BdMO Online Forum

Nadim Ul Abrar wrote:problem 16 :

In triangle $ABC$, the centroid is $G$ and $N$ is the midpoint of $CA$. The line through $G$ parallel to $BC$ meets $AB$ at $X$. Prove that $\angle AXC = \angle NGC$ if and only if angle $ACB$ is a right angle

Let the midpoint of $AB$ be $M$.
$\angle AXC=\angle NGC \Longleftrightarrow \angle BXC=\angle BGC \Longleftrightarrow B,C,G,X$ are concyclic.
$\Longleftrightarrow BCGX$ is an isosceles trapeziod.[Since $GX \parallel BC$.]
$\Longleftrightarrow BX=CG \Longleftrightarrow \frac{1}{3}AB=\frac{2}{3}CM \Longleftrightarrow MA=MB=MC \Longleftrightarrow \angle ACB=90^{\circ}$.
Nobody solved problem $15$, so I'm not posting a new one.

Tahmid Hasan wrote:Nobody solved problem $15$, so I'm not posting a new one.

Far more than 2 days have passed, so by the rules you have to post the solution(or possibly hints) and then anybody can post a new problem.

I have a solution of problem 15.Let's move on to the next one.

Problem 17:Let $ ABC$ be a triangle with $AC > AB$. The point $X$ lies on the side $BA$ extended through $A$, and the point $Y$ lies on the side $CA$ in such a way that $BX = CA$ and $CY = BA$. The line $XY$ meets the perpendicular bisector of side $BC$ at $P$. Show that $\angle{BPC}+\angle{BAC}=180$
{Source:British MO 2006]

Sol 17

Check it out

Complete Rectengle $ABCE$ and $ACBG$ .

$GB \cap CE=F$ .

Note that $AX=AY,CY=CF,\angle XAY=\angle YCF$ . So $X,Y,F$ are collinear & $XF$ the bisector of $\angle EFG$ .

Let the circumcircle of$\triangle CFB$ intersect $XF$ at $P'$ .

Now $P'B$ being equal to $P'C$ imply $P'=P$ .And $\angle BAC=\angle BFC$ Hence , Proved .

zadid xcalibured wrote:Problem 17:Let $ ABC$ be a triangle with $AC > AB$. The point $X$ lies on the side $BA$ extended through $A$, and the point $Y$ lies on the side $CA$ in such a way that $BX = CA$ and $CY = BA$. The line $XY$ meets the perpendicular bisector of side $BC$ at $P$. Show that $\angle{BPC}+\angle{BAC}=180$
{Source:British MO 2006]

Draw parallelogram $ABA'C$. Draw the internal bisector $\ell$ of $\angle BA'C$ such that $\ell \cap BA=X',\ell \cap AC=Y'$.
$\angle BA'X'=\frac{1}{2}\angle A, \angle BX'A'=180^{\circ}-\angle A'BX'-\angle BA'X'=\frac{1}{2}\angle A$.
So $\angle BA'X'=\angle BX'A' \rightarrow BX'=BA'$.
But $BA'=CA=BX$. So $BX=BX' \rightarrow X=X'$[Since they are on the same side of $B$.]
Similarly $Y=Y'$, hence $XY$ is the internal bisector of $\angle BA'C$.
$AB \neq AC \rightarrow CA' \neq BA'$.
So $P \in \odot A'BC \rightarrow \angle BA'C+\angle BPC=180^{\circ} \rightarrow \angle BAC+\angle BPC=180^{\circ}$.

Problem $18$:Points $A,B,C,D,E$ lie on a circle $\omega$ and point $P$ lies outside the circle. The given points are such that (i) lines $PB$ and $PD$ are tangent to $\omega$, (ii) $P,A,C$ are collinear, and (iii) $DE \parallel AC$. Prove that $BE$ bisects $AC$.
Source:USAJMO-2011-5

Tahmid Hasan wrote: Problem-$15$:In a square $ABCD$, let $P$ be a point in the side $CD$, different from $C$ and $D$. In the triangle $ABP$, the altitudes $AQ$ and $BR$ are drawn, and let $S$ be the intersection point of lines $CQ$ and $DR$. Show that $\angle ASB=90^{\circ}$.
Source:Cono Sur Olympiad 2012.

A little hint: Let $E=BC \cap AQ,F=AD \cap BR.$
Then $\triangle ABE \cong BCP, \triangle ABF \cong ADP$.
And $\triangle DPF \cong \triangle CPE$.

I have a solution for problem 18.So let's move on to the next one.Someone post problems.