Darn took me 40 minutes...

My solution: Let $ O $ be the circumcenter of $ \triangle AJ_bJ_c $ . From $ AB-BD=AC-CD \Longrightarrow AB+AD-BD=AC+AD-CD$ , so the tangent point of $ \odot (J_b) $ with $ AD $ coincide with the tangent point of $ \odot (J_c) $ with $ AD $ , hence $ AD $ is the common internal tangent of $ \odot (J_b) $ and $ \odot (J_c) \Longrightarrow J_bJ_c \perp AD $ . Since $ AD $ is the $ A- $ altitude of $ \triangle AJ_bJ_c $ , so $ AD, AO $ are isogonal conjugate of $ \angle J_bAJ_c $ , hence we get $ \angle BAO=\angle BAJ_b+\angle J_bAO=\angle BAJ_b+\angle DAJ_c=\tfrac{1}{2} \angle BAC $ . i.e. $ O $ lie on the bisector of $ \angle BAC $ Q.E.D

Cool, isogonal conjugates! I got it after discovering the perpendicularity and lots of angle chasing. (exploiting the angle bisectors)

Cute, but in my opinion too easy for an RMM4. [asy][asy] size(11cm); defaultpen(fontsize(9pt)); pair A = dir(130); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); draw(incircle(A, B, C), blue+dashed); draw(A--B--C--cycle, blue); pair D = foot(I, B, C); pair J_b = incenter(A, B, D); pair J_c = incenter(A, C, D); draw(circumcircle(A, J_b, J_c), heavygreen); draw(J_b--A--J_c, orange); draw(J_b--J_c, heavycyan); draw(incircle(A, B, D), heavycyan); draw(incircle(A, C, D), heavycyan); draw(D--A--I, magenta); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(50)); dot("$D$", D, dir(D)); dot("$J_b$", J_b, dir(280)); dot("$J_c$", J_c, dir(290)); /* Source generated by TSQ */ [/asy][/asy] Because $AB + CD = AC + BD$, it follows that $J_b$ and $J_c$ are tangent to the same point on $AD$. Consequently, $J_bJ_c$ and $AD$ are perpendicular. Moreover, $AD$ and $AI$ are isogonal with respect to $\angle J_bAJ_c$, and we're done. $\blacksquare$

Let the foot of the perpendicular from $J_B$ to $AD$ be $P_B$ and similarly define $P_C$. Then, \[DP_B = AD/2+(BD-AB)/2 = AD/2+(s-b-c)/2 = AD/2+(CD-AC)/2 = DP_C\]Then define $P \equiv P_B \equiv P_C$. So $AP \equiv AD$ is the $A$-altitude in $AJ_BJ_C$. Let $O$ be the circumcenter of $AJ_BJ_C$. Then, $AO$ and $AD$ are isogonal w.r.t $AJ_BJ_C$ and some trivial angle chasing shows that $AD$ and $AI$ are too. So, $O$ lies on $AI$.

Are you expecting tons of double honorable mentions? (do you think that this year's RMM is easier than IMO 2014?) Synthetically got this in 40 mins, and problem 1 is really trivial. (Maybe bronze will be in the 20s...)

The above three solutions are literally identical. Is there an another solution?

easy problem and thanks @v_enhance for your nice diagram. solution = with $AB+CD = AC+BD$ therefore incircles of both $ABD,ACD$ are tangent to $AD$ at same point. now , let incentre of $ABD$ be $J$ and incentre of $ACD$ be $K$. now $\angle KJD = 90-\angle ADB/2 $ also, by some angle chase we get , $\angle AJK = (\angle B +\angle ADB)/2$ and $\angle AKJ = 90+C/2-\angle ADB/2$ now let circumcentre of $AJK$ be $O$ than , $\angle KAO = 90-(\angle C+\angle ADB)/2$ and trivial angle chase shows $\angle KAC = \angle KAD = (\angle ADB -\angle C)/2 $ and thus , $\angle OAC = \angle KAC+\angle KAO = 90-B/2-\angle ADB/2+\angle ADB/2 - C/2 = A/2$ and hence $O$ lies on angle bisector of $\angle BAC$. hence proved how much do you rate this problem on IMO level??

aditya21 wrote: easy problem and thanks @v_enhance for your nice diagram. You're welcome aditya21 wrote: how much do you rate this problem on IMO level?? I would say a 1/4, or very easy 2/5 (say IMO 2009/2).

Who is author?

I believe it's a Russia proposal, though I don't know the exact author. I guess we'll find out once the official solutions come out on their website. (As long as I'm saying this, I'll also add that #1 is from Peru, #2 and #3 are from UK and #5 and #6 are from Bulgaria.)

v_Enhance wrote: I believe it's a Peruvian proposal... v_Enhance wrote: (As long as I'm saying this, I'll also add that #4 is from Russia...) Don't these two sentences contradict each other? From what I saw in Geoff Smith's FB post, #1 is from Peru and #4 is from Russia.

Is this even IMO level, with an average score of 6.1/7? How about problem 1 with an average of 6.8/7? I believe problem 5 should be problem 4 and problem 5 replaced with a harder one

I think this is a fair enough choice for a P4.Actually this should be a model P4.

AnonymousBunny wrote: v_Enhance wrote: I believe it's a Peruvian proposal... v_Enhance wrote: (As long as I'm saying this, I'll also add that #4 is from Russia...) Don't these two sentences contradict each other? Ack, sorry. I had it in my head this was the thread for #1. You're right, of course.

By some quick angle chasing, we see that lines $AD$ and $AI$ are isogonal in $\angle J_bAJ_c$. So, it suffices to prove that $AD$ is an altitude in triangle $J_bAJ_c$. This would follow if we prove that both the incircles are tangent to $AD$ at the same point, which follows if we show that the tangents from $D$ to both the incircles have the same length. That is easy to see using the fact that the tangent from $A$ to the incircle in triangle $ABC$ has length $s - a$, $s$ being the semiperimeter.

It is well known that the incenters of $\triangle ABD$ and $\triangle ACD$ are at the same point on $AD$. Therefore, we know that $AD$ is the altitude from $A$ to $I_BI_C$. It is easy to see that if we let $\angle BAI_B=\alpha$ and $\angle I_CAC$ =\beta,$ then we have that we have that $\angle IAI_B=(\alpha+\beta)-\alpha=\beta,$ so it follows that $AD$ and $AI$ are isogonal. But $AD$ and $AO$ are also isogonal, so we are done.

what Let $I$ be the incenter (mostly just for notation reasons - I just need a fast way to write the angle bisector) The length of the tangent from $D$ to the incircle of $\triangle ABD$ equals $\frac{c+(s-b)+AD}{2}-c=\frac{s+AD-b-c}{2}$. Since this is symmetric in $b$ and $c$, it follows that this is also the length of the tangent from $D$ to the incircle of $\triangle ACD$. Thus the incircles touch $\overline{AD}$ at the same point and $\overline{AD} \perp \overline{J_BJ_C}$. On the other hand, $\angle J_CIA=\frac{1}{2}\angle A-\angle CAJ_C=\angle J_BAJ_C-\angle DAJ_C=\angle J_BAD$, so $\overline{AD}$ and $\overline{AI}$ are isogonal and the conclusion follows. $\blacksquare$

Let $\angle A=\alpha$ etc, and let $\angle ABD=\theta$. Note that $$\angle I_BAD=\frac{\theta}{2}$$and $$\angle I_CAI=(\frac{\alpha}{2})-(\frac{\alpha-\theta}{2})=\frac{\theta}{2},$$so $AI$ and $AD$ are isogonal in $\triangle I_bAI_c$. This means that it suffices to show that $AD\perp I_bI_c$, since the $AO$-cevian is isogonal to the altitude. To do this, we will show that the foot from $I_b$ to $AD$ is the same as the foot from $I_c$ to $AD$. Let the former foot be $F_b$ and the latter be $F_c$. We have $$DF_b=\frac{DB+DA-AB}{2}$$and similarly $$DF_c=\frac{DC+DA-AC}{2},$$Thus, it suffices to show that $\frac{DB+DA-AB}{2}=\frac{DC+DA-AC}{2}$ $$AC+DB=AB+CD$$$$b+(s-b)=c+(s-c),$$which is clearly true and we are done.

Sol sketch Claim 1: The incircles of triangles $ABC, ACD$ are tangent. (Proven by standard length formulae) Claim 2: $J_BJ_C \perp AD$ From here angle chasing finishes.

Supose that $D_b, D_c$ are orthogonal projections $J_b, J_c$ on $BC;$ $E$ be orthogonal projection of $J_b$ on $AD$. We have $$DD_b = DB - D_bB = \dfrac{BC + AB - CA}{2} - \dfrac{BD + AB - DA}{2} = \dfrac{CD + DA - CA}{2} = DD_c$$We also have $DE = DD_b = DD_c,$ then $J_cE \perp AD$ or $AD \perp J_bJ_c$ at $E$. Suppose that $I$ is incenter of $\triangle ABC$. We have $$\angle{DAJ_b} = \dfrac{\angle{DAB}}{2} = \dfrac{\angle{BAC} - \angle{DAC}}{2} = \angle{IAC} - \angle{J_cAC} = \angle{IAJ_c}$$Hence $AD, AI$ are isogonal conjugate in $\angle{J_bAJ_c}$ or $AI$ passes through circumcenter of $\triangle AJ_bJ_c$