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### Re: Secondary and Higher Secondary Marathon

Posted: **Sun Jan 20, 2013 7:38 pm**

by **Tahmid Hasan**

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.

### Re: Secondary and Higher Secondary Marathon

Posted: **Sun Jan 20, 2013 9:07 pm**

by **zadid xcalibured**

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.

### Re: Secondary and Higher Secondary Marathon

Posted: **Sun Jan 20, 2013 9:16 pm**

by **zadid xcalibured**

This solution process can be used as a lemma.

### Re: Secondary and Higher Secondary Marathon

Posted: **Sun Jan 20, 2013 9:25 pm**

by **zadid xcalibured**

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 $\angle{BPC}$.

### Re: Secondary and Higher Secondary Marathon

Posted: **Sun Jan 20, 2013 11:19 pm**

by **SANZEED**

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.

### Re: Secondary and Higher Secondary Marathon

Posted: **Sun Jan 20, 2013 11:25 pm**

by **SANZEED**

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

### Re: Secondary and Higher Secondary Marathon

Posted: **Mon Jan 21, 2013 12:07 am**

by **zadid xcalibured**

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

### Re: Secondary and Higher Secondary Marathon

Posted: **Mon Jan 21, 2013 1:29 am**

by **Tahmid Hasan**

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.

### Re: Secondary and Higher Secondary Marathon

Posted: **Mon Jan 21, 2013 11:40 pm**

by **Nadim Ul Abrar**

@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

### Re: Secondary and Higher Secondary Marathon

Posted: **Tue Jan 22, 2013 12:08 pm**

by **Phlembac Adib Hasan**

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?