Do you have any other problems from the shortlist

Let $\{X\}=HI\cap AC$. On one hand we have $$1=\dfrac{XH}{XI}=\dfrac{CH}{CI}\cdot \dfrac{\sin{(90^\circ-\hat{A})}}{\sin{\widehat{ICJ}}}$$on the other hand we have $\dfrac{JI}{\sin{\widehat{ICJ}}}=\dfrac{IC}{\sin{(90^\circ-\hat{C})}}$, therefore $JI=\dfrac{CH\cos{\hat{A}}}{\cos{\hat{C}}}=AH$

According to our team leader, this problem G1 was the original problem 4 selected for IMO 2015.

, so $\triangle JIY\cong \triangle XHY$ and we are done.

Let $M = IH \cap AC$ $\angle HCB= \angle HAB= \angle HGB$. Thus $HBGC$ is cyclic. Furthermore $\angle CAH= \angle CBH = \angle CGH= \angle IJC$. By law of sines in $\triangle IMJ$ and $\triangle MAH$: $\frac{IM}{IJ} = \frac{ \sin \angle IJM}{ \sin \angle IMJ} = \frac{ \sin \angle MAH}{ \sin \angle AMH} = \frac{MH}{AH}$ $\Longrightarrow \frac{IM}{IJ} = \frac{MH}{AH}$ $\Longrightarrow IJ=AH$

Let $K$ be the midpoint of $AM$. Note that $BG\parallel AH\perp BC$ and $HG\parallel CH\perp AB$, so $\angle KJI=\angle CGI=\angle CBH=90^\circ-\angle C=\angle KAH$, so $\frac{IJ}{AH}=\frac{\sin\angle IKJ\cdot\frac{IK}{\sin\angle KJI}}{\sin\angle AKH\cdot\frac{HK}{\angle KAH}}=1$.

Since $\measuredangle CHG = \measuredangle CBG = 90^\circ$, it follows that $CHGB$ is cyclic, so \[ \measuredangle CAH = \measuredangle CBH = \measuredangle CGH = \measuredangle CGI = \measuredangle CJI = \measuredangle MJI. \]Moreover, $\measuredangle IMJ = \measuredangle JMH$, and $IM = MH$, so the Law of Sines implies $IJ = AH$.

Notice that quadrilateral $CHBG$ is cyclic. Define $X\equiv GH\cap AC$, and let $Y$ be the spiral center sending $AH$ to $IJ$. Then quadrilaterals $YAXH$ and $YXIJ$ are cyclic. Then $$\angle HYX=\angle HAX=\frac{\pi}{2}-\angle C$$which, combined with $$\angle IYX=\angle IJC=\angle IGC=\angle HGC=\angle HBC=\frac{\pi}{2}-\angle C$$gives us that $YX$ is a median and an angle bisector in $\triangle YHL$, and so $YH=YL$; we're done.

Let $Y$ be the reflection of $J$ across $X$. Note that $HI$ and $JY$ bisect each other, so $JIYH$ is a parallelogram. Furthermore, $\angle JC = \angle JYH$. Because $GCIJ$ is cyclic, $\angle IJC = \angle IGC$. Finally, $BHCG$ is cyclic because $BG$ is perpendicular to $BC$ and $\angle GHC = \angle BHC - \angle BHG = 90$ degrees. Then $\angle IGC = \angle HBC = 90 - C$. Now, $\angle IGC = \angle IJC = 90 - C$. This means $\angle JYH = 90 - C$. But note that $\angle HAC$ is $90 - C$, so that means $HA = HY = JI$, as wanted.

Let $M=HI\cap AC$. Then using $\triangle MIJ\sim \triangle MCG$ and a well-known result: $AH=BC\cdot\tfrac{cos(\angle A)}{\sin(\angle A)}$, we get: $IJ=CG\cdot\tfrac{MI}{MC}=\sqrt{BC^2+BG^2}\cdot\tfrac{MH}{MC}=\sqrt{BC^2+AH^2}\cdot\cos{\angle A}=\sqrt{AH^2\cdot\tfrac{\sin^2(\angle A)}{cos^2(\angle A)}+AH^2}\cdot\cos{\angle A}=AH$ $\blacksquare$

By Ratio Lemma on $\triangle HCI,$ we have $\dfrac{HC\sin\angle XCH}{IC\sin\angle XCI}=1.$ But by Law of Sines, $\dfrac{IJ}{\sin\angle XCI}=\dfrac{IC}{\sin\angle IJC}\implies HC\sin\angle XCH=IC\sin\angle XCI=IC\left(\dfrac{IJ\sin\angle IJC}{IC}\right)=IJ\sin\angle IJC\implies IJ=\dfrac{HC\sin\angle XCH}{\sin\angle IJC}$ Now note that since $AB||HG$ and $CH\perp AB,$ we have $CH \perp HG$ and consequently $GBHC$ cyclic. But now $90^{\circ}-\angle C=\angle HBC=\angle HGC=\angle IJC$ where the last equality is derived from $\overarc{CI}$ intercepting both angles. From here, we can finish since $\angle XCH=90^{\circ}-\angle A,$ so $IJ=\dfrac{HC\cos \angle A}{\angle C},$ but since $HC=2R\cos\angle C$ and similar for $AH$ we are done. ~G_U and wu2481632

Let $K$ be on $AC$ such that $AH\parallel IK$, then $AHKI$ is a parallelogram. Note $C,H,B,G$ lie on circle with diameter $CG\implies\angle IKJ=\angle CAH=\angle CBH=\angle CGH=\angle CJI\implies IJ=IK=AH$.

Let $G'$ be te reflection of $G $ apone $BC $. Then $AHBG $is a parallelograme. Hence $G'B \perp BC$ and $G'A \perp AC $. Hence$ G'$ is the antipode of $C $ wrt circumcircle. Hence $ \widehat {BGC}= \widehat {BAC} $ Since $AHGB $ is a parallelogram $\widehat {HGB}=\widehat {HAB}$ . This fact together with above result implies that $\widehat {IJC} =\widehat {IGB} =\widehat {CAH}$ Let $T $ be the intersection of $AC $ and $HI $. Hence from above results we have that $\widehat {HAT}= \widehat{TJI} $ . Now since $HTI $ are collinear, by applying sine rule to triangles $AHT$ and $JT $ we can get that $AH=JI $

From the condition on $G$, then we have $GB\parallel AH \perp BC$, and $GH\parallel AB \perp HC$, so $\angle GBC=\angle GHC=90^{\circ}$. So $GBHC$ is cyclic with diameter $GC$. Recall the fact that circle $(BHC)$ and circle $(ABC)$ are reflecions of each other across $BC$, so $GC=2R$, where $R$ is the circumradius of $(ABC)$. Let $M$ be the midpoint $BC$ and $O$ be the circumenter of $ABC$, then $AH=2OM=2R\cos \angle A$. It suffices to prove $IJ=2R\cos\angle A$. From cyclic quadrilateral $GCIJ$, we have $\triangle KIJ\sim \triangle KCG$, so $IJ=\frac{GC}{CK}\cdot IK=2R\cdot \frac{HK}{KC}$. But $\angle KHC =90^{\circ}$, and since $AB\parallel KH$, so $\angle HKC =\angle A$. So $\frac{HK}{KC}=\cos \angle A$. So $IJ=2R\cdot \frac{HK}{KC}=2R\cos\angle A=AH$, as we want to prove.

Denote $X \equiv HI \cap AC$ and select $Y \not\equiv X$ on $AC$ so that $HX = HY.$ We'll show that $\triangle HAY \cong \triangle IJX$, which will immediately imply the desired result. Since $\triangle HXY$ is $H$-isosceles, we have $\measuredangle HYA = \measuredangle YXH = \measuredangle JXI.$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(12cm); 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 = 974, xmax = 1016, ymin = 977, ymax = 1021; /* image dimensions */ pen ffevev = rgb(1.,0.8980392156862745,0.8980392156862745); pen evfuev = rgb(0.8980392156862745,0.9568627450980393,0.8980392156862745); pen qqzzqq = rgb(0.,0.6,0.); pen evevff = rgb(0.8980392156862745,0.8980392156862745,1.); pen efevfu = rgb(0.9372549019607843,0.8980392156862745,0.9568627450980393); pen wwqqzz = rgb(0.4,0.,0.6); filldraw((997.2317851346063,1011.8514742019076)--(990.401569616485,992.548359947194)--(1014.3203594866458,992.7758870803227)--cycle, ffevev, red); 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draw((1000.6369118508378,1008.0504086479674)--(1003.9176821640968,1017.3223099565412), blue); draw((990.5259260194581,979.4753930846811)--(1000.6369118508378,1008.0504086479674)); draw(arc((990.9370331388956,1018.8781627734236),1.2463766094394286,-48.14491154752593,-6.834831907638776), wwqqzz); draw(arc((990.9370331388956,1018.8781627734236),1.0386471745328572,-48.14491154752593,-6.834831907638776), wwqqzz); draw(arc((997.2317851346063,1011.8514742019076),1.2463766094394286,-89.45499118739333,-48.14491154752147), wwqqzz); draw(arc((997.2317851346063,1011.8514742019076),1.0386471745328572,-89.45499118739333,-48.14491154752147), wwqqzz); /* dots and labels */ dot((997.2317851346063,1011.8514742019076),dotstyle); label("$A$", (997.4106401780465,1012.2729156794516), NE * labelscalefactor); dot((990.401569616485,992.548359947194),dotstyle); label("$B$", (989.2261004427276,991.4168804148308), S * labelscalefactor); dot((1014.3203594866458,992.7758870803227),dotstyle); label("$C$", (1014.6521832752919,991.583063962756), NE * labelscalefactor); dot((997.3561415375789,998.7785073393937),dotstyle); label("$H$", (996.0396259076631,998.147314105804), N * labelscalefactor); dot((990.5259260194581,979.4753930846811),dotstyle); label("$G$", (990.0154722953725,977.9144671459029), S * labelscalefactor); dot((1000.6369118508378,1008.0504086479674),dotstyle); label("$X$", (1001.6898665371219,1007.6197763375442), NE * labelscalefactor); dot((1003.9176821640968,1017.3223099565412),dotstyle); label("$I$", (1004.0995279820381,1017.7569727609854), NE * labelscalefactor); dot((1006.9316638465494,1001.0237200764522),dotstyle); label("$Y$", (1007.0908318446927,1001.4294391773279), E * labelscalefactor); dot((990.9370331388956,1018.8781627734236),dotstyle); label("$J$", (989.8492887474473,1019.668083562126), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Also, since $B, C, G, H$ are inscribed in the circle of diamter $\overline{CG}$, we have $\measuredangle IJX = \measuredangle IJC = \measuredangle IGC = \measuredangle HGC = \measuredangle HBC = \measuredangle YAH.$ The above two paragraphs imply that $\triangle HAY$ and $\triangle IJX$ are inversely similar. But in fact since $HY = HX = IX$, these two triangles are congruent and we're done.

[asy][asy] import cse5; pathpen = black; pointpen = black; pointfontsize = 9; size(9cm); pair B = origin, C = (3.5, 3), A = (5, 0), G = (-1.5, 1.75), H = (3.5, 1.75), I = (4.74, 1.75), J = (5.23, -0.47), K = (2.88, 2.47), L = (4, 2), M = (3.5, 0), P = (4.12, 1.75); D(circumcircle(G,C,I)); D(circumcircle(G,C,H)); D("A",A,dir(225)); D("B",B,dir(-90)); D("C",C,dir(45)); D("G",G,dir(180)); D("H",H,dir(240)); D("I",I,dir(45)); D("J",J,dir(-30)); D("K",K,dir(120)); D("L",L,dir(60)); D("M",M,dir(-90)); D("P",P,dir(60)); D(A--B--C--cycle); D(C--G--I--cycle); D(K--A); D(B--L); D(C--M); D(C--J); D(G--B); D(I--J); D(G--J); markscalefactor=0.02; D(rightanglemark(H,K,C)); D(rightanglemark(C,L,H)); D(rightanglemark(P,H,C)); markscalefactor=0.05; D(anglemark(H,G,C)); D(anglemark(H,B,C)); D(anglemark(L,A,H)); markscalefactor=0.045; D(anglemark(B,C,H)); D(anglemark(B,G,H)); D(anglemark(H,A,M)); markscalefactor=0.055; D(anglemark(B,C,H)); D(anglemark(B,G,H)); D(anglemark(H,A,M)); [/asy][/asy] We have that the quadrilateral $CIJG$ is cyclic. Call $P$ the midpoint of $HI$. Then $\triangle{IJP} \sim \triangle{CGP}$ which yields $IJ = \frac{IP}{CP}(GC)$. Call $K, L, M$ the feet of the altitudes from $A$ to $BC$, $B$ to $CA$, and $C$ to $AB$ respectively. We then angle chase to prove the following lemma. $\triangle{GCH} \sim \triangle{ALH}$. Proof: We have that $\angle{BCH} = 90^{\circ} - \angle{CHK} = 90^{\circ} - \angle{AHM} = \angle{HAM}$. Then $\angle{BGH} = \angle{HAM}$ because $BGHA$ is a parallelogram. So $\angle{BGH} = \angle{BCH}$ and $BGCH$ is cyclic, which implies $\angle{CGH} = \angle{CBH}$. Since $\angle{ALH} = \angle{BKH}$, $ALKB$ is cyclic and $\angle{CBH} = \angle{KBH} = \angle{LAH}$. We get $\angle{CGH} = \angle{LAH}$, but then $\angle{CHG} = \angle{HLA} = 90^{\circ}$. So, $\triangle{GCH} \sim \triangle{ALH}$ by AA similarity. By the lemma, we have $GC = \frac{CH}{HL}(AH)$. We finish by noting that right triangles $CHP$ and $CLH$ are similar, which leads to $$IJ = \frac{IP}{CP}(GC) = \frac{HP}{CP} \cdot \frac{CH}{HL}(AH) = \frac{\frac{CP}{CH}(HL)}{CP} \cdot \frac{CH}{HL}(AH) = AH.$$

My solution: Since $GH\parallel AB $ and $BJ\parallel AH$ we have $\angle CHG=\angle GBC=90^{\circ} $ so $BHCG$ is cyclic. Now consider $S$ to be the intersection point of $HI$ and $AC$ and let $M$ be the symmetric of $A$ with respect to $S$.Obviously $IMHA$ is parallelogram so $IM\parallel AH$ and by angle chasing $\angle AMI=\angle CAH=\angle HBC=\angle HGC=\angle IJM\Rightarrow IJ=MI$ but from parallelogram $IMHA$ we have $MI=AH$ which gives the result.

Take $X=HI \cap AC$. First, we show that $BHCG$ is cyclic. This follows from $\angle HGB = \angle HAB = \angle HCB$. This implies $HI \parallel AB$. Also, we can get that $\angle IJC = \angle IGC = \angle HBC = 90-C$. Now we have $\frac{JI}{IX}= \frac{\sin A}{\cos C} = \frac{AH}{HX}$, and from $IX=HX$ we are done.

$Z = HI \cap AC.$ First note that $BHCG$ is cyclic. $\frac{BG}{GC}$ = $Cos A$ $\frac{IJ}{GC}$=$\frac{IZ}{ZC}$=$\frac{HZ}{ZC}$ = $CosA$ Therefore we get $IJ=BG=AH$ We are done

We aim to compute $IJ$ in terms of simpler segments and angles. From the Sine Law in the circumcircle of $CIJG$ we have \[ IJ = \frac{CI \sin\angle ICJ}{\sin \angle IJC} = \frac{CI\sin\angle ICJ}{\sin \angle IGC}. \]\noindent The Sine Law in triangles $CIX$ and $CHX$ yields \[ CI\sin \angle ICJ = CI\sin \angle ICX = IX\sin \angle IXC = XH\sin \angle CXH \]Now from the right triangle $CHX$ and the Sine Law in triangle $ACH$ we obtain $XH\sin \angle CXH = XH \cdot \frac{CH}{CX} = CH\sin \angle ACH = AH\sin \angle CAH$. Therefore $\displaystyle IJ = \frac{AH\sin \angle CAH}{\sin \angle IGC}$ and so it suffices to show that $\angle HGC = \angle HBC$. But the latter is equivalent to $BHCG$ being cyclic, which is true since $\angle BGH = \angle BAH = \angle BCH$ from the parallelogram $ABGH$.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(6cm); 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 = -10.671234287465047, xmax = 9.697683134003627, ymin = -10.168212879537606, ymax = 11.095090426513261; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); pen qqttcc = rgb(0,0.2,0.8); pen qqwwzz = rgb(0,0.4,0.6); /* draw figures */ draw((-5.02069455901368,-2.6484270805675583)--(4.174956772253968,-2.668816551361056), linewidth(0.7) + ccqqqq); draw((-5.02069455901368,-2.6484270805675583)--(-2.370601880966813,3.7958459758698724), linewidth(0.7) + ccqqqq); draw((-2.370601880966813,3.7958459758698724)--(4.174956772253968,-2.668816551361056), linewidth(0.7) + ccqqqq); draw((-5.02069455901368,-2.6484270805675583)--(-5.02905260967416,-6.417907929837968), linewidth(0.7) + ccqqqq); draw((-2.3789599316272936,0.026365126599462108)--(-2.370601880966813,3.7958459758698724), linewidth(0.7) + ccqqqq); draw((-0.1693425039129801,5.399527421039835)--(-5.02905260967416,-6.417907929837968), linewidth(0.7) + qqttcc); draw(circle((-1.963875234604572,-0.7704552596972081), 6.425654363729697), linewidth(0.7) + qqwwzz); draw((-3.9384279570619523,5.3442953933020325)--(-0.1693425039129801,5.399527421039835), linewidth(0.7) + ccqqqq); draw((-3.9384279570619523,5.3442953933020325)--(-2.370601880966813,3.7958459758698724), linewidth(0.7) + ccqqqq); draw((-5.02905260967416,-6.417907929837968)--(4.174956772253968,-2.668816551361056), linewidth(0.7) + ccqqqq); /* dots and labels */ dot((-5.02069455901368,-2.6484270805675583),dotstyle); label("$B$", (-5.953090931395261,-2.753465322339407), NE * labelscalefactor); dot((4.174956772253968,-2.668816551361056),dotstyle); label("$C$", (4.533325884084516,-2.8400187957697365), NE * labelscalefactor); dot((-2.370601880966813,3.7958459758698724),dotstyle); label("$A$", (-3.32763556970454,3.4937893958847322), NE * labelscalefactor); dot((-2.3789599316272936,0.026365126599462108),linewidth(4pt) + dotstyle); label("$H$", (-2.362100835081225,-0.791586591251946), NE * labelscalefactor); dot((-5.02905260967416,-6.417907929837968),linewidth(4pt) + dotstyle); label("$G$", (-5.56457935318749,-7.352287836045756), NE * labelscalefactor); dot((-1.2741512177701368,2.7129462738196484),linewidth(4pt) + dotstyle); label("$X$", (-0.8906917862215897,2.6994035037713306), NE * labelscalefactor); dot((-0.1693425039129801,5.399527421039835),linewidth(4pt) + dotstyle); label("$I$", (-0.11171052506060646,5.757626231642962), NE * labelscalefactor); dot((-3.9384279570619523,5.3442953933020325),linewidth(4pt) + dotstyle); label("$J$", (-4.352830724714849,5.728775073832852), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Cute. Let $X = \overline{AC} \cap \overline{GH}$. Note that, \begin{align*} \frac{AH}{\sin (180 - A)} &= \frac{XH}{\sin (90 - C)}\\ \iff AH &= \frac{XH \cdot \sin A}{\cos C} \end{align*}Also we have, \begin{align*} \frac{IJ}{\sin A} &= \frac{IX}{\sin (\angle IJC)}\\ \iff IJ &= \frac{IX \cdot \sin A}{\sin (\angle IJC)} \end{align*}Then equating it suffices to show, \begin{align*} \cos C &= \sin (\angle IJC)\\ &= \sin(\angle HGC) \end{align*}We claim that $\angle HGC = 90 - C$ which would prove the claim. To show this proceed with complex numbers and let $(ABC)$ be the unit circle. We find $h = a+b+c$ and $g = b + h - a$. Then it suffices to show, \begin{align*} \arg\left(\frac{c-a}{h-a} \right) &= \arg\left(\frac{h-g}{c-g} \right)\\ \iff \left( \frac{c-a}{h-a} \right) &/ \left(\frac{h-g}{c-g} \right) \in \mathbb{R} \end{align*}Expanding we may see, \begin{align*} \frac{c-a}{h-a} \cdot \frac{c-g}{h-g} &= \frac{c - a}{b + c} \cdot \frac{-2b}{a-b} \end{align*}Then it suffices to check that, \begin{align*} \frac{c-a}{b+c} \cdot \frac{-2b}{a-b} &= \overline{\left( \frac{c-a}{b+c} \cdot \frac{-2b}{a-b} \right)}\\ &= \frac{1/c - 1/a}{1/b + 1/c} \cdot \frac{-2/b}{1/a - 1/b}\\ &= \frac{ab - bc}{ac + ab} \cdot \frac{-2a}{b - a}\\ &= \frac{c- a}{b + c} \cdot \frac{-2b}{a - b} \end{align*}as desired. Thus $\angle HGC = 90 - C$ and therefore $\cos C = \sin(\angle HGC)$ as desired. $\square$

[asy][asy] import olympiad; import cse5; defaultpen(fontsize(10pt)); usepackage("amsmath"); usepackage("amssymb"); usepackage("gensymb"); usepackage("textcomp"); size(8cm); pair MRA (pair B, pair A, pair C, real r, pen q=anglepen) { r=r/2; pair Bp=unit(B-A)*r+A; pair Cp=unit(C-A)*r+A; pair P=Bp+Cp-A; D(Bp--P--Cp,q); return A; } pointpen=black+linewidth(2); pen polyline=linewidth(pathpen)+rgb(0.6,0.2,0); pen polyfill=polyline+opacity(0.1); pen angleline=linewidth(pathpen)+rgb(0,0.4,0); pen anglefill=angleline+opacity(0.4); markscalefactor=0.01; size(12cm); pair A=dir(130),B=dir(210),C=dir(330); // filldraw(A--B--C--cycle,polyfill,polyline); D(A--B--C--cycle,polyline); pair H=orthocenter(A,B,C); pair G=B+H-A; pair P=extension(G,H,A,C); pair I=2*P-H; pair J=2*foot(circumcenter(G,C,I),A,C)-C; pair X=H+I-J; D(B--G--H--A,polyline); D(H--I--J--A); D(circumcircle(G,C,I)); D(H--X); D("A",D(A),dir(65)); D("B",D(B),B); D("C",D(C),C); D("H",D(H),dir(-41)); D("G",D(G),S); D("P",D(P),dir(190)); D("I",D(I),unit(I-H)); D("J",D(J),unit(J-H)); D("X",D(X),unit(A+C)); [/asy][/asy] Let $P = IH \cap AC$. Construct point $X$ such that $A$ is the midpoint of $\overline{JX}$. Now $JIXH$ is a parallelogram as its diagonals bisect each other. Therefore it suffices to prove $AH = HX$. Now note by Reim that $GHXC$ is cyclic, and that $\angle GBC = \angle GHC = 90^\circ$, so $GBHC$ is cyclic. But this implies \[\angle HAX = 90^\circ - \angle ACB = \angle HBC = 180^\circ - \angle HXC = \angle AXH\]so we are done.

Observe that $CH \perp GH$, as $AB \parallel GH$ and $CH \perp AB$. Therefore $GBHC$ is cyclic, and in particular we obtain that $\triangle CAD \sim \triangle CGH$, as \[\angle CGH = \angle CBH = \angle DAC.\]Now we have that by the sine law; \begin{align*} \frac{IJ}{MI} &= \frac{\sin \angle IMJ}{\sin \angle IJM} \\ &= \frac{\sin \angle AMH}{\sin \angle IGC} \\ &= \frac{AH \sin \angle HAM}{MH \sin \angle HAM} \\ \implies \frac{IJ}{MI} &= \frac{AH}{MH} \\ \implies IJ &= AH, \end{align*}as desired. $\blacksquare$

Notice that $\triangle MIJ \sim \triangle MCG$ with ratio $\frac{MI}{CM} = \frac{HM}{CM} = \cos A$. So to show that $IJ=AH=2R \cos A$, it suffices to show that $CG = 2R$. This follows as for $C'$ the $C$-antipode, $B$ is the midpoint of $\overline{C'G}$ and $\overline{BC} \perp \overline{GC'}$.

solved with arnovs and orons BHCG is cyclic because <GBC = <BHC = 90. Subsequently, it's easy to see that <EJI = <IGC = <HBC = <DAC = <FAE. Since HE = EI and <AEH = 180 - <AEI LoS easily gives AH = JI.

Solved with megarnie. Let $D = BH \cap AC$, $M$ be a point on $AC$ such that $D$ is the midpoint of $JM$. By construction, $JIMH$ is a parallelogram. Claim: $GBHC$ is cyclic. Proof. Since $BG \parallel AH$, and $CH \perp AB$, we have $CH \perp GH$. Thus, $\angle GHC = \angle GBC = 90^{\circ}$, and the conclusion follows. $\blacksquare$ Angle chasing, we have \begin{align*} \angle HGC &= \angle HBC \\ &= \angle HAC \end{align*}However, $GJIC$ is cyclic, so \begin{align*} \angle HGC &= \angle IJC \\ &= \angle JMH \\ &= \angle HAC \end{align*}Thus, $\Delta HAM$ is isoceles, and $HA = HM = IJ$, as desired.

since $AH$ is parallel to $BG$, $BG$ is perpendicular to $BC$ since $CF$ is perpendicular to $AB$, $CF$ is perpendicular to $BH$ thus, $GBHC$ are concyclic so $HBC=HGC$, and since $CGJI$ is cyclic, $HGC=CJI$ additionally, we easily get that $HAC=HBC$ let $M$ be the midpoint of $HI$ by law of sines, $\frac{HM}{\sin{HAM}}=\frac{AH}{\sin{AMH}}$ and $\frac{IM}{\sin{IJM}}=\frac{IJ}{\sin{IMJ}}$, but $HAM=IJM$ and $AMH=180-JMI$ so $AH=IJ$

Let $HG \cap AC=D$, Now by parallelogram condtion we get, $AHCG$ cyclic $\implies \angle HAC = \angle HGC = \angle AJI \implies AH \parallel IJ \\ \implies \angle AHD = \angle AHI = \angle HIJ = \angle DIJ \\ implies \triangle ADH \cong \triangle IDJ$ due to equality of all angles with $HD=ID$ and Conclusion follows

First notice that in the complex plane we have $g= a +b + c + b - a = 2b + c$, so $g - c = 2b$ so $GC$ has length $2R$ where $R$ is the radius of the circumcircle. Let $HI$ meet $AC$ at $X$. Now, by similar triangles $XIJ$ and $XCG$, we know that $\frac{IJ}{GC} = \frac{XI}{XC} = \frac{XH}{XC} = \cos A$. Thus $IJ = 2R \cos A = AH$ by reflecting the orthocenter, and we are done.

\[ \textbf{IMOSL 2015/G1} \] Let \( GH \) $\cap$ \( AC = k \). Let perpendiculars from \( I \) to \( BC \) meet \( AC \) at \( L \). Let perpendiculars from \( I \) to \( BC \) meet \( AC \) at \( L \). Reflect \( A \) in \( BH\) to get \( M \). Claim 1: \( GM \perp AC \) Proof: \[ AH = HM = BG \quad \text{and} \quad \angle BMH = \angle BMA - \angle HMA= \angle A-\angle 90- \angle C=90-\angle B=\angle BGH\]imply that $GBH$ is an isosceles trapezium. This gives us $GM \parallel BH,$ and the claim follows. Claim 2: \( KJ = KM \) Proof: Since \( C, I, J, \) and \( G\) are cyclic, we get: \[ CK \cdot KJ = IK \cdot KG= KH\cdot KG \]\[ \implies KJ = \frac{KH \cdot KG}{CK} =KG \frac{KH}{CK}=KG \cos A=KM \] This also implies that \( IJMH \) is a parallelogram. Thus, we have: \[ \angle IJL = \angle IJM = \angle JMH = 90 - \angle C \]\[ =\angle HAA = \angle JL \] Hence, \( IJ = IL = AH \), using the fact that \( IJHA \) is a parallelogram.

Due to the law of sines, it suffices to show that $\angle IJX = \angle XAH$. This follows since \[\angle IJX = \angle IGC = \angle HBC = \angle CAH,\]where the second step follows since $\overline{CH} \perp \overline{AB} \parallel \overline{IG}$, giving us cyclic quadrilateral $HBGC$.

Let $HI \cap AC = M$. As $\measuredangle GBC = 90^{\circ} = \measuredangle GHC$ we get $GBHC$ is cyclic. Hence $$ \measuredangle CAH= \measuredangle CBH=\measuredangle CGH=\measuredangle CGI=\measuredangle CJI= \measuredangle MJI $$Addditionally, $ \measuredangle IMJ= \measuredangle JMH $, and $ IM= MH$. Then,. by law of sines, we are done

Construct $P,Z$ with $\overrightarrow{AH}=\overrightarrow{BG}=\overrightarrow{IP}=\overrightarrow{ZC}$. In particular $Z$ is the $B$ antipode on $(ABC)$. Then \[\measuredangle IJP=\measuredangle IGC=\measuredangle ABZ=\measuredangle ACZ=\measuredangle JPI,\]so $IJ=IP=AH$ as desired.