I think it is very easy for the second problem of imo.

The circle throw K, L, M is tangent to PQ iff <MLK=<QMK=<AQP and <MKL=<PML=<APQ iff triangles APQ and MKL are similar iff AP/AQ = MK/ML = BQ/PC iff AP·PC = AQ·BQ (Let E, F the interseptions of QP with the circuncircle) iff FP·EP = EQ·QF iff EQ=FP iff O is on the perpendicular bisector of QP iff OQ=OP

Yes it is indeed an easy problem for the IMo like last year's Geo problem number 1. Anyway, I hae exactly the same solution as m.candales and probably itis a common solution. He had just lef t a few steps I feel: That is $ AQ\dot QB = EQ\dot QF$ AND $ AP\dot PC = EP\dot PF$ and since $ AQ %Error. "dotQB" is a bad command. = AP %Error. "dotPC" is a bad command. $, we get that that $ EQ\dot QF = EP\dot PF$. This year's IMO DAy 1 is easy.

The circle $ \Gamma$ is tangent to the line $ PQ$ if and only if $ \measuredangle MLK = \measuredangle QML$. Since $ MK$ is parallel to $ AB$, it follows that $ \measuredangle AQP = \measuredangle MLK$. Since $ MK$ and $ ML$ are middle lines in triangles $ \triangle PQB$ and $ \triangle PQC$, respectively, it follows that $ \measuredangle PAQ = \measuredangle LMK = \alpha$. Finally, the triangles $ \triangle APQ$ and $ \triangle MKL$ are similar. Using side ratios, we get \[ \frac{AP}{AQ}=\frac{MK}{ML}=\frac{BQ}{PC},\] or equivalently $ AP \cdot PC = AQ \cdot BQ$. Consider the triangle $ \triangle ABO$ and apply Stewart's theorem for cevian $ OQ$ \[ OQ^2 = R^2 \cdot \frac{BQ}{AB} + R^2 \cdot \frac{AQ}{AB} - AQ \cdot BQ = R^2 - AQ \cdot BQ.\] Analogously, \[ OP^2 = R^2 - AP \cdot PC,\] and the result follows.

Nice Alexilic though very similar to mine and candales. Anyway, is an easy trigonometric solution possible? I am trying for one still.

Same solution as above; just a silly remark. After proving that $ AP \cdot PC = AQ \cdot BQ$ there is no need for Stewart or some other geometric construction. Just note that $ AP \cdot PC = \text{power of}\ P\ \text{wrt. the circumcircle} = R^{2} - OP^{2}$. Similarly, $ AQ \cdot BQ = R^{2} - OQ^{2}$. Thus, $ OP = OQ$.

Valentin Vornicu wrote: Agr_94_Math wrote: That makes the problem actually useless infact. There is nothing invovled int he problem. I shall be surprised if some contestant doesnt solve this problem. I will bet you that more than 50% of the contestants won't get 7 points on this problem I also think so, because they may cut points for the case when $ K=L$. I think most of the participants didn't consider this case, at least no one in our team didn't mention this

I think complex numbers can solve this... But it would require a lot of computation. With complex numbers, we may just need to prove $ p\overline{p}=q\overline{q}$ if you take the circumcircle of ABC for the unit circle... But it's really a lot of computation.

cefer wrote: Valentin Vornicu wrote: Agr_94_Math wrote: That makes the problem actually useless infact. There is nothing invovled int he problem. I shall be surprised if some contestant doesnt solve this problem. I will bet you that more than 50% of the contestants won't get 7 points on this problem I also think so, because they may cut points for the case when $ K = L$. I think most of the participants didn't consider this case, at least no one in our team didn't mention this I doubt they will cut points for this. Since the statement clearly says that $ \Gamma$ is the CIRCLE (and not line) which passes through $ K$, $ L$, $ M$, there is no need to study degenerate cases. Indeed, if the problem said that $ \Gamma$ is the circumcircle of triangle $ KLM$ maybe it would have given some space for interpretation (since triangles are more likely to be considered degenerated in some cases).

A better question might be: What percent of competitors do you predict to get at least $ 5$ points?

somebody asked for a simple trig solution: if we let $ AP=x$ and $ AQ=y$ from the similarity of $ KLM$ and $ APQ$ (i don't remember the order of the points) we get that $ \frac{x}{y}=\frac{c-y}{b-x}$ (or something like this, i don't remember well), or equivalently $ bx-x^2=cy-y^2$.. from law of cosines on triangles $ AOP$ and $ AOQ$ we get that we have to prove that $ x^2+R^2-2xR\cos(90-B)=y^2+R^2-2yR\cos(90-C)$, which is what he had above, and we're done..

orl wrote: Let $ ABC$ be a triangle with circumcentre $ O$. The points $ P$ and $ Q$ are interior points of the sides $ CA$ and $ AB$ respectively. Let $ K,L$ and $ M$ be the midpoints of the segments $ BP,CQ$ and $ PQ$. respectively, and let $ \Gamma$ be the circle passing through $ K,L$ and $ M$. Suppose that the line $ PQ$ is tangent to the circle $ \Gamma$. Prove that $ OP = OQ.$ Proposed by Sergei Berlov, Russia I think the following configuration is true Quote: Let $ ABC$ be a triangle with circumcentre $ O$. The points $ P$ and $ Q$ are interior points of the sides $ CA$ and $ AB$ respectively. Let $ K,L$ and $ M$ be the midpoints of the segments $ BP,CQ$ and $ PQ$. respectively, and let $ \Gamma$ be the circle passing through $ K,L$ and $ M$. Suppose that $ N$ is the second intersection of $ \Gamma$ and $ PQ$. Prove that $ ON\perp PQ$. [geogebra]ceebac8d3c1e0ce42fe10433c26b1b917d7e2f47[/geogebra] When $ M\equiv N$, we conclude that $ OM$ is the perpendicular bisector of the line segment $ PQ$, so the result follows.

To back up Valentin- I spent a long time on this one not seeing anything, and messing with some variants of coordinates. Then I drew in the angles at M, saw the similarity and Power of a Point, and finished in a minute. If you're tuned into geometry and see the right idea, its simple and quick. If you're not, you can flounder for hours getting nowhere. If you want to try complex numbers, I recommend putting the origin at the circumcenter $ O'$ of $ KLM$. You then want to prove that $ O$ is on the line $ O'M$.

jmerry wrote: If you want to try complex numbers, I recommend putting the origin at the circumcenter $ O'$ of $ KLM$. You then want to prove that $ O$ is on the line $ O'M$. How much is it different from putting the origin into O' and not into O? I saw only difference in working with k,l,m instead of a,b,c... Can you explain it to me?

@jmerry, please tell me a complex numbers solution. If you work from the origin of the circumcentre of $ KLM$, then are you not making it complicated? If you take the origin as $ O$ itself, then just have to prove that two points $ P$ and $ Q$ have the same distance from the origin whihc I feel is ismpler to work upon.

It's a lot easier to build $ A,B,C$ from $ K,L,M$ than the other way around. $ Q$ and $ P$ being equidistant from $ M$ on opposite sides of the tangent is particularly nice. Of course, I didn't actually go through with the solution; I'm recommending that path because it looks like it might work to me.

Quote: An easy extension. Let $ ABC$ be a triangle with the circumcircle $ C(O,R)$ . Condider the points $ P\in (AC)$ , $ Q\in (AB)$ and a point $ M\in (PQ)$ for which $ MQ = m\cdot MP$ , $ m\ne 0$ . Let $ K\in PB$ , $ MK\parallel AB$ and $ L\in QC$ , $ ML\parallel AC$ . Suppose that the line $ PQ$ is tangent to the circumcircle of $ \triangle MKL$ . Prove that $ \boxed {\ OQ^2 - m\cdot OP^2 = R^2(1 - m)\ }$ . Proof. Observe that $ \left\|\begin{array}{ccc} \frac {MK}{QB} = \frac {PM}{PQ} = \frac {1}{m + 1} & \implies & MK = \frac {1}{m + 1}\cdot QB \\ \\ \frac {ML}{PC} = \frac {QM}{QP} = \frac {m}{m + 1} & \implies & ML = \frac {m}{m + 1}\cdot PC\end{array}\right\|$ and $ \left\|\begin{array}{c} \widehat {KLM}\equiv\widehat {QMK}\equiv\widehat {PQA} \\ \\ \widehat {LKM}\equiv\widehat {PML}\equiv\widehat {QPA}\end{array}\right\|\ \implies$ $ \triangle KLM\ \sim\ \triangle PQA\ \implies\ \frac {MK}{AP} = \frac {ML}{AQ}\ \implies$ $ \frac {1}{m + 1}\cdot\frac {QB}{PA} = \frac {m}{m + 1}\cdot\frac {PC}{QA}$ $ \implies$ $ QB\cdot QA = m\cdot PA\cdot PC$ $ \implies$ $ OQ^2 - R^2 = m\left(OP^2 - R^2\right)$ $ \implies$ $ OQ^2 - m\cdot OP^2 = R^2(1 - m)$ .

Generalization: Extend PQ to meet the circumcircle at X and Y. Let N be the midpoint of XY. Prove that K, L, M, and N are concyclic. Note that this implies the problem statement; I took the test for practice and came across this lemma, but couldn't prove it.

Statistically, this problem is easier than problem 4. It took me longer to solve problem 4 as well. I think problem 2 and 4 should be interchanged.

Notice that $MK \parallel BQ$ and $ML \parallel PC$ by the midline theorem(on triangles $\triangle APB$ and $\triangle AQC$). And by the tangent condition we have $\angle MKL = \angle PML = \angle QPA$ and analogously $\angle AQP = \angle MLK$ so $\triangle APQ \sim \triangle MKL$. Notice that $MK = \frac{QB}{2}$ and $ML = \frac{PC}{2}$. By similar triangle ratios we get that $ML \cdot AP = MK \cdot AQ$ so it follows that $AP \cdot PC = AQ \cdot QB$ so $P$ and $Q$ have the same power wrt $(ABC)$. Then it follows that $OP = OQ$ as desired.

extend $PQ$ to meet the circle at $R$ and $S$ first, $AB$ and $MK$ are parallel, $AC$ and $ML$ are parallel, so $LMK=PAQ$, $APM=PML$, and $AQM=QMK$ since $PQ$ is tangent to $KLM$, $PML=MKL$ and $MKL$ is similar to $APQ$ also, $PQ=2MK$ and $PC=2ML$, so power of a point yields that $RP=SQ$, and since $OR=OS$, $ORP=OSQ$, and $RP=SQ$, we get that $OP=OQ$

Motivation Notice that $MK || QB$ and $ML || PC$. Now let's make use of the tangency condition: $\angle{KLM}=\angle{QMK}=\angle{PQA}$. Since $\angle{LMK}=\angle{QAP}$, we know that $\triangle{KLM} \sim \triangle{PQA}$. Note that to prove $OP=OQ$, we can prove $MO \perp PQ$ or $Pow_{w}{P}=Pow_{w}{Q}$. Let's examine the latter since it may be complicated to prove $M,O',O$ are collinear ($O'$ is the center of $\Gamma$). Solution Continuing on, we see that we need to prove $AQ*QB=AP*PC$. From $\triangle{KLM} \sim \triangle{PQA}$, we know that $\frac{KM}{LM}=\frac{PA}{QA}$, so $KM*QA=LM*PA$. Utilizing the midpoint condition, we see that $2*KM*QA=QB*QA=2*LM*PA=PC*PA$, and we are done.

Let \( X \) be the reflection of \( Q \) across \( K \) and \( Y \) the reflection of \( P \) across \( L \). We need to prove \(\triangle APX \sim \triangle AQY\) to show that \(\frac{AP}{PX} = \frac{AQ}{QY}\). Since \(\triangle QBY \cong \triangle PXC\), it follows that \(\angle AQY = \angle XPA\). Using the extended Law of Sines and angle congruences, we find \(\sin \angle BAX \sin \angle YAB = \sin \angle CAY \sin \angle XAC\), leading to \(\angle BAX = \angle CAY\). Thus, \(\triangle APX \sim \triangle AQY\), confirming \(OP = OQ\)

We let $\omega$ be the circumcircle of ($ABC$). Our solution is based off the following lemma. $\Delta AQP \sim \Delta MLK$ Since, $PQ$ is tangent to $\Gamma$ so, $\angle KLM = \angle KMQ$ and $\angle LKM = \angle LMP$. We know that \begin{align*} \frac{PM}{PQ} = \frac{PK}{PB} = \frac{1}{2}\\ \frac{QM}{QP} = \frac{QL}{QC} = \frac{1}{2} \end{align*}So, $MK \parallel QB$ and $ML \parallel PC$ (By Similarity). From this we get, \begin{align*} &\angle AQP = \angle AQM = \angle KMQ = \angle KLM \\ &\angle APQ = \angle APM = \angle LMP = \angle LKM \end{align*} So, $\Delta AQP \sim \Delta MLK$. This gives us \begin{align} \frac{AQ}{ML} = \frac{AP}{MK} \end{align}Since $\Delta QML \sim \Delta QPC$ and $\Delta PMK \sim \Delta PQB$ (Proved in the lemma), we get \begin{align} \frac{ML}{PC} = \frac{MK}{BQ} = \frac{1}{2} \end{align} From multiplying Eq 1 and Eq 2 \begin{align*} &\Rightarrow \frac{AQ}{PC} = \frac{AP}{BQ} \\ &\Rightarrow AQ \cdot BQ = AP \cdot PC \end{align*} So, \begin{align*} &\Rightarrow \text{Pow}_{\omega}(Q) = AQ \cdot BQ = AP \cdot PC = \text{Pow}_{\omega}(P) \\ &\Rightarrow R^2 - OQ^2 = \text{Pow}_{\omega}(Q) = \text{Pow}_{\omega}(P) = R^2 - OP^2 \end{align*} Hence, $OQ = OP$.

Note that $\overline{MK}$ is parallel to $\overline{QB}$ and that $\overline{ML}$ is parallel to $\overline{PC}$. In fact, $MK = \frac 12 QB$ and $ML = \frac 12 PC$ by similar triangles. We will use this later. Note also that $\angle MLK = \angle QMK = \angle AQP$ and $\angle LMK = \angle QAP$ from the two sets of parallel lines. Thus, by AA similarity $\triangle MLK \sim \triangle AQP$. This means that$$\frac{ML}{AQ} = \frac{MK}{AP}\implies \frac{2\cdot ML}{AQ} = \frac{2\cdot MK}{AP}\implies \frac{CP}{AQ} = \frac{BQ}{AP}$$. Therefore, $CP\cdot AP = AQ\cdot BQ$, so $P$ and $Q$ have the same power with respect to $(ABC)$. Finally, simply note that$$\text{Pow}_{ABC}(P) = R^2 - OP^2 = \text{Pow}_{ABC}(Q) = R^2-OQ^2\implies OP = OQ$$. $\blacksquare$

Let us call the circumcircle of $\triangle ABC$ $\omega_1$, we first prove that $Pow_{\omega_1}P = Pow_{\omega_1}Q $ since $K,L,M$ are midpoint of $BP,CQ,PQ$ respectively, thus by converse of Thales theorem, $MK \parallel AB$ and $ML \parallel AC$ and $$\angle AQP = \angle QMK = \angle MLK$$$$\angle APQ = \angle PML = \angle MKL$$therefore $\triangle APQ \sim \triangle MKL$ and $$\frac{AP}{AQ} = \frac{MK}{ML}$$Now, in $\triangle PQB$ $K$ is the midpoint of $BP$ and $MK \parallel QB$ thus by midpoint theorem, $MK = 2QB$. similarly in $\triangle QBC$ $L$ is the midpoint of $CQ$ and $ML \parallel PC$ thus by midpoint theorem, $ML = 2PC$. Thus $$\frac{QB}{PC} = \frac{MK}{ML} = \frac{AP}{AQ}$$$$AP \cdot PC = AQ \cdot QB$$that is $Pow_{\omega_1}P = Pow_{\omega_1}Q $ Now, By Power of point $Pow_{\omega_1}P = OP^2 - R^2 $ and $Pow_{\omega_1}Q = OQ^2 - R^2 $ thus $$OP^2 - R^2 = OQ^2 - R^2$$$$OP = OQ$$

solved with Upwgs_2008

Using the tangency condition, we get the angle equalities \[\measuredangle MKL = \measuredangle PML = \measuredangle QPA, \quad \measuredangle MLK = \measuredangle QMK = \measuredangle PQA\] to give $\triangle APQ \sim \triangle MKL$. Now we use complex numbers, setting $(ABC)$ as the unit circle, and letting all points $X$ represent the complex number $x$. We can now rewrite our similarity as \[\frac{\overline p - \overline a}{\overline q - \overline a} = \frac{k - m}{\ell - m} = \frac{\frac{b+p}{2} - \frac{p+q}{2}}{\frac{c+q}{2} - \frac{p+q}{2}} = \frac{b-q}{c-p}\]\[\implies (c \overline p - \overline ac + \overline ap) - |p|^2 = (b \overline q - \overline ab + \overline aq) - |q|^2.\] Notice that $P$ lying on chord $AC$ of the unit circle gives us \[a+c = p + ac \overline p \implies 1 + \overline ac = \overline ap + c \overline p,\] and analogously with $Q$ on chord $AB$, our equation can be rewritten as \[1 - |p|^2 = 1 - |q|^2 \implies OP = OQ \quad \blacksquare\]

Handwritten proof I submitted for $OTIS$ application.

storage

We angle chase. Claim: $AP\cdot CP = AQ\cdot BQ$. Lemma: $\triangle{APQ}$ is oppositely similar to $\triangle{MKL}$. We directed-angle chase (all angles are in directed angles). We know that $\measuredangle{MKL} = \measuredangle{PML}$ by the Tangent Lemma. Since $AC$ is parallel to $ML$ (since $M$ and $L$ are midpoints), $\measuredangle{PML} = \measuredangle{MPA}$, which is equal to $\measuredangle{GPA}$. Since $\measuredangle{GPA} = -\measuredangle{APG}$, $$\measuredangle{MKL} = -\measuredangle{APG}.$$Similarly, $$\measuredangle{MLK}=-\measuredangle{AGP}.$$By AA similarity, the lemma is proven. Now, we know that $$\frac{AQ}{AP} = \frac{ML}{MK}\implies AQ\cdot MK = AP\cdot ML$$due to the similar triangles. Multiplying both sides by $2$, we have $AQ\cdot QB = AP\cdot PC$ (since $M, K,$ and $L$ are midpoints), which is the claim. We know that $P$ and $Q$ have the same power in respect to $(ABC)$, so $OP^2-r^2=OQ^2-r^2$, where $r$ is the circumradius of $\triangle{ABC}$. The conclusion follows.