IMO Marathon

[sourav das]

Problem 17: Solution

Let $BC\cap AD=O$ then given condition implies $O,K,L$ are collinear. Given conditions also implies $BC$ is common tangent to circles $ABP,CQD$. Note that a homothety with center $O$ sends circles $ABP$ to circle $DQC$. Let $KL\cap \odot ABP =\{P,H\}$ and $KL \cap \odot \{G,Q\}$. Then property of homothety implies: $\frac{OH}{OQ}=\frac{OP}{OG}$...(i) . Power of point implies $(OB^2)(OC^2)=(OH.OP)(OQ.OG)=OP^2.OQ^2$ (using (i))
That means $OB.OC=OP.OQ$ Power of point theorem implies $P,Q,B,C$ are concyclic.

Anyone can take my turn.

[*Mahi*]

Problem $16$: Let $ABC$ be an acute triangle. $(O)$ is circumcircle of $\triangle ABC$. $D$ is on arc $BC$ not containing $A$. Line $\ell$ moves through $H$ ($H$ is the orthocenter of $\triangle ABC$) cuts $\odot ABH,\odot ACH$ again at $M,N$ respectively.
(a). Find $\ell$ such that the area of $\triangle AMN$ is maximum.
(b). $d_{1},d_{2}$ are the lines through $M$ perpendicular to $DB$, the line through $N$ perpendicular to $DC$ respectively. $d_{1}$ cuts $d_{2}$ at $P$. Prove that $P$ moves on a fixed circle as $\ell$ varies.
Source: Vietnam 2013-6.

Now I look like a total slowpoke out here :/ sorry for the late.

Part A:

$(\triangle AMN) = (\triangle AHM)+ (\triangle AHN) = \frac 1 2 AH \cdot HM \cdot \sin \angle AHM + \frac 12 AH \cdot HN \cdot \sin \angle AHN = \frac 1 2 AH \cdot MN \ cdot \angle AHN$, which is maximized when $\sin \angle AHN$ is maximized or equal $1$, or $\angle AHN = 90^\circ$, or $\ell || BC$.

Part B:

Let $AI || DB$ and $I$ be the second intersection point with $ \bigcirc ABH$. define similarly point $J$ with $AJ || DC$ and $J$ on $\bigcirc ACH$ . Now by symmetry over $AB$, $AI = DB$, and similarly $AJ = DC$, and also $\angle IAJ = \angle BDC = 180^\circ - \angle A$.
Now two facts:
(i) $A$ is on the perpendiculur bisector of $MN$ (chase some angles to prove $\angle AMN = \angle ANM$)
(ii) $\angle MAN = 2 \angle A$ (chase some more angles).
Now, as $MP \perp AI$ and $NP \perp AJ$, so $\angle MPN = \angle A = \frac 12 \angle MAN$, which along with fact (i) implies that $A$ is the circumcentre of $\triangle PMN$. So $AI$ is the perpendicular bisector of $PM$ and $AJ$ is the perpendicular bisector of $PN$.
So, $\angle AIP + \angle AJP = \angle AIM + \angle AJM $ (reflection on $AI,AJ$) = $180^\circ$ (chase a little angles).
So, $P$ is on the fixed circle $\bigcirc AIJ$ [proved]

[Tahmid Hasan]

My solution sketch for $17$: Let $AP \cap DQ=R,BP \cap CQ=S,AD \cap BC=T$. $T,K,L$ are collinear. Manelaus on $\triangle TDQ$, line $AP$ and $\triangle TCQ$, line $BP$ implies $RS \parallel CD$. Also $PRQS$ is concyclic.
So $\angle QCB=\angle BCD-\angle DCQ=\angle RPS-\angle RSQ=\angle QPB \Rightarrow BCPQ$ is concyclic.

[*Mahi*]

Given conditions also implies $BC$ is common tangent to circles $ABP,CQD$

I don't get it

$\angle CQD = \angle ABC = \angle CDO$, so $BC$ touches $\bigcirc CQD$, and similarly for $\bigcirc APC$.
[Edit: looks like that part was removed but keeping the post for general inquiry]

[Tahmid Hasan]

Given conditions also implies $BC$ is common tangent to circles $ABP,CQD$

I don't get it

$\angle CQD = \angle ABC = \angle CDO$, so $BC$ touches $\bigcirc CQD$, and similarly for $\bigcirc APC$.
[Edit: looks like that part was removed but keeping the post for general inquiry]

It was just alternate segment theorem, sorry for noticing the fact at first . When I realized it, I removed the part.

[sourav das]

Sorry to everyone for skipping without explanation . Post a problem now (Please)...

[Tahmid Hasan]

Problem $18$: Let $ABCD$ be a cyclic quadrilateral such that the triangles $BCD$ and $CDA$ are not equilateral. Prove that if the Simson line of $A$ with respect to $\triangle BCD$ is perpendicular to the Euler line of $\triangle BCD$, then the Simson line of $B$ with respect to $\triangle ACD$ is perpendicular to the Euler line of $\triangle ACD$.
Source: Romania TST-2012-Day-1-2.
Note: Complex numbers provides immediate solution, I'd like to see a synthetic one.

[zadid xcalibured]

Pure Euclidean proof for $16.a$
All the $\triangle{AMN}$ are similar.So the area is maximum when $AM$ and $AN$ are radii of the respective circles.That is $MN$ is parallel to $BC$.

[Nadim Ul Abrar]

Problem $\boxed {19 }$
Points $C,M,D$ and $A$ lie on line $l $ in that order with $CM = MD$. Circle $\omega $ is tangent to line
at $A$. Let $B$ be the point on $\omega$ that is diametrically opposite to $A$. Lines $BC$ and $BD$ meet $\omega$ at
$P$ and $Q$. Prove that the lines tangent to $\omega$ at $P$ and $Q$ and line $BM$ are concurrent.

[Tahmid Hasan]

Problem $\boxed {19 }$
Points $C,M,D$ and $A$ lie on line $l $ in that order with $CM = MD$. Circle $\omega $ is tangent to line
at $A$. Let $B$ be the point on $\omega$ that is diametrically opposite to $A$. Lines $BC$ and $BD$ meet $\omega$ at
$P$ and $Q$. Prove that the lines tangent to $\omega$ at $P$ and $Q$ and line $BM$ are concurrent.

$\angle BPQ=\angle BAQ=90^{\circ}-\angle ABQ=\angle BDC$. Similarly $\angle BQP=\angle BCD$.
So $\triangle BPQ,\triangle BDC$ are directly similar.[Be careful of the orientations!]
Hence $BM$ is the $B$-symmedian of $\triangle BPQ$, so the tangents at $P,Q$ and $BM$ concur according to 'Lemma-1 Yufei Zhao'.
Note: Seems a little to easy to be in 'IMO marathon'!
Problem $20$: In the triangle $ABC$, $\angle B$ is greater than $\angle C$. $T$ is the midpoint of the arc $BAC$ of the circumcircle of $ABC$ and $I$ is the incenter of $ABC$. $E$ is a point such that $\angle AEI=90^\circ$ and $AE\parallel BC$. $TE$ intersects the circumcircle of $ABC$ for the second time in $P$. If $\angle B=\angle IPB$, find the angle $\angle A$.
Source: Iran TST-2008-10.