Do you mean J,I are excenters of excircles inscribed in angles ABH and ACH ?

@Math-lover123, no tricky wording here. J is simply the B-excenter of ABH and I is simply the C-excenter of ACH. The concyclicity is true for any H on BC, not necessarily the foot of the A-altitude as the problem states. See two problems about cyclic quadrilateral (problem 1), incenters and cyclic, etc. The proof is analogous for excircles.

nice problem let $X,S$ be feet from $I,J$ to $BC$ then. $XP=BX-BP=((BH+BA+AH)-(BC+BA-CA))/2=(AH-CH+AC)/2=HS=SJ$ and similarly $SP=XI$ and since $\angle IXP=\angle JSP=90$ then $JSP\cong IXP$ so $\angle IPX=\angle SJP=90-\angle SPJ$ so $\angle IPJ=90=2*45=\angle IHJ$ therefore $IHPJ$ is cyclic.

First of all we've $\frac {Cos(A-B)/2-Cos C/2}{Cos (A-B)/2-Cos A/2}+\frac{Cos B}{Cos(B-C)/2}{2Sin B/2}=2Cos C/2$. Now note $\frac {S-c}{CI}=\frac {Cos(A-B)/2-Cos C/2}{Cos (A-B)/2-Cos A/2}$ and $\frac{Cos B}{Cos(B-C)/2}{2Sin B/2}=\frac {Cos B/2}{Sin(B+A/2)}\times \frac {BH}{AH}=\frac {Cos(C/2+\alpha)}{Sin(\alpha)}$ now also we had $\frac {Cos(C/2-\theta)}{Sin(\theta)}=\frac {S-c}{CI}$. Where $\angle{HIA}=\theta,\angle{PJC}=\alpha$. Now so we've $\frac {Cos(C/2+\alpha)}{Sin(\alpha)}+\frac {Cos(C/2-\theta)}{Sin(\theta)}=2Cos C/2$ and that implies $\frac {Cos(C/2+\alpha)}{Sin(\alpha)}=\frac {Cos(C/2+\theta)}{Sin(\theta)}$ now so we've $Cot (\alpha)=Cot(\theta)$ and certainly that implies $\theta=\alpha$ and so done.

In fact, $H$ can be any point on $BC$. For this, if we denote $I', J'$ as the projections of $I, J$ onto $BC$, then we have to prove that $II'P\sim PJJ'$. Moreover, assume without loss of generality that $BH<BP$. Then the problem is equivalent to proving that(after a few simplifications) \[4xybc\sin B \sin C = ((c+t)^2-x^2)((b+t)^2-y^2)\] After using cosine rule and simplifying, we have to prove \[bc\sin B \sin C = t^2(1+\cos\alpha)(1-\cos\alpha)\] where $\angle AHB = \cos\alpha$. This is obvious by sine rule in triangles $AHB, AHC$. Previously, I had copied the question wrong and got the following result: In triangle $ABC$, $H$ is any point on $BC$ and the $A$-excircle touches $BC$ at $P$. Also, $I, J$ are the $H$-excentres of triangle $ABH, ACH$ respectively. Then $IHPJ$ is cyclic.

Proposed by Mehdi E'tesami Fard

leader wrote: nice problem let $X,S$ be feet from $I,J$ to $BC$ then. $XP=BX-BP=((BH+BA+AH)-(BC+BA-CA))/2=(AH-CH+AC)/2=HS=SJ$ and similarly $SP=XI$ and since $\angle IXP=\angle JSP=90$ then $JSP\cong IXP$ so $\angle IPX=\angle SJP=90-\angle SPJ$ so $\angle IPJ=90=2*45=\angle IHJ$ therefore $IHPJ$ is cyclic. Please, why $XP=BX-BP$ ? Please cite the theorems you used, thank you .

It is not a theorem, but I'll explain it for you Since $B,P,X$ lie on $BC$ on this order, i.e, are collinear, it follows that $BX=BP+PX$ and the result follows.

Oh yeah I copied wrong .. But why $ XP=BX-BP=((BH+BA+AH)-(BC+BA-CA))/2 $ ?

Dear tuvie, Please explain the steps which seledur wrote.Because i did not understand too. Thanks...

I still cant believe this, I spent an hour solving the wrong problem! If you are confused, the incircle tangency $P$ is actually the excircle tangency... You have to be kidding me. Here is a quick solution for any point: Suppose $X, Y$ are the tangent points of $(J), (I)$ with $BC$. We get $PX = BP + BX = \dfrac{AB + AF - BF - AB + AC + BC}{2} = \dfrac{AC + AF + FC}{2} = FY$. Since $FJX \sim FIY \implies \dfrac{JX}{FY}=\dfrac{FX}{IY} \implies \dfrac{JX}{PX}=\dfrac{PY}{IY} \implies DJX \sim DIY$ so done. That was a really really bad typo seriously.

can someone tell me what is excenter?

@gearss: Look at the first image from: http://en.wikipedia.org/wiki/Incircle_and_excircles_of_a_triangle

Dear Mathlinkers, for more precision, you can see http://perso.orange.fr/jl.ayme vol.5 Le theoreme de Feuerbach-Ayme p.10-13 Sincerely Jean-Louis

Let $I',J'$ be projection of $I$ and $J$ onto line $BC$. Then , by properties of $Ex-circle$. We see that, $BJ' = \frac{AB+BH+AH}{2}$ and $BP = \frac{AB+BC-AC}{2} \implies PJ' = \frac{AH+AC-HC}{2}$. Now, $II' = \frac{AH \cdot HC}{AC+CH-AH}$. Now using Pythagoras theorem, one see that $PJ'=II'$. Similarly, $PI'=JJ'$. Thus we have $\triangle PII' \cong PJJ'$ Thus $\angle IPJ = 90^{\circ}$ and easy to see that $\angle IHJ = 90^{\circ}$. So $IHPJ$ is a cyclic quadrilateral. and $II'$ is its diameter. So we are done! $\Box$.

$BJ' = \frac{AB+BH+AH}{2}$ Sorry but how do you get this ?

gearss wrote: can someone tell me what is excenter? Just google it!!!

Let $M$ be the midpoint of $IJ$ and $X,Y$ be the projection of $I ,J$ onto $BC$ After getting that $YP=CH , XP=YH$ then $YP.YH=XP.XH$ thus $Y , X $has the same power wrt $(PHI)$ let its center be $M'$ then $XM'=YM'$ thus $M'$ is the intersection of the perpindacular bisectors of$ XY $and$ IH $and we have $M$ is on the perpindacular bisector of $XY$ thus $M=M'$

G #2/100 Solved a slightly different variation of the problem where $D$ is the $A$ intouch of the triangle $ABC$ instead Let the projections of $X$ and $Y$ onto $BC$ be $N$ and $M$ respectively Firstly, observe that $X$ and $Y$ are the $B$ and $C$ excenters of $\triangle ABP$ and $\triangle ACP$ respectively $$NP=BN-BP=\frac{AB-BP+AP}{2}$$$$MD=CM-CD= \frac{AP+AC+PC}{2}-\frac{AC+BC-AB}{2}=\frac{AB+AP-BP}{2}=NP$$ By symmetry, we also have $MP=ND$ So, $MP^2+NP^2=ND^2+MD^2$ $YP^2+PC^2=YD^2+DX^2$ $$YD^2+DX^2=YX^2$$$$\angle{YDX}=90$$ Hence, $YDPX$ concyclic

Just use one of the circle theorema that is one of the angle of the quardarateral equals to the other side of the opposite angle.

well if you missed the insane length solution you can also reduce to serbia 2018 and use MMP!!!! yay!!!!

Relabel $I, J$ as $X, Y$ and $H$ as $D$. Let $S$ be the foot from $Y$ onto $\overline{BC}$ and $T$ the foot from $X$. Now note that it suffices to show that $$SD^2 - SP^2 = TP^2 - TD^2.$$Note that \begin{align*} SD &= \frac 12\left(AP+AC-PC+2PD\right) \\\ SP &= \frac 12\left(AP+AC-PC\right) \end{align*}so \begin{align*} SP^2 - SD^2 &= (SD-SP)(SP+SD) \\ &= PD \cdot (AP+AC-CD) \\ &= PD(b+c-s) \end{align*}which is symmetric in $b$ and $c$. Hence done!

Silly Problem

Let $E$ and $F$ be the foot of altitude from $I$ and $J$ respectively Let $M$ be the midpoint of $DP$, and let $N$ be the midpoint of $EF$, let $H$ be the midpoint of $IJ$ Claim 1: $M \equiv N$

Now this implies the conclusion as $HM$ becomes the midline in trapezoid $AJFE$.

The LaTeX gave me brain damage... Note that $\angle XPY$ is a right angle, so $(PXY)$ has diameter $\overline{XY}$. Hence, it suffices to prove Let $S$ and $R$ be the feet of the altitudes from $X$ and $Y$ to $\overline{BC}$, respectively. Hence, we are able to write \begin{align*} DX^2 &= XS^2+DS^2 \\ DY^2 &= YR^2+DR^2 \\ PX^2 &= XS^2 + PS^2 \\ PY^2 &= YR^2+PR^2. \end{align*} This means it is sufficient to prove \[DR^2+DS^2 = PR^2+PS^2\]\[\iff DS^2-PS^2 = PR^2-DR^2.\] WLOG let $R,D,P,S$ lie on $\overline{BC}$ in that order. Notice that $X$ is the definition of the $B$-excenter of $\triangle ABP$. This means that \begin{align*} DS+PS &= BS+PS-BD \\ &= \frac{1}{2}(AB+AP+BP)+\frac{1}{2}(AB+AP-BP)-\frac{1}{2}(AB+BC-AC) \\ &= \frac{1}{2} (AB+AP+BP+AB+AP-BP-AB-BC+AC) \\ &= \frac{1}{2} (AB+AC-BC+2AP). \end{align*} Similarly, $Y$ is the $C$-excenter of $\triangle ACP$, so \begin{align*} PR+DR &= CR+PR-CD \\ &= \frac{1}{2}(AC+AP+CP)+\frac{1}{2}(AC+AP-CP) - \frac{1}{2}(AC+BC-AB) \\ &= \frac{1}{2} (AC+AP+CP+AC+AP-CP-AC-BC+AB) \\ &= \frac{1}{2} (AB+AC-BC+2AP). \end{align*} Thus, \begin{align*} DS+PS&=PR+DR \\ \iff DS^2-PS^2 &=(DS-PS)(DS+PS) \\ &= DP(DS+PS) \\ &= DP(PR+DR) \\ &= (PR-DR)(PR+DR)=PR^2-DR^2. \ \square \end{align*}

Since $\angle XPY = 90^{\circ}$, we only need to prove that $XYD = 90^{\circ}$, or equivalently, $XY^2 = YD^2 + XD^2$. Since $XY^2 = XP^2 + YP^2$, it suffices to prove that $XP^2 + YP^2 = XD^2 + YD^2$. Let $K, L$ be the feet of altitudes from $Y, X$ to $BC$, respectively. Then, we have $CK = \frac{AC + CP + AP}{2}, BL = \frac{AB + BP + AP}{2}$ and $BD = \frac{AB + BC - AC}{2}$. From this, one can verify that $KD = LP$, immediately implying $XP^2 + YP^2 = XD^2 + YD^2$. $\blacksquare$

This problem is true for arbitrary point $H$ lies on $BC$. Relabel point $I, J$ of the problem as $J_1, J_2$. Let $I$ be incenter of $\triangle ABC$ and $D_1, D_2$ be orthogonal projections of $J_1, J_2$ on $BC$. Suppose that $J$ is midpoint of $J_1J_2$. Since $\angle{J_1HJ_2} = 90^{\circ},$ we have $JH = \dfrac{J_1J_2}{2}$. To prove $P \in (J_1J_2H),$ we need to show that $J$ lies on perpendicular of $HP$. But $J$ also lies on perpendicular of $D_1D_2$ then it suffice to prove that $D_1H = D_2P$. Indeed, we have $D_2P = BD_2 - BP = \dfrac{AB + BH + AH}{2} - \dfrac{AB + BC - CA}{2} = \dfrac{CA + AH - CH}{2} = CD_1 - CH = D_1H$. So it means that $J$ lies on perpendicular of $HP$. Hence $JH = JP = \dfrac{J_1J_2}{2}$ or $P \in (J_1J_2H)$

So this problem is actually true for all $H,$ so let the point on $BC$ be $P,$ and the A-intouch point be $D.$ Note that $X,Y$ are the excenters of $\triangle ABP, \triangle CPA,$ respectively. Let $K$ be the foot from $X$ to $BC,$ and $L$ the foot from $Y.$ Then note that we simply need to show $KL=PK$ by Pythagorean theorem. Let $DP=x.$ Then because $BD=\frac{a-b+c}{2}$ and $BK=\frac{AB+BP+PA}{2}, CL=\frac{AC+CP+PA}{2},$ we are done by lengths$.\blacksquare$