Hi could you please send all the problems from this olympiad

We have to prove that for any points \(M , N\) outside the \(\odot ABC\) on rays \(AB , AC\) such that \(BM = CN\), there exists a point \(D\) on \(\odot ABC\) through which \(\odot AMN\) crosses. So, actually what we have to prove is that \(\odot ABC\) and \(\odot AMN\) are not tangential, they intersect each other at two points, \(A , D\). Draw a line \(PQ\) tangent to \(\odot ABC\) at \(A\). Therefore by Alternate Segment theorem, \(\angle PAB = \angle ACB.\) --- \([1]\) But, \(AB \ne AC\) and \(BM = CN.\) => \(\dfrac{AB}{BM} \ne \dfrac{AC}{CN}\) => \(BC\) is not parallel to \(MN\) => \(\angle ACB \ne \angle ANM\) --- \([2]\) The two equations (1) and (2), tells that \(\angle PAB \ne \angle ANM\) => \(PQ\) is not tangent to \(\odot AMN\). Therefore, \(\odot AMN\) is also not tangent to \(\odot ABC.\)

Choose $D$ such that $\triangle BCD$ is isosceles with $|BD|=|CD|$ and $D$ lies on the same side as $A$ respective $BC$. Now for arbitrary points $M, N$ with the given conditions we have $|BD|=|DC|$, $|BM|=|CN|$ and \[ \angle MBD = 180^{\circ}-\angle DBA = 180^{\circ}-\angle DCA = \angle NCD.\]So by SAS we have $\triangle BMD \cong \triangle CND$ and therefore $\angle DMA = \angle DNA$, so the points $D,A,M,N$ are concyclic. $\hfill \square$

We have done many times problem with the same idea but we have to add a condition that ''$M,N$ lies resp. on the lines $AB,AC$ (not necessary rays) on the same side wrt $BC$ ''or ''$M,N$ lies resp. on the lines $AB,AC$ on the different side wrt $BC$ '' there is different point $D$ for either pencil. the proof is based on the similarity; let $ M_1,N_1 $ two points satisfying the first added condition .$ M_2,N_2$ two other points satisfying the second added condition s.t. $BM_1=BM_2,CN_1=CN_2 .$ consider $D$ the intersection of $\odot (AM_1N_1),\odot (AM_2N_2) \left[\text{or} \odot (AM_1N_2),\odot (AM_2N_1) \right]$.it s the center of similarity that send $M_1\to M_2,N_1\to N_2 \left[\text{or} M_1\to N_2,M_2\to N_1 \right]$ so it send$B$ to $C$ (preserve the midpoint) but it preserve the distance so it 's a rotation thus $D$ is on the $BC$-bisector any pair of points $M,N$ that satisfy in addition the first condition ( or the second condition ) will divide $M_1M_2 ,N_1N_2 \ \left[ \ \text{ or} M_1M_2 ,N_2N_1 \ \right]$ resp. in the same ratio so the similarity will send $M$ to $N$ .therefore $\odot(AMN)$ pass through $D $ RH HAS

Let $O$ be the circumcenter of $(ABC)$ and let $\ell$ be the line through $A$ and $O$. It is well-known that if a circle intersects another in one point and is not tangent to it, it must have at least a second intersection with this circle. Hence it is enough to show that $(AMN)$ is not tangent to $ABC$. Notice that a circle is tangent to $(ABC)$ at $A$ if and only it center lies on $\ell$, so in fact, we only need to show that if a point $O_1\in\ell$ is the center of a circle $\gamma$ passing through $A$ and $M$, then this cirlce does not pass through $N$. Clearly, this point $O_1$ is unique when $M$ is fixed, as it is the intersection of $\ell$ and the perpendicular bisector of $AB$, so let $O_1$ be fixed. By taking a homothety at $A$ with scale factor $\frac{AO_1}{AO}=\frac{AM}{AN}$, circle $(ABC)$ gets mapped to circle $(AMN')$. As homotheties preserve parallel lines, $N'$ is the point on $AC$ such that $MN'\parallel BC$. By the intercept theorem $\frac{|CN'|}{|BM|}=\frac{|AC|}{|AB|}<1$, so $|CN'|<|BM|=|CN|$, so $N'\neq N$, as desired.

$\measuredangle CDN=\measuredangle BDM$, and $\measuredangle DMB=\measuredangle DMA=\measuredangle DNA=\measuredangle DNC$. So $\triangle MBD\cong \triangle NCD$ by AAS congruency. This gives $BD=CD$, which implies $D$ lies on the perpendicular bisector of $BC$ irrespective of how $M$ and $N$ are chosen as long as they satisfy the problem's criteria. We are done!

Kawaii spirsim!! Consider some $(M,N), (M', N')$ satisfying the condition. Let $(ABC) \cap (AMN) \setminus A = P, (ABC) \cap (AM'N') \setminus A = P'$. Note that as $P$ is the miquel point of $BMNC$, The spiral centre sending $BM \to CN$ is precisely $P$. But by Gliding Lemma this is also the spiral centre sending $BM' \to CN'$, i. e. $P'$. Thus $P = P'$ as desired. $\square$

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maths_enthusiast_0001 wrote: Assume FTSOC there exists no such $D \neq A$ on $(ABC)$. This implies $(ABC)$ and $(AMN)$ are tangent at $A$. Hence by degenerate reim's, $BC \parallel MN$. Thus $\Delta ABC \sim \Delta AMN \implies \frac{AB}{AM}=\frac{AC}{AN} \implies \frac{AB}{BM}=\frac{AC}{CN} \implies \boxed{AB=AC}$ $\left(\because \vert BM\vert=\vert CN\vert\right)$. But it is given that $\vert AB\vert \ne \vert AC\vert$ thus, contradiction! $\blacksquare$ ($\mathcal{QED}$) This is incorrect- how do you know the point $D$ is not different for different choices of $M$ and $N$?

AshAuktober wrote: This is incorrect- how do you know the point $D$ is not different for different choices of $M$ and $N$? I have fixed any $M,N$ satisfying the given conditions($M \in \overrightarrow{AB},N \in \overrightarrow{AC},\vert BM\vert=\vert CN\vert$). Then proved that there exists such a point $D$ for all such $M,N$. I don't understand?

maths_enthusiast_0001 wrote: I have fixed any $M,N$ satisfying the given conditions($M \in \overrightarrow{AB},N \in \overrightarrow{AC},\vert BM\vert=\vert CN\vert$). Then proved that there exists such a point $D$ for all such $M,N$. I don't understand? To be clear, the statement is equivalent to showing that all the possible circle $(AMN)$ have the same common chord with $(ABC)$.

AshAuktober wrote: To be clear, the statement is equivalent to showing that all the possible circle $(AMN)$ have the same common chord with $(ABC)$. ok got it, thanks for pointing it out.