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Source: Bulgarian IMO TST 2005, Day 2, Problem 2

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58 posts

#1 h Jul 7, 2013, 9:28 AM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ , $AC \not= BC$ , be an acute triangle with orthocenter $H$ and incenter $I$ . The lines $CH$ and $CI$ meet the circumcircle of $\bigtriangleup ABC$ at points $D$ and $L$ , respectively. Prove that $\angle CIH = 90^{\circ}$ if and only if $\angle IDL = 90^{\circ}$

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109 posts

#2 h Jul 7, 2013, 11:15 AM • 2 Y

Y by Adventure10, Mango247

Easy with trigonometry. Let $R=\frac{1}{2}, \angle BAC=2\alpha, \angle ABC=2\beta, \angle BCA=2\gamma$ .
Wlog $\alpha \geq \beta$ . Use angle chasing to get $\angle HCI=\alpha - \beta$ and $\angle AIC= \beta + 90^\circ$ . We have the equalities $CI= 2\sin\alpha\sin\beta$ and $CH=\cos 2\gamma$ .
$\angle CIH=90^\circ$ iff $CI=CH\cos\angle HCI$ iff $2\sin\alpha\sin\beta = (-2\cos (\alpha+\beta) ^2 +1)\cos (\alpha-\beta)$ iff $2\cos (\alpha+\beta) ^2 \cos (\alpha-\beta)=\cos (\alpha-\beta)-2\sin\alpha\sin\beta = \cos (\alpha+\beta)$ iff $2 \cos (\alpha+\beta) \cos (\alpha-\beta)=1$ .

$\angle IDL=90^\circ$ iff $LD=LI\cos\angle DLC$ iff $\sin (\alpha-\beta) = AL\cos (2\beta + 90^\circ - 2\alpha)$ iff $\sin (\alpha-\beta) = \cos (\alpha+\beta) \sin (2\alpha-2\beta)$ iff $2 \cos (\alpha+\beta) \cos (\alpha-\beta)=1$ .

Both statements are equivalent to the same statement, done.

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#3 h Jul 20, 2013, 1:44 PM • 1 Y

Y by Adventure10

PP. Let acute $\triangle ABC$ with circumcircle $w=C(O,R)$ , incenter $I$ , orthocenter $H$ and $c<b$ . The lines $AH$ and $AI$ meet
again $w$ at $D$ and $L$ respectively. Prove that $IH\perp IA\iff \cos B+\cos C=1\iff$ $r=R\cos A\iff$ $DI\perp DL$ .

Proof. Denote $\{L,N\}=LO\cap w$ , $P\in AD\cap BC$ and $S\in AI\cap BC$ . Therefore:

$\blacktriangleright\ DI\perp DL\iff$ $m\left(\widehat{DIL}\right)=B-C\iff$ $m\left(\widehat{ADI}\right)=\frac {B-C}{2}\iff$ $N\in DI\iff$ $ADLN$ is an isosceles trapezoid $\iff$ $IO\parallel BC\iff$ $\boxed{r=R\cos A}$ .

$\blacktriangleright\ IH\perp IA\iff$ $HPSI$ is cyclically $\iff AI\cdot AS=AH\cdot AP\iff$ $\frac {b+c}{2s}\cdot\frac {4bcs(s-a)}{(b+c)^2}=2Rh_a\cos A\iff$ $\frac {bc(b+c-a)}{b+c}=bc\cos A\iff$ $b+c-a=(b+c)\cos A\iff$ $(b+c)(1-\cos A)=a\iff$ $2\sin^2\frac A2(\sin B+\sin C)=\sin A\iff$ $\boxed{\cos B+\cos C=1}$ .

Remark. The relation $\cos A+\cos B+\cos C=1+\frac rR$ is well-known. Thus, $\boxed{\cos B+\cos C=1\iff r=R\cos A}$ .

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#4 h Jul 21, 2013, 12:51 PM • 2 Y

Y by Adventure10, Mango247

Note that $D$ is the reflection of $H$ in $AB$ . Let the circle $\Gamma$ through $IDH$ intersect $DL$ at $I'$ and $IL$ at $E$ . If $\angle CIH = 90$ , then by a bit of an angle chase we get that $I'$ is in fact the reflection of $I$ over $AB$ . Let $X$ be the point where the incircle touches $AB$ . Then $X$ is the circumcentre of $\Gamma$ . For this part, I unfortunately did it with trig (short just show $I$ is the midpoint of $CE$ ) so I hope to find it soon. Hence, $I'I$ bisects $\angle EID$ and so $\angle LDI = 90$ since $\angle LDE = \angle IDH$ .

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#5 h Mar 18, 2015, 10:00 AM • 2 Y

Y by Adventure10, Mango247

$\angle{IDL}=90^{\circ} \Leftrightarrow \angle{DIL}=A-B$ and \angle{DCI}=\frac{A-B}{2} \Leftrightarrow CI=ID.Thus $I$ lies on the perpendicular bisector of $CD$ ,so $OI \perp CD \Leftrightarrow r=RcosC$

$\angle{CIH}=90^{\circ}$ . Let $X$ and $Y$ be the feet of altitudes from $B$ and $A$ respectively.Then $CXHIY$ is cyclic with diameter $CH \Leftrightarrow XI=IY$ .Let $E$ and $F$ be the touchpoints of the incircle with $BC$ and $AC$ respectively.Then $ \triangle{IYE} \cong \triangle{IXF} \Leftrightarrow EY=XF \Leftrightarrow ccosA-(s-a)=(s-b)-ccosB \Leftrightarrow cosA+cosB=1 \Lefrightarrow r=RcosC$.

Thus $\angle{IDL}=90^{\circ} \Leftrightarrow \angle{CIH}=90^{\circ} \Leftrightarrow r=RcosC$

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#6 h May 20, 2016, 12:26 PM • 1 Y

Y by Adventure10

Please check if this solution is correct as I am not too sure

Let $CE$ be altitude, $CF$ be internal angle bisector, $M$ midpoint of $AB$ , $N$ the midpoint of arc $\overarc{ACB}$ , $IH\cap AB\equiv P$ .
First assume $\angle CIH=90^{\circ}$ , then $C,I,E,P$ are cyclic. Then $\angle FCE=\angle FPI=\angle FPD\implies F,C,P,D$ cyclic. Note $C,F,M,N$ cyclic so $\angle CDP=\angle CFP=\angle LNC=180^{\circ}-\angle CDL\implies L,D,P$ are collinear. Thus from here $D$ is the touchpoint of the $C-$ mixtillinear circle with $(ABC)$ , hence well-known that $N,I,D$ are collinear and $\angle IDL=90^{\circ}$ .

If $\angle IDL=90^{\circ}$ ,now define $P\equiv DL\cap AB$ , then $D,P,N,M$ cyclic. Then $\angle DPM=\angle DNM=\angle DCL\implies D,P,C,F$ cyclic. Using the same link as above $\angle CIP=90^{\circ}\implies C,I,E,P$ cyclic, then $\angle HPB=\angle DPB=\angle DCF=\angle IPB\implies \angle CIH=90^{\circ}$ .

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435 posts

#7 h Aug 19, 2021, 10:17 PM

Y by

Virgil Nicula wrote:

PP. Let acute $\triangle ABC$ with circumcircle $w=C(O,R)$ , incenter $I$ , orthocenter $H$ and $c<b$ . The lines $AH$ and $AI$ meet
again $w$ at $D$ and $L$ respectively. Prove that $IH\perp IA\iff \cos B+\cos C=1\iff$ $r=R\cos A\iff$ $DI\perp DL$ .

Proof. Denote $\{L,N\}=LO\cap w$ , $P\in AD\cap BC$ and $S\in AI\cap BC$ . Therefore:

$\blacktriangleright\ DI\perp DL\iff$ $m\left(\widehat{DIL}\right)=B-C\iff$ $m\left(\widehat{ADI}\right)=\frac {B-C}{2}\iff$ $N\in DI\iff$ $ADLN$ is an isosceles trapezoid $\iff$ $IO\parallel BC\iff$ $\boxed{r=R\cos A}$ .

$\blacktriangleright\ IH\perp IA\iff$ $HPSI$ is cyclically $\iff AI\cdot AS=AH\cdot AP\iff$ $\frac {b+c}{2s}\cdot\frac {4bcs(s-a)}{(b+c)^2}=2Rh_a\cos A\iff$ $\frac {bc(b+c-a)}{b+c}=bc\cos A\iff$ $b+c-a=(b+c)\cos A\iff$ $(b+c)(1-\cos A)=a\iff$ $2\sin^2\frac A2(\sin B+\sin C)=\sin A\iff$ $\boxed{\cos B+\cos C=1}$ .

Remark. The relation $\cos A+\cos B+\cos C=1+\frac rR$ is well-known. Thus, $\boxed{\cos B+\cos C=1\iff r=R\cos A}$ .

why is $AI = \frac {b+c}{2s}$ and $AS = \frac {4bcs(s-a)}{(b+c)^2}$ ????

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#8 h Aug 20, 2021, 2:19 PM

Y by

Assume $\angle IDL=90^\circ$ . We claim that $I$ lies on the circle with diameter $CH$ .

Note that $\angle IDL=90^\circ$ implies by $\sqrt{bc}$ invert that circumcenter of $ABC$ lies on $AG$ , where $G$ is the $C$ -excircle touchpoint with $AB$ . It is well-known that $M$ , the midpoint of $AB$ , is the midpoint of $EG$ also, where $E$ is the incircle touchpoint with $AB$ .

Let $E'$ be the antipode of $E$ wrt $(I)$ , note that $E'$ lies on $CG$ , thus $IM\parallel CG\equiv CO$ . Also, $OM\parallel CH$ , thus if $K=IM\cap CH$ , then $\frac{1}{2}CH=OM=CK\implies K$ is the midpoint of $CH$ . Also as $CO,CH$ are isogonal wrt $\angle ACB$ , we get that $\angle HKI=\angle HCO=2\angle HCI\implies CK=CI$ . We conclude that $I$ lies on the circle with diameter $CH$ .

The other direction is exactly the same but in reverse.

$[asy] size(12cm); defaultpen(fontsize(10pt));pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen org=magenta; pen heavy=heavymagenta; pair O,C,A,B,H,I,D,L,K,M,E,G,e; O=(0,0);C=dir(129.9);A=dir(205.6);B=dir(334.4);path w=circumcircle(A,B,C); H=orthocenter(A,B,C);I=incenter(A,B,C);D=2foot(C,A,B)-H;L=intersectionpoints(w,C--100I-99C)[1];M=midpoint(A--B);K=midpoint(C--H); E=foot(I,A,B);G=extension(C,O,A,B);e=2I-E; draw(w,heavyblue);draw(A--B--C--cycle,heavymagenta); draw(C--D,heavymagenta);draw(C--G,heavymagenta);draw(circumcircle(C,I,H),heavyblue+dashed);draw(K--M,heavymagenta);draw(C--L,heavymagenta);draw(E--e,heavymagenta); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$H$",H,dir(H)); dot("$I$",I,dir(I)); dot("$D$",D,dir(D)); dot("$L$",L,dir(L)); dot("$M$",M,dir(M)); dot("$K$",K,dir(K)); dot("$E$",E,dir(E)); dot("$G$",G,dir(G)); dot("$E'$",e,dir(e)); dot("$O$",O,dir(90)); [/asy]$

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