BdMO Online Forum

Time for a new problem.
Problem $31$: Given a trapezoid $ABCD (AD \parallel BC)$ with $\angle ABC > 90^\circ$. Point $M$ is chosen on the lateral side $AB$. Let $O_1$ and $O_2$ be the circumcenters of the triangles $MAD$ and $MBC$, respectively. The circumcircles of the triangles $MO_1D$ and $MO_2C$ meet again at the point $N$. Prove that the line $O_1O_2$ passes through the point $N$.
Source: International Zhautykov Olympiad 2013-1.

Actually $\triangle{MO_{1}D}$ and $\triangle{MO_{2}C}$ are spirally symmetric.So are $\triangle{MO_{1}O_{2}}$ and
$\triangle {MDC}$.Now assuming the contrary we can show a contradiction.

This solution process can be used as a lemma.

Problem $32$:Consider $\omega$ to be the circumcircle of $\triangle{ABC}$.$D$ is the midpoint of arc $BAC$ and $I$ is the incentre.Let $DI$ intersect $BC$ at $E$ and $\omega$ again at $F$.Let $P$ be a point on the line $AF$ such that $PE$ is parallel to $AI$.Prove that $PE$ is the bisector of $\angle{BPC}$.

Tahmid Hasan wrote:Time for a new problem.
Problem $31$: Given a trapezoid $ABCD (AD \parallel BC)$ with $\angle ABC > 90^\circ$. Point $M$ is chosen on the lateral side $AB$. Let $O_1$ and $O_2$ be the circumcenters of the triangles $MAD$ and $MBC$, respectively. The circumcircles of the triangles $MO_1D$ and $MO_2C$ meet again at the point $N$. Prove that the line $O_1O_2$ passes through the point $N$.
Source: International Zhautykov Olympiad 2013-1.

I have solved it using angle chasing.
$\angle O_{1}ND=\angle O_{1}MD=\angle O_{1}DM=\frac{1}{2}(180^{\circ}-\angle MO_{1}D)=90^{\circ}-\angle BAD$
$=180^{\circ}-\angle BAD-90^{\circ}=\angle ABC-90^{\circ}=\angle MBC-90^{\circ}=90^{\circ}-\frac{1}{2}\angle MO_{2}C= \frac{1}{2}(180^{\circ}-\angle MO_{2}C)=\angle O_{2}CM=\angle O_{2}MC$
$=\angle O_{2}NC$.
Also $\angle DNM+\angle CNM=180^{\circ}-\angle MO_{1}D+MO_{2}C$
$=180^{\circ}-2\angle BAD+360^{\circ}-2\angle ABC=540^{\circ}-2\times 180^{\circ}=180^{\circ}$ so $D,N,C$ are on a single line. Thus $O_{1},N,O_{2}$ will also be collinear.

@Zadid vai,you haven't posted the source of Problem $\boxed {32}$

Well,this is problem 3 of geomtry problem set of BDMC 2012.

SANZEED wrote:@Zadid vai,you haven't posted the source of Problem $\boxed {32}$

It's an Iran TST-2012 problem given in the last camp pset. That time I made a really complicated solution using harmonic division with a little help from Fahim vai, later I made another one- a lot simpler. Here it is-
Let $D'$ be the midpoint of arc $BC$ not containing $A$. So $DD' \perp BC$.
Let $DD' \cap PE=Q$. So $\angle PFD=\angle AD'D=\angle PQD \Rightarrow PFQD$ is concyclic.
So $PE.EQ=FE.ED=BE.CE \Rightarrow PBQC$ is concyclic. Since $BQ=CQ, PE$ bisects $\angle BPC$.
Note: This problem holds for any point in segment $AD'$ not particularly $I$.
It's very late at night now, I don't have any good problem to post. Maybe tomorrow.

@Tahmid . Love you bro .

Problem $\boxed {33}$

Let $ABC$ be a right triangle ($\angle C=90$),let $CD$ be the hight and let $K$ be a point in the triangle such that $AK=AC$.Prove that the diametar of the curcumcircle of triangle $ABK$ that contains point $A$ is perpendicular to $DK$

Source : Serbia 1990

Nadim Ul Abrar wrote:@Tahmid . Love you bro .

Problem $\boxed {33}$

Let $ABC$ be a right triangle ($\angle C=90$),let $CD$ be the hight and let $K$ be a point in the triangle such that $AK=AC$.Prove that the diametar of the curcumcircle of triangle $ABK$ that contains point $A$ is perpendicular to $DK$

Source : Serbia 1990

Can you please provide a figure?
And are you sure this statement is correct and you didn't make any typo?