513), linewidth(1.) + qqffff); draw((-1.7693050841942273,-0.8248099356998294)--(-1.1147577444415655,-3.5258920276887893), linewidth(1.) + zzffff); draw((-1.1147577444415655,-3.5258920276887893)--(0.5862308824283616,-0.76526659691777), linewidth(1.) + zzffff); draw((0.5862308824283616,-0.76526659691777)--(-1.7693050841942273,-0.8248099356998294), linewidth(1.) + zzffff); draw((-0.5307229291187352,-3.6204024562374513)--(0.5862308824283616,-0.76526659691777), linewidth(1.) + qqffff); draw((-0.5307229291187352,-3.6204024562374513)--(-1.7693050841942273,-0.8248099356998294), linewidth(1.) + qqffff); draw((-0.5307229291187352,-3.6204024562374513)--(-1.1142121489034864,-0.8082504684989615), linewidth(1.) + ffdxqq); draw(circle((-0.5620359734415863,-1.9621036033460575), 1.6585944628647706), linewidth(1.) + eqeqeq); draw((-1.1147577444415655,-3.5258920276887893)--(-0.5307229291187352,-3.6204024562374513), linewidth(1.) + zzffff); /* dots and labels */ dot((-0.7345095425602122,1.637179242402999),linewidth(4.pt) + dotstyle); label("$A$", (-0.6920124967348659,1.671566252061225), NE * labelscalefactor); dot((3.9603296372618035,-0.4114778542466089),linewidth(4.pt) + dotstyle); label("$B$", (3.9969915841162575,-0.32934497209507185), NE * labelscalefactor); dot((-0.5307229291187352,-3.6204024562374513),linewidth(4.pt) + dotstyle); label("$C$", (-0.4899002518705933,-3.5429296654370033), NE * labelscalefactor); dot((-2.5705820908832644,-0.8199178443234377),linewidth(4.pt) + dotstyle); label("$D$", (-2.531233924999746,-0.7436750740668303), NE * labelscalefactor); dot((-1.6781553930806994,-4.440260208581232),dotstyle); label("$N$", (-1.6419400475969468,-4.3412730326508795), NE * labelscalefactor); dot((-1.3716065955608117,-2.465968948071211),linewidth(4.pt) + dotstyle); label("$M$", (-1.3286660680573243,-2.380784257467437), NE * labelscalefactor); dot((-1.1147577444415655,-3.5258920276887893),linewidth(4.pt) + dotstyle); label("$I_1$", (-1.0760257619769837,-3.441873543004867), NE * labelscalefactor); dot((-1.7693050841942273,-0.8248099356998294),linewidth(4.pt) + dotstyle); label("$I_2$", (-1.7328905577858695,-0.7436750740668303), NE * labelscalefactor); dot((0.5862308824283616,-0.76526659691777),linewidth(4.pt) + dotstyle); label("$I_3$", (0.6217170948829058,-0.6830414006075486), NE * labelscalefactor); dot((-1.1142121489034864,-0.8082504684989615),linewidth(4.pt) + dotstyle); label("$X$", (-1.0760257619769837,-0.7234638495804031), NE * labelscalefactor); dot((-0.08,-0.8351237657055474),linewidth(4.pt) + dotstyle); label("$I$", (-0.03514770092598004,-0.7537806863100439), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $H$ denote that orthocenter. And let $X$ be the intersection of $g$ with the tangent from $C$ to $(I_3)$. Claim 1: $CX$ is tangent to $(I_3)$. Proof: since we have that $ABCD$ and $ABCX$ are tangential quadrilaterals, then by pitot theorem this means that $ADCX$ is tangential too. $\blacksquare$ Now this means that $CI_2$ and $CI_3$ bisect $\angle XCD$ and $XCB$. Claim 2: $I_1I_2I_3C$ is cyclic. Proof: small angle chase. $\blacksquare$ Lastly since $C$ lies on $(I_1I_2I_3)$ then we show that $g$ is the Steiner line of $C$ wrt $\triangle I_1I_2I_3$. This implies that $H$ lies on $g$ hence we are done

Let $\omega_1,\omega_3$ be the incircles of $\triangle ABM, \triangle AND$ respectively and let the tangent from $C$ to $\omega_1$ hit $g$ at $E$, then by excentral Pitot on $ABCE$ and Pitot on $ABCD$: \[AE-EC=AB-BC=AD-DC\]Which implies that $CE$ is also tangent to $\omega_3$ and since $CE$ is tangent to both $\omega_1,\omega_3$ we in fact get that $E$ is insimillicenter of $\omega_1, \omega_3$. Now: \[\angle I_1CI_3=\angle I_1CE+\angle ECI_3=\frac{\angle BCD}{2}=\angle I_1I_2I_3\]Which implies that $I_1I_2CI_3$ is cyclic, but from symetry on the angle bisectors we get that $g$ is the $C$-steiner line in $\triangle I_1I_2I_3$ therefore it's well-known that it goes through the orthocenter of $\triangle I_1I_2I_3$ thus we are done .

remarkable solved with cursed_tangent1434, stillwater_25 and rjp08

monge is pretty dope Claim: $(I_1I_2I_3C)$ is cyclic. Let $\omega_{\square}$ denote the incircle of $\square$. We have by monge $C$ lies on the other internal tangent not equal to $AM$ between $\omega_{\triangle ADN}$ and $\omega_{\triangle ABM}$. Therefore: $$\angle I_1 C I_3 = \tfrac{1}{2} \angle BCD = \tfrac{1}{2} (\angle NMC + \angle MNC) = \angle II_2I_3$$Implying cyclicity. Now we consider the $C$-simson line of $\triangle I_1 I_2 I_3$, say with points $J$, $H$, and $I$, on $I_1I_2$, $I_2I_3$, and $I_3I_1$ respectively. Since $I_1I_2$ and $I_1I_3$ bisect $\angle NMC$, and $\angle CBA$ respectively. The simson line is the $C$-Midline of $\triangle C (CJ \cap AN) (CI \cap AN)$. Therefore $AN$ is the $C$-Steiner line and the orthocenter lies on it.

Denote the incenter of $ABCD$ by $I$. $B, I_1, I$ and $D, I_3, I$ are collinear. Let $J$ be the isogonal conjugate of $I$ in $\triangle AI_1I_3$. Because $AI$ is the internal bisector of $\angle BAD$, $\angle I_1AI=\angle IAB-\angle I_1AB=\angle I_1AI_3-\angle I_1 AK=\angle KAI_3$ , so $J \in g$ . Let $\odot JI_1I_3$ meet $g$ again at $H$ . $\angle I_1I_3H=\angle I_1 JH=\angle I_1AM-\angle AI_1J =(\frac{\pi}{2} - \angle II_1M)-\angle II_1I_3 = \frac{\pi}{2}-\angle I_3I_1I_2 $, so $I_3H\perp I_1I_2$. Similarly $I_1H\perp I_3I_2$, so $H$ is the orthocenter of $\triangle I_1I_2I_3$ and the proof is complete.

Let $\omega_1, \omega_2, \omega_3$ be the incircles of triangles $ABM$, $MNC$, $NDA$, respectively. It suffices to show that $g$ is the Steiner Line for some $P \in (I_1 I_2 I_3)$. We do this by showing that the reflections of $g$ over $I_1 I_2$, $I_2 I_3$, and $I_3 I_1$ concur at a point on $(I_1 I_2 I_3)$. Claim. The three aforementioned lines concur at point $C$. Proof. Since $I_1 I_2$ bisects $\angle AMB$ and $I_2 I_3$ bisects $\angle MNC$, the reflections of $g$ over $I_1 I_2$ and $I_2 I_3$ are $BC$ and $CD$, respectively. We wish to show that the reflection of $g$ about $I_3 I_1$ goes through $C$. Denote this as $g'$. Note that $g$ is a common internal tangent to $\omega_1$ and $\omega_3$. Hence, $g'$ will be the second common tangent. We now only need to show that the tangent to $\omega_1$ from $C$ is also tangent to $\omega_3$. Let $E=g \cap g'$. Note that $ABCE$ and $ABCD$ are tangential quadrilaterals; then by Pitot, \begin{align*} AB+CE&=BC+AE \\ AB+CD &= BC+AD \\ \implies CE+AD&=AE+CD, \end{align*}which implies $ADCE$ is tangential as well by Converse Pitot. This shows that the tangent to $\omega_1$ through $C$ is tangent to $\omega_3$ as well, finishing the claim. It only remains to show $C$ lies on $(I_1 I_2 I_3)$. We have \begin{align*} \angle I_1 C I_3 = \angle I_1 CE + \angle EC I_3 &= \frac{\angle ECM + \angle ECD}{2} \\ =\frac{1}{2} \angle BCD = 90^{\circ} -\frac{1}{2}&\angle MCN = 180^{\circ} - MI_2 N=I_1I_2I_3, \end{align*}showing $CI_1I_2I_3$ and cyclic and finishing the proof.