Let CP meet the circumcircle of triangle ABC at P', then <P'BP=<P'BA+<PBA=<PCA+<PBA=<PBC+<PCB=<P'PB. Thus P'P=P'B. Since <P'=<A, so we have $\angle{BPC}=90^{o}+\frac{A}{2}=\angle{BIC}$. Thus B, P, I, C lie on the same circle (Let's say with centre $D$). But it is a well-known result that $D$ lies on line $AI$, since we have $\angle{DIP}\leq 90^{o}$, so $\angle{AIP}\geq 90^{o}$. Hence $AP \geq AI$ with equality holds iff $I=P$.

(Almost) same solution as mine. This was very easy..... Seems as tough this year will be a year with many many honorable mentions.....

Ok…………………………………………………………………. We have: $ \angle PBA + \angle PCA = \angle PBC + \angle PCB$ If we add $ \angle PBC + \angle PCB$, to both side of the equality we have: $ 2(\angle PBC + \angle PCB) = \angle PBA + \angle PCA + \angle PBC + \angle PCB = \angle B + \angle C$ then, it means that $ \angle PBC + \angle PCB$, is fixed for each $ P$. then we conclude that $ \angle BPC$ is also fixed, and it is clear that, the angle is equal to$ \angle BIC$ then the locus of $ P$ is arc$ BIC$. We know that this arc is tangent to $ AB,AC$ at $ B,C$ respectively. And also know that $ AI$ goes through the center of circle $ BIC$. Then $ I$ is the nearest point of circle to $ A$. And for each point $ P$ on the circle we have: $ AI < AP$ and if $ P$ is $ I$, it is clear that $ AP = AI$. How ever it was nice and simple.

[spam]Finally a geo problem from IMO which I can solve. It is a very nice restate of the fact that the intersection $J$ of the perpendicular bisector of $BC$ with $AI$ satisfies $JB=JI=JC$.[/spam]

You don't need to construct the point P', Shobber, but of course that works. I like the idea of having an easy question for problem 1, but isn't this too simple?

just prove that BIPC is cyclic than let intersection of AI and circle aroun BIPC be V and AP and that circle be W. Than using angles we can prove that IBV is 90 from which we conclude IV is diameter.(2 radius).Than IV>=PW and we use potention of point A,we have AI*AV=AP*AW and if we assume that AI>AP we easy get a contradiction.From that we conclude AP>=AI and easy solve the cases of equality I hope this is enough for 7

The problem is too much easy at IMO stage. My solution is: Actually P is a point on the circum circle of BIC and of course inside ABC. Let P is on arc BI. To prove, $AP \geq AI$ it is sufficient to prove $\angle{AIP}\geq \angle{API}$. $\angle{AIP}=360-\angle{AIC}-\angle{PIC}=360-(90+B/2)-(180-\angle{PBC})=90-B/2+\angle{PBC}$ $\angle{API}=360-\angle{APB}-\angle{BPI}=360-\angle{APB}-(180-\angle{BCI})=180-\angle{APB}+C/2$ $\angle{AIP}\geq \angle{API}$ implies, $90-B/2+\angle{PBC}\geq 180-\angle{APB}+C/2$ implies, $90-B/2+\angle{PBI}+\angle{IBC}\geq 180-\angle{APB}+C/2$ implies, $\angle{PBI}+\angle{APB}\geq 90+C/2$ implies, $\angle{PBI}+\angle{APB}\geq \angle{AIB}$ But, $\angle{APB}\geq \angle{AIB}$ Since P is inside AIB. So the above inequality is true. Equality when P is I and then $\angle{PBI}=0$

What happened to my attached picture? It was there when posted the proof but now it is gone!

As I read "triangle triviality" I thought that it just can't happen at the IMO; but in a while I realized it was true - the only facts I used to solve this problem are very, very basic things like sum of angles in a triangle

My solution was as shobbers solution but at the end i have used inequalities defferently. I hope my solutin is also true Abdurashid

I had a much longer solution to this, not sure if that's a good thing though.. probably isn't. Most people here seem to have got this one right, so the limits for medals are likely to be higher this year than last.

i couldn't believe that this is an imo problem when i saw it.it only took me 10 minutes to solve it(and i'm not good at geometry).i imagine all of you guys did it in....3 min

It was a $\le$10 minute problem for me, too. My take on the problem: First, we note that $P$ must be a point on circle $BIC$; $\angle PBA+\angle PCA-\angle PBC-\angle PCB=\angle CBA+\angle BCA-2\angle PBC-2\angle PCB$ $=(180^\circ-\angle BAC)-2(180^\circ-\angle BPC)=2\angle BPC-\angle BAC-180^\circ$. This is zero if $P=I$ (by the original form), so $\angle BPC=\angle BIC$ and $BPIC$ is cyclic. It took a lot less time to notice the fact than to write this out; anyone that has done olympiad angle chasing should see the cyclic quadrilateral immediately. Now, we want to show that $I$ is the closest point to $A$ on circle $BIC$. Equivalently, the line $AI$ passes through the center of circle $BIC$. The center must lie on the perpendicular bisector of $BC$, which meets the bisector $AI$ of angle $BAC$ at $A'$ on the circumcircle of $ABC$ (the midpoint of arc $BC$). We want to show that $A'$ is the center of circle $BIC$. Chasing some angles, $\angle IA'C=\angle AA'C=\angle ABC=2\angle IBC$. Similarly, $\angle IA'B=2\angle ICB$. The only point other than $I$ making these (signed) angles is the center of circle $BIC$, so $A'$ is that center and we are done.

ooooooooooooops! I can't beleive it was IMO problem,really easy.(It took me 30 seconds to solve) we have $\angle ABP+\angle PCA=\angle PBC+\angle PCB=90^{\circ}-\frac{A}{2}\rightarrow \angle BPC=90^{\circ}+\frac{A}{2}$ hence points $B,P,I,C$ are on a circle,we know that $AI$ pass through circumcenter of $BIC$ therefore $AI\le AP$

Smart finishing Amir!

Another way: $B,C,I,P$ are concyclic and we prove that circle with center $A$ and radius $AI$ is tangent to the circumcircle of $\Delta BCI$ - the end

Consider the extreme case where $P = I$, and draw the lines containing $BP$ and $CP$. Let one of the lines be fixed. A little rotation of the other in one sense (by an angle of, suppose $\Delta$) disturbs the harmony of the sum, so we have to "fix" this by rotating the other the opposite sense (the same variation $\Delta$). This shows us that $BCPI$ might be cyclic to preserve the harmony of the sum (of the angles). Now, let $O$ be the center of the circle passing through $BCPI$, notice that $A,I,O$ are collinear. So the easiest way of going from $A$ to the center of the circunference is passing through $I$ and then going to the center, because any other walk, we could conclude, by the triangular inequality: $AP+R > AI+R \Rightarrow AP>AI$ =] I'll write it better later...

Hi! My solution shortly: At first I prove easy fact: if O - interrior point of triangle ABC and $A_{1}\in BC, B_{1}\in AC, C_{1}\in AB$ - points s.t. $OA_{1}= OB_{1}= OC_{1}$ then $OA_{1}\geq r$ where r - incircle's radius. Let circumcentre of $\triangle A_{1}B_{1}C_{1}$ = O and $OA_{1}$ = R. Then $R\geq r$. We see P is on circumcircle of $\triangle C_{1}B_{1}O$ because $\angle C_{1}OB_{1}= \angle C_{1}PB_{1}$. So $AP=\frac{R}{\cos(\frac{\angle C_{1}OB_{1}}{2})}=\frac{R}{\cos(\frac{\angle C+\angle B}{2})}$. On the other hand, $AI=\frac{r}{\cos(\frac{\angle C+\angle B}{2})}$. So $R\geq r$ means $AI \leq AP$

Even I did not participate and had a training in IMO level since last IMO i have solved three problems! But this one was a kind of Regional olympiad level problem! But I am still sure that there will be UZB team member that wont get HM.

Observe that from angle chasing, $P$ lies on the circle passing through $B, I, C$ . Also from the incentre-excentre lemma, this circle is externally tangent to that with centre $A$ passing through $I$ at $I$, giving us the required relation. $\square$

The LHS and RHS sum to $180^\circ-\angle BAC$ so the LHS and RHS are both $90^\circ-\frac{\angle BAC}{2}$ so $$\angle BPC=90^\circ+\frac{\angle BAC}{2}=180^\circ-\frac{\angle ABC+\angle ACB}{2}=\angle BIC$$so $P$ must lie on $(BIC)$. Let $O$ be the center of $(BIC)$. Then, we just need to prove that $I$ lies on $AO$. We see that $BO=OC$ and $$\angle BOC=2(180^\circ-\angle BIC)=\angle ABC+\angle ACB=180^\circ-\angle BAC$$so $O$ is the midpoint of arc $BC$ of $(ABC)$. Then, $\angle BAO = \angle BAI = \frac{\angle BAC}{2}$, as desired.

I haven't seen a solution with this approach yet (comparing angle sizes to deduce side lengths), so I'm wondering if this would get me full points (or any at all). First we claim that $P$ does not lie in the interior of $\Delta IBC$. For the sake of contradiction, suppose otherwise. Then, we have $$\begin{cases} \angle PBA > \angle IBA \\ \angle PCA > \angle ICA \end{cases}$$It follows that $\angle PBA + \angle PCA > \dfrac{1}{2} (\angle ABC + \angle ACB)$. However, we have $$\begin{cases} \angle PBC < \angle IBC \\ \angle PCB < \angle ICB \end{cases}$$which leads to $\angle PBC + \angle PCB < \dfrac 12 (\angle ABC + \angle ACB)$, contradiction. Now, WLOG suppose that $P \notin \overline{AI}$ lies on the half-plane of $AI$ that contains $B$, or $\angle PBC > \angle IBC$ and $\angle PCB < \angle ICB$. The given condition is equivalent to \begin{align*} \angle ABI - \angle PBI + \angle ACI + \angle PCI &= \angle IBC + \angle PBI + \angle ICB - \angle PCI \\ \dfrac 12 \angle B + \dfrac 12 \angle C - \angle PBI + \angle PCI = \dfrac 12 \angle B + \dfrac 12 \angle C + \angle PBI - \angle PCI\\ \angle PCI = \angle PBI \end{align*}It follows that $BPIC$ is a cyclic quadrilateral. Consider \begin{align*} \angle AIP &= \angle AIB - \angle PIB &= 90^{\circ} + \dfrac 12 \angle ACB - \angle PIB &= 90^{\circ} + \angle ICB - \angle PCB &= 90^{\circ} + \angle ICP \end{align*}Since we are considering the case $A,I,P$ are not collinear, $\Delta AIP$ is a nondegenerate triangle, and $\angle ICP > 0^{\circ}$. It follows that $\angle AIP > 90^{\circ}$, and it is well known that the side opposite to the largest angle of an obtuse triangle has greatest length, so it must be true that $AP > AI \, \forall P \notin \overline{AI}$. The only remaining case is $P \in \overline{AI}$. If $AP = AI$, then we must have $P=I$. Moreover, it is not hard to see that if $P \neq I$, then the left side of the given condition is strictly less than the right side, yielding a contradiction. Thus, $AP = AI$ only if $P=I$, and checking we see that the given condition holds. $\blacksquare$

We have: $\angle PBA + \angle PBC = \angle ABC$ $\angle PCA + \angle PCB = \angle ACB$ $\Rightarrow \angle PBA + \angle PCA + \angle PBC + \angle PCB = \angle ABC + \angle ACB$ $\Rightarrow \angle PBC + \angle PCB = \frac{\angle ABC + \angle ACB}{2}$ $\Rightarrow \angle BPC = 180 - \frac{\angle ABC + \angle ACB}{2} = \angle BIC$ Therefore, $BIPC$ is cyclic. By the incenter-excenter lemma, if $O$ is the center of circle $(BIPC)$, then $A$, $I$, $O$ are collinear. Since $OI = OP$, by the triangle inequality we know $AP \geq AI$. Equality occurs when $P = I$. $\blacksquare$

We have $$\angle PBA + \angle PCA = \angle PBC + \angle PCB \implies (\angle ABC - \angle PBC)+(\angle BCA - \angle PCB) = \pi- \angle BPC$$This gives us $\pi - \angle BAC = 2(\pi-\angle BPC) \implies \angle BPC = \frac{\pi}{2} + \frac{\angle BAC}{2} = \angle BIC$. As a result, $BPIC$ is cyclic. Drawing $(BPIC)$ and setting the center to $O$, and applying the triangle inequality on $APO$, we have $AP+PO>AO=AI+AO \iff AP>AI$. Hence, equality may only occur when $P=I.$

An easy problem! (Not written a formal proof tho) We know that $\angle IBP = \angle ICP$ so point $P$ lies on $(IBC)$. ALSO, $(IBC)$ is tangent to circle with radius $AI$ since $A, I, L$ are collinear. (Here $L$ is the center of $(IBC)$) So equality only holds when $P$ lies on $I$.

Amir.S wrote: ooooooooooooops! I can't beleive it was IMO problem,really easy.(It took me 30 seconds to solve) we have $\angle ABP+\angle PCA=\angle PBC+\angle PCB=90^{\circ}-\frac{A}{2}\rightarrow \angle BPC=90^{\circ}+\frac{A}{2}$ hence points $B,P,I,C$ are on a circle,we know that $AI$ pass through circumcenter of $BIC$ therefore $AI\le AP$ Brilliant Ending!!

$BPIC$ is cyclic then the rest is trivial.

Clearly $BPIC$ is cyclic, so the unique closest point on the circle to $A$ is the intersection of segment $AM_A$ and the circle (since $M_A$ is the center), where $M_A$ is the arc midpoint, but this is just $I$, so we are done.

$ $

Solved with some help... my first IMO geo btw We have \begin{align*}\angle PBC+\angle PCB = & \ \angle PBA+\angle PCA \\ \angle ABC+\angle ACB = & \ 2 \left (\angle PBA+\angle PCA\right) \\ \frac 12\left (180^\circ-\angle BAC\right) = & \ 180^\circ -\angle PAB-\angle APB+180^\circ-\angle PAC-\angle APC \\ 90^\circ-\frac 12 \angle BAC = & \ 360^\circ-\left(\angle APB+\angle APC\right )-\angle BAC \\ 90^\circ+\frac 12 \angle BAC= & \ \angle BPC \\ \angle BIC = & \ \angle BPC.\end{align*}Now let $I_A$ be the $A$-excenter of $\triangle ABC$. Then $BICI_A$ is cyclic, by the Incenter Excenter Lemma. Let the center of $(BICI_A)$ be $O$. Since $$180=\angle BI_AC+\angle BIC=\angle BI_AC+\angle BPC,$$it follows that $BPCI_A$ is cyclic as well. So, $P$ must lie on $(BCI_A)$. So by the triangle inequality we have $$OA \leq AP+PO \implies OI+IA \leq AP+OI \implies IA \leq AP,$$and the equality obviously holds if and only if $I=P$, as desired. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9.36cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -1.41, xmax = 1.71, ymin = -1.04, ymax = 0.7; /* image dimensions */ pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); /* draw figures */ draw(circle((0.,0.), 0.5), linewidth(0.8) + yellow); draw((-0.2995682831586557,0.4003234239029431)--(-0.4,-0.3), linewidth(0.8)); draw((-0.2995682831586557,0.4003234239029431)--(0.4,-0.3), linewidth(0.8)); draw((-0.4,-0.3)--(0.4,-0.3), linewidth(0.8)); draw(circle((0.,-0.5), 0.447213595499958), linewidth(0.8) + red); draw((-0.4,-0.3)--(-0.14119247578894345,-0.07565970639050944), linewidth(0.8) + green); draw((-0.4,-0.3)--(0.14119247578894334,-0.9243402936094904), linewidth(0.8) + green); draw((0.14119247578894334,-0.9243402936094904)--(0.4,-0.3), linewidth(0.8) + green); draw((-0.14119247578894345,-0.07565970639050944)--(0.4,-0.3), linewidth(0.8) + green); draw((-0.2995682831586557,0.4003234239029431)--(0.14119247578894334,-0.9243402936094904), linewidth(0.8)); draw((-0.2995682831586557,0.4003234239029431)--(0.,-0.05278640450004202), linewidth(0.8) + blue); /* dots and labels */ dot((-0.2995682831586557,0.4003234239029431),linewidth(4.pt)); label("$A$", (-0.2838752322507354,0.43122566059848144), NE * labelscalefactor); dot((-0.4,-0.3),linewidth(4.pt)); label("$B$", (-0.384104276037203,-0.26636848415533076), NE * labelscalefactor); dot((0.4,-0.3),linewidth(4.pt)); label("$C$", (0.4177280742545375,-0.26636848415533076), NE * labelscalefactor); dot((-0.14119247578894345,-0.07565970639050944),linewidth(4.pt) + uuuuuu); label("$I$", (-0.12350876219238732,-0.04185542607364408), NE * labelscalefactor,uuuuuu); dot((0.,-0.5),linewidth(4.pt) + uuuuuu); label("$O$", (0.016811899108667254,-0.46682657172826536), NE * labelscalefactor,uuuuuu); dot((0.14119247578894334,-0.9243402936094904),linewidth(4.pt) + uuuuuu); label("$I_A$", (0.15713256040972184,-0.8917977173828866), NE * labelscalefactor,uuuuuu); dot((0.,-0.05278640450004202),linewidth(4.pt) + uuuuuu); label("$P$", (0.016811899108667254,-0.021809617316350626), NE * labelscalefactor,uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Let $M$ be the midpoint of arc $\widehat{BC}$ of $(ABC)$. The conditions imply $$\angle BPC = 90^\circ + \frac{\angle{A}}{2},$$implying that $P$ lies on the circle centered at $M$ passing through $B$ and $C$. Then by the triangle inequality, $$AP + PM \geq AM \implies AP \geq AI,$$with equality when $P = I$.

We now that BPIC-cyclic by angle chasing.And Circle with center A an radius AI is tangent to (BIC) (easy by tangent from I to (BIC))And AP>AI when P!=I.

Observe that $$\angle PBA + \angle PCA +\angle PBC + \angle PCB = 180^\circ - \angle A \implies \angle PBA + \angle PCA = 90^\circ - \frac12 \angle A \implies \angle BPC = 90^\circ + \frac12 \angle A.$$But it is well-known that this is equal to $\angle BIC,$ therefore $B, P, I, C$ are concyclic with center $O.$ But by the Incenter-Excenter Lemma, $AI$ passes through $O,$ therefore $AP \geq AI$ is obvious, with equality iff $P=I.$ QED