This is also from the ISL. We have $PA^2+PD^2=PD^2+PE^2+EA^2=PB^2+PE^2=PD^2+PE^2+BD^2$ $\Rightarrow AE=BD$. In the same way we show that $CE=BF,\ AF=CD$, so we can see that $D, E, F$ are the projections of $I_a,I_b,I_c$ on $BC,CA,AB$ respectively. The conclusion follows.

yes thats like the only solution by the way, how did you know that was from ISL 2003? [Moderator edit: The problem can be extended: Let $ABC$ be a triangle, and $P$ a point in the plane. Let $D$, $E$, $F$ be the feet of the perpendiculars from the point $P$ to the lines $BC$, $CA$, $AB$, respectively. Furthermore, let $I_{a}$, $I_{b}$, $I_{c}$ be the excenters of triangle $ABC$. Then, prove that $AP^{2}+PD^{2}=BP^{2}+PE^{2}=CP^{2}+PF^{2}$ holds if and only if the point $P$ is the circumcenter of triangle $I_{a}I_{b}I_{c}$. ]

Oh, I have the ISL problems .

grobber wrote: Oh, I have the ISL problems . We (the romanian committee) gaved the problems to the students at RoMOP to be studied. Of course, as these problems were already used in some competitions, it is ok to put them online. All other ISL problems are still not to be disclosed

I only have whatever you gave to the students at the training camp, because actually, that's how I got the problems: I know quite a few of those students .

grobber wrote: This is also from the ISL. We have $PA^2+PD^2=PD^2+PE^2+EA^2=PB^2+PE^2=PD^2+PE^2+BD^2$ $\Rightarrow AE=BD$. In the same way we show that $CE=BF,\ AF=CD$, so we can see that $D, E, F$ are the projections of $I_a,I_b,I_c$ on $BC,CA,AB$ respectively. The conclusion follows. I had same but didn't see why the conclusions follow, anyone can explain?

PP (IMO 2003). Let $ABC$ be a triangle and $P$ a point in the interior of this triangle. Let $D$ , $E$ , $F$ be the feet of the perpendiculars from the point $P$ to the lines $BC$, $CA$, $AB$ respectively. Assume that $AP^{2}+PD^{2}=BP^{2}+PE^{2}=CP^{2}+PF^{2}$ . Furthermore, let $I_{a}$, $I_{b}$, $I_{c}$ be the excenters of triangle $ABC$ . Show that the point $P$ is the circumcenter of triangle $I_{a}I_{b}I_{c}$ . Proof. Observe that $PA^2+PD^2=$ $PE^2+EA^2+PD^2$ and $PB^2+PE^2=$ $PD^2+DB^2+PE^2$ $\Longrightarrow$ $AE=BD$ . In the same way we show that $CE=BF$ and $AF=CD$ . Prove easily that $D$ , $E$ , $ F$ are the projections of $I_a$ , $I_b$ , $I_c$ on $BC$ , $CA$ , $AB$ respectively, i.e. $AE=BD=s-c$ , $AF=CD=s-b$ , $BF=CE=s-a$ . Denote the projection $F'=\mathrm{pr}_{AB}I_a$ . Observe that $BF'=BD$ and $b=(s-a)+(s-c)=BF+BF'=\mathrm{pr}_{AB}[PI_a]=PI_a\cdot\sin\widehat{PI_aF'}=PI_a\cdot\sin B$ $\implies$ $PI_a=\frac {b}{\sin B}=2R$ , where $R$ is the length of the circumcenter for $\triangle ABC$ . In conclusion, $PI_a=PI_b=PI_c=2R$ , i.e. the point $P$ is the circumcenter of the triangle $I_{a}I_{b}I_{c}$ . Otherwise. Prove easily that the triangle $I_aPI_b$ , $I_bPI_c$ and $I_cPI_a$ are isosceles , i.e. $PI_a=PI_b=PI_c$ . Observe that in $\triangle I_aPI_b$ exists the relation $\frac {PI_a}{\sin\frac C2}=\frac {I_aI_b}{\sin C}$ . Therefore, $PI_a=\frac {I_aI_b}{2\cos\frac C2} =$ $\frac {r_a+r_b}{2\cos^2\frac C2}=$ $\frac {\frac {S}{s-a}+\frac {S}{s-b}}{\frac {2s(s-c)}{ab}}=$ $\frac {abcS}{2s(s-a)(s-b)(s-c)}=2R$ . In conclusion, $PI_a=PI_b=PI_c=2R$ . Otherwise. If $A'$ , $B'$ , $C'$ are the intersections of $AI$ , $BI$ , $CI$ respectively with the circumcircle of $\triangle ABC$ , then $A'B'C'\sim I_aI_bI_c$ , where $I\in A'I_a\cap B'I_b\cap C'I_c$ and $B'C'\parallel I_bI_c$ , $C'A'\parallel I_cI_a$ , $A'B'\parallel I_aI_b$ (homothety). Since $2\cdot IA'=II_a$ obtain that the length of the circumradius for $\triangle I_aI_bI_c$ is equally to $2R$ , where $R$ is the length of the circumradius for $\triangle A'B'C'$ , i.e. the circumradius of $\triangle ABC$ . See and here.

Suppose $P$ is the circumcentre of $\triangle I_aI_bI_c$ (which is the isogonal conjugate of the orthocentre of $\triangle I_AI_bI_c$). We have $\triangle ABC$ is the orthic triangle of $\triangle I_aI_bI_c$, so $I_aP \perp BC$ and similaarly for others. Therefore, $D, E, F$ is just the nagel triangle of $\triangle ABC$, which means $BD = AE = s-c$. So, by pythagorean theorem we have $AP^2 - PE^2 = AE^2 = BD^2 = BP^2 - PD^2$. So, done.

Sorry, where do you know that $D, E,$ and $F$ are the points where the excircles touch the sides of triangle $ABC$ ?

rizkydarmawan wrote: Sorry, where do you know that $D, E,$ and $F$ are the points where the excircles touch the sides of triangle $ABC$ ? Note that $\triangle ABC$ is just the orthic triangle of $\triangle I_AI_BI_C$. Since $P$ is the circumcentre, which is the isogonal conjugate of the orthocentre, and $\triangle ABC$ is the cevian triangle of the orthocentre, $I_AP \perp BC$ hence we deduce that it is the nagel triangle Note that i mentioned this in my proof...

This problem reappeared in India TSTs 2004.I cannot get why they repeated the same problem...... Bye... Sayantan....

That's because most of the countries use problems from the ISL for their TST's (which is why the ISL is kept strictly confidential until the next IMO).

Lemma : $P$ is the Bevan point of triangle $\triangle ABC$ $AP^2 + PD^2 = BP^2 + PE^2$ $\implies AE^2 + PE^2 + PD^2 = BP^2 + PE^2$ $\implies AE^2 = BD^2$ $\implies AE = BD$ Similarly, $CP=PA$ and $CE = BF$ $\implies D,E,F$ are the extouch points of $\triangle ABC$ Now since, $PD,PE,PF$ are perpendiculars to sides $BC,CA,AB$ $\implies P$ is the Bevan point, that is the circumcentre of the extriangle. QED

Seems to be original so far based on this thread?

Storage. ISL 2003 G3 wrote: Let $ABC$ be a triangle and let $P$ be a point in its interior. Denote by $D$, $E$, $F$ the feet of the perpendiculars from $P$ to the lines $BC$, $CA$, $AB$, respectively. Suppose that \[AP^2 + PD^2 = BP^2 + PE^2 = CP^2 + PF^2.\]Denote by $I_A$, $I_B$, $I_C$ the excenters of the triangle $ABC$. Prove that $P$ is the circumcenter of the triangle $I_AI_BI_C$. Proposed by C.R. Pranesachar, India $$\begin{cases} AP^2+PD^2=BP^2+PE^2\implies AE=BD \\ AP^2+PD^2=CP^2+PF^2\implies AF=CD \\ BP^2+PE^2=CP^2+PF^2\implies BF=CE\end{cases}$$So from here we get that $AD,BE,CF$ bisects the perimeter of the triangle $ABC$. So, $\{AD,BE,CF\}$ are the $A,B,C\text{ Nagel Cevians}$ respectively. Hence, $\overline{P-D-I_A},\overline{P-F-I_C},\overline{P-E-I_B}$. After that a bit of angle chasing shows that $PI_A=PI_B=PI_C\implies P$ is the $\text{Bevan Point}$ or Circumcenter of the Excentral Triangle of $\triangle ABC$ i.e $I_AI_BI_C$.

This seemed too easy for G3

In fact, in this solution, I will not even refer to $I_a, I_b, I_c,$ as the problem is equivalent to demonstrating that $P$ is the reflection of $I$ over $O,$ where all centers are defined wrt $\triangle ABC$. It is easy to see by doing pairwise subtraction that $AE=BD, DC=AF, EC=FB$. Notice that this essentially shows that the points $D,E,F$ are the extouch points, or that $PF$ is the reflection of the perpendicular from $I$ to $AB$ over the perpendicular from $O$ to $AB$, in other words, they are isotomic conjugates of the intouch points(aka the extouch points). A similar thing hold for $PE,$ and therefore, because the corresponding perpendicular of $P$ are reflections of the perpendiculars from $I$ over the perpendiculars from $O,$ we get that $P$ is the reflection of $I$ over $O$.

AnonymousBunny wrote: That's because most of the countries use problems from the ISL for their TST's (which is why the ISL is kept strictly confidential until the next IMO).

Confidential to the common people and contestants . Since trainers know the problems. My Sol : (I was dumb tho) Bit of length chasing to get $AE=BD,BF=CE,AF=CD$. Thus they are extouch points and now angel chase.

Lemma1 : Let incircle touch BC at D' and A-excircle touch BC at D. CD = BD'. Proof : Let AC = b, AB = c, BC = a, AB+BC+CA/2 = p. Let A-excircle meet AC and AB at S and K. AS = AS+AK/2 = AB+AC+CD+BD/2 = p CD = AS - AC = p - b BD' = p - b Lemma2 : Let A-excircle touch BC at D, B-excircle touch AC at E and C-excircle touch AB at F. AF = CD, AE = BD and BF = CE. Proof : Let incircle touch BC at D', AC at E' and AB at F' CD = BD' = BF' = AF BD = CD' = CE' = AE BF = AF' = AE' = CE. Lemma3 : Only 3 points like D,E,F exists on segments of ABC such that AF = CD, AE = BD and BF = CE. Proof : let assume there are also points X,Y,Z such that AX = CY,AZ = BY and BX = CZ. AX = AF + n so CY = CD + n so BY = BD - n so AZ = AE - n so CZ = CE + n so BX = BF + n witch gives contradiction. AP^2 + PD^2 = AE^2 + PE^2 + PD^2 and BP^2 + PE^2 = BD^2 + PD^2 + PE^2 so AE = BD. same way we can prove AF = CD and BF = CE. From lemma2 and lemma3 we have D,E,F are extouch points in triangle ABC so PDIa, PEIb and PFIc are collinear. ∠PIaIb = 90 - (90 - ∠A/2) = ∠PIbIa ---> PIa = PIb. same way we can prove PIb = PIc P is circumcenter of IaIbIc. we're Done.

P. Since $AP^2+PD^2=BP^2+PE^2,$ we have $AP^2-PE^2=BP^2-PD^2.$ This simplifies to $AE=BD.$ Similarly, $CD=AF,BE=FB.$ Note that letting $AB,BC,CA$ be constants, we have unique values $AE,CD,FB.$ This fixes $D,E,F.$ Note that when $D,E,F$ are exactly the $A,B,$ and $C$ excenter touchpoints on $BC,CA,AB$, respectively, these equalities before satisfy. Thus, $I_BE\perp AC,I_AD\perp BC,I_CF\perp AB.$ This implies that $I_BEP,I_CFP,I_ADP$ are all collinear. Note that $ABC$ is the orthic triangle of $\triangle I_AI_BI_C$ so \[\angle CI_BP=90^\circ-\angle I_BCE=90^\circ-\angle I_BI_CI_A=\angle BI_BI_C.\]This is true for the other vertices of $\triangle I_AI_BI_C$ as well. It follows that $P$ is the isogonal conjugate of $I,$ so while $I$ is the orthocenter, $P$ is the circumcenter.

First, we will prove that the circumcircle of $I_AI_BI_C$ satisfies this property. We will then show that there exists only one point satisfying this property, finishing the problem. Let $\angle I_BI_AI_C = A$, $\angle I_AI_CI_B = C$, $\angle I_CI_BI_A = B$. Then $\angle I_API_B = 2C$, so $\angle I_BI_AP = 90 - C$, while since it is well-known that $ABC$ is the orthic triangle of $I_AI_BI_C$, $I_CI_BCB$ is cyclic, so $\angle BCI_A = C$ and $\angle CI_AD = 90 - C$. Therefore, $I_A$, $D$ and $P$ are collinear. The same argument works to prove that $I_B$, $E$ and $P$ are collinear and that $I_C$, $F$ and $P$ are collinear. Now, note that $I_AC = \frac{DC}{cos(C)}$, $I_CI_A = \frac{I_AC}{cos{A}}$, so $DC = I_AI_Ccos(A)cos(C)$. In the same way, $I_CA = AFcos(A)$, $I_CI_A = I_CAcos(C)$, so $AF = I_AI_Ccos(A)cos(C)$. This implies that $AF = DC$. Note that $(AF)^2 = (DC)^2$ implies that $(AF)^2 + (PD)^2 = (PC)^2$, so $(AF)^2 + (PF)^2 + (PD)^2 = (CP)^2 + (PF)^2$, so since $(AF)^2 + (PF)^2 = (AP)^2$, $(AP)^2 + (PD)^2 = (CP)^2 + (PF)^2$. A similar argument works to prove the other two equalities, so the circumcircle of $I_AI_BI_C$ satisfies this property. Now, note on the other hand that by doing the previous argument backwards we are forced to have $AF = DC$, $BF = CE$, $AE = BD$. Note that this implies that $AF - AE = AB - AC$, $BF - BD = BA - BC$, $CD - CE = CB - CA$. Then, note that $BD = BF + BC - BA$, $CD = BC - (BF + BC - BA) = BA - BF$, $CE = CA + BA - CB - BF$, $AE = CA - (CA + BA - CB - BF) = CB + BF - BA$, $AF = BA - AC + CB + BF - BA = CB + BF - AC$, $BF = AB - (CB + BF - AC) = AB + AC - CB - BF$, so $2BF = AB + AC - BC$, and $BF = \frac{AB + AC - BC}{2}$. Therefore, the position of $F$ on line $AB$ is fixed. In the same way, the positions of $D$ and $E$ are fixed, so $P$, the intersection of the perpendiculars to the sides that the three points are on at those points, is fixed, as required.

Clearly, we have $$BF = \sqrt{BP^2 - PF^2} = \sqrt{CP^2 - PE^2} = CE.$$Analogously, we find $CD = AF$ and $AE = BD$. Thus, the $A$-excircle touches $BC$ at $D$, the $B$-excircle touches $CA$ at $E$, and the $C$-excircle touches $AB$ at $F$, implying $\overline{I_ADP} \perp BC$ and similarly for the other two excenters. Now, let $k_a, k_b, k_c$ denote denote the tangents to $(I_AI_BI_C)$ at $I_A, I_B, I_C$ respectively. It's well-known that $ABC$ is the orthic triangle of $I_AI_BI_C$, so $I_BI_CBC$ is cyclic with diameter $I_BI_C$. Hence, $k_a \parallel BC \perp \overline{I_ADP}$ follows from Reim's. Similarly, we deduce $k_b \perp \overline{I_BEP}$ and $k_c \perp \overline{I_CFP}$, which finishes. $\blacksquare$ Remark: This problem is way too easy for a G3.

i'm sorry, what? Note that \[ AE^2 = AP^2 - PE^2 = BP^2 - PD^2 = BD^2 \implies AE = BD \]and similarly $BF = CE$, $CD = AF$ - so $D$, $E$, $F$ are the corresponding excircle touchpoints on $BC$, $CA$, $AB$ respectively. In particular, we deduce $I_AD$, $I_BE$, $I_CF$ concur at $P$. Hence $P$ is the Bevan Point of triangle $ABC$, which is known to be the circumcenter of $I_AI_BI_C$. $\square$

Use Pythagorean Theorem to remove $P$ from the image. \[BD^2-AE^2=CD^2-AF^2\]\[CE^2-BF^2=AE^2-BD^2\]\[AF^2-CD^2=BF^2-CE^2\] which implies all sides are $0$. So $P$ is the Bevan point of $\triangle ABC$, done.

$\color{magenta} \boxed {\textbf{SOLUTION G3}}$ $$AP^2 - PE^2 = BP^2 - PD^2 \implies AE = BD=y$$Similarly, $AF=CD=x$ and $BF=CE=z$ Let, $\textbf{A-excircle}$ meets $BC$ at $D'$ So, $$CD'=s-AC=s-y-z=x=CD \implies D' \equiv D$$$\implies I_A,D,P$ are collinear and $I_AP \perp BC$ Simlarly, $I_BP \perp AC \equiv E$ and $I_CP \perp AB \equiv F$ We need to show, $PI_A=PI_B=PI_C$ $$\angle PI_AI_C=\angle DI_AB=\frac{1}{2} \angle ABC=\angle BI_CF=\angle I_AI_CP \implies PI_A=PI_C$$Similar work gives desired $\blacksquare$

storage

Claim: $D, E, F$ are the $A-, B-$ and $C-$ extouch points, respectively. Proof: Observe from the Pythagorean Theorem that $$(CD^2 + DF^2) + PF^2=CP^2 + PF^2=AP^2 + PD^2 = (AF^2 + FD^2) +PD^2$$$$\implies CD = AF.$$Similarly, $BF = CE, BD = AE$. This is enough to prove the claim. $\square$ Redefining $D,E,F$ as the extouch points, we have $P = I_AD \cap I_BE \cap I_CF$. It suffices to show this point is the circumcentre of $\Delta I_AI_BI_C$, which may be done by an angle chase. $\square$