SL 2015 G1: Prove that IJ=AH

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315 posts

#1 h Jul 7, 2016, 6:31 PM • 14 Y

Y by doxuanlong15052000, mathmaths, Davi-8191, Rarvic, thczarif, lilavati_2005, Purple_Planet, samrocksnature, megarnie, Adventure10, Mango247, ItsBesi, Rounak_iitr, MS_asdfgzxcvb

Let $ABC$ be an acute triangle with orthocenter $H$ . Let $G$ be the point such that the quadrilateral $ABGH$ is a parallelogram. Let $I$ be the point on the line $GH$ such that $AC$ bisects $HI$ . Suppose that the line $AC$ intersects the circumcircle of the triangle $GCI$ at $C$ and $J$ . Prove that $IJ = AH$ .

This post has been edited 2 times. Last edited by v_Enhance, Jul 7, 2016, 7:46 PM
Reason: Use official SL wording

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tastymath75025

3223 posts

tastymath75025

#3 h Jul 7, 2016, 6:55 PM • 25 Y

Y by the9th123821, rafayaashary1, wu2481632, anantmudgal09, sa2001, naw.ngs, pad, thczarif, myh2910, Mathasocean, Purple_Planet, mathlogician, mathleticguyyy, AllanTian, animath_314159, samrocksnature, anonman, math31415926535, megarnie, veirab, Jalil_Huseynov, Adventure10, Mango247, sabkx, MS_asdfgzxcvb

solution

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the9th123821

194 posts

the9th123821

#4 h Jul 7, 2016, 7:07 PM • 3 Y

Y by samrocksnature, Adventure10, Jack_w

Do you have any other problems from the shortlist

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Aiscrim

409 posts

Aiscrim

#6 h Jul 7, 2016, 7:10 PM • 4 Y

Y by samrocksnature, Adventure10, Mango247, Acorn-SJ

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 $JI=\dfrac{CH\cos{\hat{A}}}{\cos{\hat{C}}}=AH$

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TelMarin

248 posts

TelMarin

#7 h Jul 7, 2016, 7:14 PM • 3 Y

Y by Imayormaynotknowcalculus, samrocksnature, Adventure10

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

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droid347

2679 posts

droid347

#8 h Jul 7, 2016, 7:17 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

Let $X$ be on $AC$ so $AH=HX$ . Let $Y=IH\cap AC$ and we have $\angle YXH=90^{\circ}-\angle C$ . By the circle we have $\angle IJY=\angle HGC$ and it's not hard to find $\angle HGC=90^{\circ}-\angle C$ with trig, so $\triangle JIY\cong \triangle XHY$ and we are done.

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Math_CYCR

431 posts

Math_CYCR

#9 h Jul 7, 2016, 7:22 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

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 $\triangle MAH$ :

$\frac{IM}{IJ} = \frac{ \sin \angle IJM}{ \sin \angle IMJ} = \frac{ \sin \angle MAH}{ \sin \angle AMH} = \frac{MH}{AH}$

$\Longrightarrow \frac{IM}{IJ} = \frac{MH}{AH}$

$\Longrightarrow IJ=AH$

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ABCDE

1963 posts

ABCDE

#10 h Jul 7, 2016, 7:27 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

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 $\frac{IJ}{AH}=\frac{\sin\angle IKJ\cdot\frac{IK}{\sin\angle KJI}}{\sin\angle AKH\cdot\frac{HK}{\angle KAH}}=1$ .

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v_Enhance

6870 posts

v_Enhance

#12 h Jul 7, 2016, 7:36 PM • 9 Y

Y by MeowX2, akjmop, Purple_Planet, Aryan-23, jchang0313, v4913, samrocksnature, Adventure10, Mango247

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 $IJ = AH$ .

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EulerMacaroni

851 posts

EulerMacaroni

#13 h Jul 7, 2016, 7:54 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

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.

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PepsiCola

85 posts

PepsiCola

#14 h Jul 7, 2016, 8:06 PM • 2 Y

Y by samrocksnature, Adventure10

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.

This post has been edited 4 times. Last edited by PepsiCola, Jul 7, 2016, 10:45 PM
Reason: Figured out how to do angles too!

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mjuk

196 posts

mjuk

#15 h Jul 7, 2016, 8:44 PM • 2 Y

Y by samrocksnature, Adventure10

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:
$IJ=CG\cdot\tfrac{MI}{MC}=\sqrt{BC^2+BG^2}\cdot\tfrac{MH}{MC}=\sqrt{BC^2+AH^2}\cdot\cos{\angle A}=\sqrt{AH^2\cdot\tfrac{\sin^2(\angle A)}{cos^2(\angle A)}+AH^2}\cdot\cos{\angle A}=AH$ $\blacksquare$

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Generic_Username

1088 posts

Generic_Username

#16 h Jul 7, 2016, 10:28 PM • 5 Y

Y by wu2481632, mathzhang, samrocksnature, Adventure10, Mango247

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

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buzzychaoz

178 posts

buzzychaoz

#17 h Jul 8, 2016, 1:06 AM • 3 Y

Y by samrocksnature, electrovector, Adventure10

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 $CG\implies\angle IKJ=\angle CAH=\angle CBH=\angle CGH=\angle CJI\implies IJ=IK=AH$ .

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samithayohan

41 posts

samithayohan

#19 h Jul 8, 2016, 2:46 AM • 3 Y

Y by samrocksnature, Adventure10, Mango247

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 $AH=JI$

This post has been edited 1 time. Last edited by samithayohan, Jul 8, 2016, 3:11 AM
Reason: Latex error

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Scilyse

386 posts

Scilyse

#130 h Feb 8, 2024, 9:51 PM

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$[asy] import olympiad; import cse5; defaultpen(fontsize(10pt)); usepackage("amsmath"); usepackage("amssymb"); usepackage("gensymb"); usepackage("textcomp"); size(8cm); pair MRA (pair B, pair A, pair C, real r, pen q=anglepen) { r=r/2; pair Bp=unit(B-A)*r+A; pair Cp=unit(C-A)*r+A; pair P=Bp+Cp-A; D(Bp--P--Cp,q); return A; } pointpen=black+linewidth(2); pen polyline=linewidth(pathpen)+rgb(0.6,0.2,0); pen polyfill=polyline+opacity(0.1); pen angleline=linewidth(pathpen)+rgb(0,0.4,0); pen anglefill=angleline+opacity(0.4); markscalefactor=0.01; size(12cm); pair A=dir(130),B=dir(210),C=dir(330); // filldraw(A--B--C--cycle,polyfill,polyline); D(A--B--C--cycle,polyline); pair H=orthocenter(A,B,C); pair G=B+H-A; pair P=extension(G,H,A,C); pair I=2*P-H; pair J=2*foot(circumcenter(G,C,I),A,C)-C; pair X=H+I-J; D(B--G--H--A,polyline); D(H--I--J--A); D(circumcircle(G,C,I)); D(H--X); D("A",D(A),dir(65)); D("B",D(B),B); D("C",D(C),C); D("H",D(H),dir(-41)); D("G",D(G),S); D("P",D(P),dir(190)); D("I",D(I),unit(I-H)); D("J",D(J),unit(J-H)); D("X",D(X),unit(A+C)); [/asy]$

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.

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MagicalToaster53

159 posts

MagicalToaster53

#131 h Feb 28, 2024, 1:42 AM

Y by

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. $\blacksquare$

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HamstPan38825

8857 posts

HamstPan38825

#132 h Mar 11, 2024, 6:28 PM

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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 $\overline{BC} \perp \overline{GC'}$ .

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GoodMorning

826 posts

GoodMorning

#133 h Mar 13, 2024, 7:30 PM • 4 Y

Y by CT17, pi271828, megarnie, OronSH

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.

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blueberryfaygo_55

339 posts

blueberryfaygo_55

#134 h May 8, 2024, 9:55 PM • 1 Y

Y by megarnie

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.

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AngeloChu

470 posts

AngeloChu

#135 h May 10, 2024, 1:31 PM

Y by

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 $AH=IJ$

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Alex9100

2 posts

Alex9100

#136 h May 10, 2024, 2:31 PM

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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

This post has been edited 5 times. Last edited by Alex9100, May 10, 2024, 2:36 PM
Reason: Incomplete

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ezpotd

1251 posts

ezpotd

#137 h Sep 17, 2024, 4:56 PM

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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.

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SimplisticFormulas

84 posts

SimplisticFormulas

#138 h Nov 20, 2024, 8:36 PM

Y by

$\textbf{IMOSL 2015/G1}$
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.

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ihatemath123

3441 posts

ihatemath123

#139 h Nov 27, 2024, 12:34 AM

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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 $HBGC$ .

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kotmhn

57 posts

kotmhn

#140 h Dec 16, 2024, 6:51 AM

Y by

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

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OronSH

1728 posts

OronSH

#141 h Dec 26, 2024, 9:01 PM

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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.

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GoodGuy2008

2 posts

GoodGuy2008

#142 h Mar 7, 2025, 5:21 AM

Y by

ok so this is my first isl problem! Make sure to support my goal(I will solve one isl problem for each like that my account gets).
$[asy] import olympiad; import geometry; size(6cm); pair A,B,C,H,G,K,I,J,P; A=dir(120); B=dir(210); C=dir(330); H=orthocenter(A,B,C); G=(B+H-A); K=intersectionpoint(line(G,H),line(A,C)); I=(2K-H);J=intersectionpoints(circle(G,C,I),line(A,C))[1];P=(2K-J); draw(A--B--C--cycle,blue); draw(B--G--H--A,blue); draw(H--I,blue);draw(circle(G,C,I),purple);draw(A--J,blue); draw(H--P--I--J--cycle,orange); draw(arc(circumcenter(G,B,C),C,G),deepred); draw(G--C--H,blue); dot(A);dot(B); dot(C);dot(H); dot(G);dot(K); dot(I);dot(J);dot(P); label(A,"$A$",N); label(B,"$B$",NW); label(C,"$C$",NE); label(G,"$G$",SW); label(H,"$H$",W); label(P,"$P$",NE); label(K,"$K$",E); label(I,"$I$",NE); label(J,"$J$",NW); [/asy]$
Let $K$ be the intersection of $GH$ and $AC$ and let $P$ be the reflection of $J$ w.r.t $K$ . Note that $HPIJ$ is parallelogram and since $JICG$ is cyclic, we have $\measuredangle IGC=\measuredangle IJC=\measuredangle JPH$ thus $GHPC$ is cyclic and so $\measuredangle GPC=\measuredangle GHC$ and since $CH\perp AB\parallel GH$ we have that $\measuredangle GBC=90=\measuredangle GHC=\measuredangle GPC$ and thus $B$ lies on $(CPHG)$ . Since $BH\perp AC\perp GP$ we have $BH\parallel GP$ which implies that $GBHP$ is an equilateral trapezoid and thus $BG=HP$ . Hence $AH=BG=HP=IJ$ and we are done.

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maths_enthusiast_0001

131 posts

maths_enthusiast_0001

#143 h Mar 26, 2025, 2:33 PM

Y by

Very simple

(also this is India TST 2016 P1)

Define $D,E,F$ as feet of altitudes from $A,B,C$ to $BC,CA,AB$ respectively.
We have $\angle{BAD}=(90^{\circ}-B)$ . Since $ABGH$ is a parallelogram, $\angle{BGH}=\angle{BAH}=\angle{BAD}=(90^{\circ}-B)$ . Also $\angle{BCH}=\angle{BCF}=(90^{\circ}-B)$ implying that, $\angle{BGH}=\angle{BCH}$ . Thus, $BGCH$ is a cyclic quadrilateral.
Now, $\angle{IJC}=\angle{IGC}=\angle{HGC}=\angle{HBC}=(90^{\circ}-C) \implies \boxed{\angle{IJC}=(90^{\circ}-C)}$ . Let $K=AC \cap HI$ .
Also as $BG \parallel AH, \angle{GBC}=90^{\circ} \implies \angle{GHC}=\angle{GBC}=90^{\circ} \implies \angle{KHC}=90^{\circ}$ . Also $\angle{KCH}=\angle{ACH}=(90^{\circ}-A)$ thus, $\angle{HKC}=A$ or $\boxed{\angle{IKJ}=A}$ .
So, we have $\angle{IKJ}=A$ and $\angle{IJK}=(90^{\circ}-C)$ . By sine rule in $\Delta IJK$ we have, $\boxed{\frac{IJ}{\sin A}=\frac{KI}{\cos C}}$ .
Also, $\angle{HAK}=(90^{\circ}-C)$ and $\angle{AKH}=180^{\circ}-\angle{HKC}=(180^{\circ}-A)$ . Again by sine rule we have, $\boxed{\frac{AH}{\sin A}=\frac{HK}{\cos C}}$ .
As $HK=KI$ , we have $\boxed{IJ=AH}$ as desired. $\blacksquare$ ( $\mathcal{QED}$ )

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bjump

994 posts

bjump

#144 h Mar 26, 2025, 5:15 PM • 1 Y

Y by imagien_bad

Slick.

Reflect $A$ over the $B$ -Altitude to $A'$ .
$\angle HBC = 90^\circ-\angle C = 180^\circ - (90+\angle C) = 180^\circ- \angle HA'C$ Therefore $(HA'CB)$ is cyclic. Reims gives $IJ \parallel A'H$ this combined with $HD=DI$ means that $IJHA'$ is a parallelogram so $IJ = A'H = AH$ .

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