Another one which can be solved with polars and poles, huh? $N$ is on $BC$ and $B'C'$, which are the polars of $A'$ and $A$ respectively (wrt the incircle). This means that $AA'$ is the polar of $N$. In the same way we find that $BB'$ is the polar of $M$, so $P=AA'\cap BB'$ must be the pole of $MN$, so $PI\perp MN$ (it's well-known that the polar of a point wrt a circle is perpendicular to the line joining the point to the center of the circle). This time the solution is suitable, because the problem was proposed to the seniors.

I knew that you would solve it in a jippy with polars andrei negut did it in the same way.

I think we can do it by vectors. Choose any point as the origin, the vector IP will be proporional to u*OA+v*OB+w*OC, MN to (v-w)OA+(w-u)OB+(u-v)OC u-a-b,v=b-c,w=c-a. By multiplying them you must get 0. But still, does there exist a pure geometrical solution?

The problem is taken from Kvant, where a pole-polar solution was given (I think). However, almost all the contestants who solved it, used complex numbers or vectors and a lot of computations

Valentin Vornicu wrote: The problem is taken from Kvant, where a pole-polar solution was given (I think). However, almost all the contestants who solved it, used complex numbers or vectors and a lot of computations Almost is well said. I didn't solve it during the contest. I'm such an idiot not computing it with complex numbers. It's really not that hard. It's NOTHING SMART in computing but you get 7 points for it. I was afraid (in the contest) that I will run out of time if I start computing. But when I saw the results of Rom TST I was very very very ... sad (and Valentin can confirm this) because I didn't have the courage to compute. Well that's that. There is allways next year (No there isn't :( ) P.S. A piece of advice : When you see a problem which can be computed with vectors, complex numbers etc. DO IT otherwise you MIGHT regret that you didn't LATER.

yes, that is a well-known fact, at least for some of us

Please iura, would you explain why the vector $\vec{IP}$ is proportional to what you say?

Valentin Vornicu wrote: Let $I$ be the incenter of the non-isosceles triangle $ABC$ and let $A',B',C'$ be the tangency points of the incircle with the sides $BC,CA,AB$ respectively. The lines $AA'$ and $BB'$ intersect in $P$, the lines $AC$ and $A'C'$ in $M$ and the lines $B'C'$ and $BC$ intersect in $N$. Prove that the lines $IP$ and $MN$ are perpendicular. If you do not like polar map. You can also invert with the inccircle. More interesting fact: the two Soddy points are on IP (so this is called the Soddy line) and is perpendicular to the Georgon (spelling?) line. Use nice inversions solve all this. (I have a short written piece on this.) The computational proofs of this also appeared on American Math Monthly.

I have that $\vec{MN}$ is collinear with $(b-c)OA+(c-a)OB+(a-b)OC$ and $\vec{IP}$ is collinear with $\sum[(b+c)(a+c-b)(a+b-c)-2a^2(b+c-a)]OA$, but I can't see why do we have $MN\cdot IP=0$. Can someone help me, please, I'm not very common with vectors?

sbos wrote: I have that $\vec{MN}$ is collinear with $(b-c)OA+(c-a)OB+(a-b)OC$ and $\vec{IP}$ is collinear with $\sum[(b+c)(a+c-b)(a+b-c)-2a^2(b+c-a)]OA$, but I can't see why do we have $MN\cdot IP=0$. Can someone help me, please, I'm not very common with vectors? sbos, how did you find those formulas? I used barycentrics, got the first one right, but the second one is very messy... I think that a straightforward check of $\overrightarrow{MN} \cdot \overrightarrow{IP}=0$ works, i.e. take $O=C$, expand everything, use the cosine theorem etc.; now it's just a potentially monstrous identity to check (if you have some computer algebra software, you can do it; otherwise i think it would a waste of time). Perhaps others can find a nicer approach to proving this... Related question: Say $\overrightarrow{m}=x\overrightarrow{OA}+y\overrightarrow{OB}+z\overrightarrow{OC}$ and $\overrightarrow{n}=t\overrightarrow{OA}+u\overrightarrow{OB}+v\overrightarrow{OC}$. Are there any necessary/sufficient symmetric conditions on $x,y,z,t,u,v$ in order to have $\overrightarrow{m} \cdot \overrightarrow{n}=0$? [edit nr. 2] I have found in Paul Yiu's "Introduction to Triangle Geometry" a way to do this: consider the barycentrics of the infinite points of $IP$ and $MN$; then check that $\sum S_A f f' = 0$, where $S (f:g:h)$ and $T(f':g':h')$ are the infinite points I was talking about, and $S_A = \frac12 \left( b^2+c^2-a^2 \right)$ etc; thus, the problem reduces to proving $\sum f(a,b,c) =0$ (I can't find the paper where I wrote $f$ right now). [edit nr. 1] just wondering: is there really a solution using only vectors or complex numbers? It seems impossible to get the perpendicularity from what I obtain

I note the points $X\in AA'\cap IN,\ Y\in BB'\cap IM$. From the wellknown relations $IN\perp AA',\ IM\perp BB'$ results $IX\cdot IN=IY\cdot IM=r^2$, i.e. the line $MN$ is the image of the circumcircle with the diameter $[IP]$ through the inversion with the pole I and the constant $r^2$. Therefore, $IP\perp MN.$

The lines A'A passes through the intersection A of the tangents CA, AB to the circumcircle (I) of the contact triangle $\triangle A'B'C'$ at the vertices B', C', hence, it is its A'-symmedian. Likewise, B'B is its B'-symmedian, C'C its C'-symmedian and the 3 symmedians concur at its symmedian point P. The tangent CA to the circumcircle (I) of the $\triangle A'B'C'$ at the vertex B' meets the opposite side C'A' at the center M of its B'-Apollonius circle (M), the tangent BC at the vertex A' meets the opposite side B'C' at the center N of its A'-Apollonius circle (N) and the tangent AB at the vertex C' meets the opposite side A'B' at the center K of its C'-Apollonius circle (K). The B'- and A'-Apollonius circles (M), (N) meet at 2 points $Q_1, Q_2$ and by transitivity of equivalence, the C'-Apollonius circle (K) also passes through these 2 points. Hence, their centers M, N, K are collinear and $Q_1Q_2 \perp MN$ is their common radical axis. The circumcenter $I \in Q_1Q_2$ of the contact triangle $\triangle A'B'C'$ lies on this radical axis, because the circumcircle (I) is perpendicular to all 3 Apollonius circles (M), (N), (K). Let the A'-symmedian meet the circle (I) again at A''. Then $A''A \equiv A'A$ is the A''-symmedian of the triangle $\triangle A''B'C'$ and the triangles $\triangle A'B'C', \triangle A''B'C$ also have the common circumcircle (I) and the A'-, A''-Apollonius circle (N). Hence, the common symmedian $A'A \equiv A'A''$ is the radical axis of the circles (I), (N). Likewise, the symmedians B'B and C'C are radical axes of the circle pairs (I), (M) and (I), (K). By transitivity of equivalence, the power of the symmedian point P to all 3 Apollonius circles is the same as the power of this point to the circle (I), hence, $P \in Q_1Q_2$ lies on their radical axis. As a result, $IP \equiv Q_1Q_2 \perp MN$. The line IP is the Brocard axis of the triangle $\triangle A'B'C'$.

What is the Brocard axis?

Dear Armpist and Mathlinkers, I discovered the problem now... ABC and A'B'C' are perspective and bilogic. Accordling to Sondat's theorem, we are done... See for example: httrp://perso.orange.fr/jl.ayme vol. 1 Le theorem de Sondat Thank you again, dear Armpist... Sincerely Jean-Louis

Let AA’ and BB’ intersect 2nd time the circle I at Q and R respectively; since the tangents at B’ and C’ concur on AQ, A’B’QC’ is harmonic, therefore the tangents at A’ and Q intersect on B’C’, hence at N. similarly, AC, A’C’ and the tangent at R concur at M. Now, I has equal power to the circles (M, MB’) and (N,NA’) – IB’ and IA’ are tangents to these circles, but they are inradii of DABC, and the same for P, since PA’.PQ = PB’.PR (from the power of P to (A’B’C’), hence PI is the radical axis of the 2 circles centered at M and N, consequently PI and MN are perpendicular. Best regards, sunken rock

Here my solution: Let $ AA' \cap (I)=A'' $ and $ BB' \cap (I)=B'' $.We know that $ A''A' $ and $ B''B' $ are polars of $ M $ and $ N $ , respectively.If $ B''A'' \cap A'B'=T $ and $ A'B'' \cap A''B'=S $ , then $ {I,P,T,S} $ is orthocentric system.Now, apply Pascal's theorem to non-convex hexagon $ A''B''A'A'B'A'' $ , we get $ A'A' \cap A''A''=N $,$ B''A' \cap A''B'=S $ and $ B'A' \cap B''A''=T $ are collinear.Similarly, we have $ M \in ST $.So $ IP \perp ST=MN $.DONE !

Valentin Vornicu wrote: The problem is taken from Kvant, where a pole-polar solution was given (I think). However, almost all the contestants who solved it, used complex numbers or vectors and a lot of computations Can you tell me exactly which Kvant have you taken this question? I mean which year's Kvant magazine?

Dear Mathlinkers, also at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=383421 Sincerely Jean-Louis

Poles and Polars are back! Valentin Vornicu wrote: Let $I$ be the incenter of the non-isosceles triangle $ABC$ and let $A',B',C'$ be the tangency points of the incircle with the sides $BC,CA,AB$ respectively. The lines $AA'$ and $BB'$ intersect in $P$, the lines $AC$ and $A'C'$ in $M$ and the lines $B'C'$ and $BC$ intersect in $N$. Prove that the lines $IP$ and $MN$ are perpendicular. Alternative formulation. The incircle of a non-isosceles triangle $ABC$ has center $I$ and touches the sides $BC$, $CA$ and $AB$ in $A^{\prime}$, $B^{\prime}$ and $C^{\prime}$, respectively. The lines $AA^{\prime}$ and $BB^{\prime}$ intersect in $P$, the lines $AC$ and $A^{\prime}C^{\prime}$ intersect in $M$, and the lines $BC$ and $B^{\prime}C^{\prime}$ intersect in $N$. Prove that the lines $IP$ and $MN$ are perpendicular. Say $AA' \cap \odot(I)=A'',BB' \cap \odot(I)=B''$ Clearly $A''B'A'C',B''A'B'C'$ are harmonic. So the tangents to the incircle through $A',A''$ meet at $N$,So $AA'$ is the polar of $N$ and similarly $BB'$ is the polar of $M$,by La Hire $MN$ is the polar of $AA' \cap BB'=P$ w.r.t $\odot(I)$ so $IP \perp MN$ $\blacksquare$

We use poles and polars wrt to the incircle Note that: $a \equiv B'C' \implies a \equiv N$ by La Hire’s Theorem $n \equiv A \implies P \in n \implies p \equiv N$ Similarly $P \in m \implies p \equiv M$ Then: $MN \equiv p \implies IP \perp MN$.$\blacksquare$