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Geometry Marathon : Season 3

Posted: Thu Jan 05, 2017 11:42 pm
by nahin munkar
$\Re$evived $\Re$ules :
Let's revive geo marathon (after 6 yrs only :mrgreen: ) . The rules will be almost same as before, just enhance solving time duration for 2 days . The difficulty level should be around G1-G5 compared with ISL(IMO Shortlist). Solver will post his own solution of the former problem within 2 days from the time of posting that one & also add a new problem. (Otherwise, if the problem is remained unsolved for 2 days, the proposer will provide that's solution & jump to next ). Thus, this marathon will move forward.

N.B.
1. Don't forget to type the problem number.
2. Try to use LaTeX code
3. Be sure about your provided problem.
Let's make a trip to the world of geometry :D Here we go....................

Problem 1
$ \triangle ABC,$ ,a right triangle with $ \angle A = 90^0 $, is inscribed in circle $ \Gamma.$ Point $ E$ lies on the interior of arc $ {BC}$ (not containing $ A$) with $ EA>EC.$ Point $ F$ lies on ray $ EC$ with $ \angle EAC = \angle CAF.$ Segment $ BF$ meets $ \Gamma$ again at $ D$ (other than $ B$). Let $ O$ denote the circumcenter of triangle $ DEF.$ Prove that $ A,C,O$ are collinear

Re: Geometry Marathon : Season 3

Posted: Fri Jan 06, 2017 12:19 am
by Raiyan Jamil
$\text {Solution of Problem 1:}$

Let $AB$ meet $CE$ at $X$. SInce $\angle A=90^o$ and $\angle EAC= \angle CAF$,$\Rightarrow$ $(X,C;E,F)=-1$. And $B(X,C;E,F) \Rightarrow ADCE$ is a harmonic quadrilateral. So, if the perpendicular bisector of $DE$ intersects $AC$ at $Y$, then $Y$ is also the intersection of $D$ and $E$ tangents.So,$DY=EY$.

Now it is enough to prove $EY=FY$. Here,
$\angle CYE=180^o-(\angle ECY+\angle CEY)$
$=180^o-(\angle ACF+\angle EAC)$
$=180^o-(\angle ACF+\angle CAF)$
$= \angle AFC$
$ \Rightarrow AEYF$ is cyclic.
So, since $AY$ is the angle bisector of $\angle EAF$,$Y$ is the midpoint of arc $EF$ of circle $AEYF$ not containing $A \Rightarrow EY=FY$ as desired.

Re: Geometry Marathon : Season 3

Posted: Fri Jan 06, 2017 11:31 am
by tanmoy
nahin munkar wrote:Problem 1
$ \triangle ABC,$ ,a right triangle with $ \angle A = 90^0 $, is inscribed in circle $ \Gamma.$ Point $ E$ lies on the interior of arc $ {BC}$ (not containing $ A$) with $ EA>EC.$ Point $ F$ lies on ray $ EC$ with $ \angle EAC = \angle CAF.$ Segment $ BF$ meets $ \Gamma$ again at $ D$ (other than $ B$). Let $ O$ denote the circumcenter of triangle $ DEF.$ Prove that $ A,C,O$ are collinear
My Solution:
Our goal is to show that $A,F,O,E$ are concyclic. Because then as $OE=OF$, $O$ lies on the internal bisector of $\angle EAF$ i.e. on $AC$.

We will show that $\angle FAE+\angle EOF=180^{\circ}$.

Let $\angle EAF=2x$ $\Rightarrow$ $\angle EAC=x=\angle EBC=\angle EDC$. Also, $\angle FDC=90^{\circ}$(Since $BC$ is the diameter of $(ABC)$.)

Then $\angle EDF=90^{\circ}+x$

Now,$\angle EOF=360^{\circ}-2\angle EDF=360^{\circ}-2(90^{\circ}+x)=180^{\circ}-2x=180^{\circ}-\angle FAE$.
Done.

Re: Geometry Marathon : Season 3

Posted: Fri Jan 06, 2017 11:52 am
by tanmoy
Problem 2
In $\triangle ABC$, $\angle ABC=90^{\circ}$. Let $D$ be any point on side $AC$, $D \neq A,C$. The circumcircle of $\triangle BDC$ and the circle with center $C$ and radius $CD$ intersect at $D,E$. Let $F$ be a point on side $BC$ so that $AF \parallel DE$. $X$ is another point on $BC$(Different from $F$) so that $XB=BF$. The circumcircle of $\triangle BDC$ and the circumcircle of $\triangle AXC$ intersect at $C,Y$.
Prove that $Y,F,D$ are collinear.

Re: Geometry Marathon : Season 3

Posted: Fri Jan 06, 2017 3:54 pm
by Raiyan Jamil
$\text{Solution of Problem 2:}$

Let $FD$ and $AB$ meet at $Y'$. Now, $\angle DBF=\angle DBC=\angle DEC=\angle EDC=\angle FAC=\angle FAD \Rightarrow BFDA$ is cyclic $\Rightarrow FDA=Y'DA=90^\circ$.
Now, $\angle BY'D=\angle AY'D=90^\circ-\angle BAD=\angle BCD \Rightarrow Y'$ lies on circle $BCD$.
Now, it's left to prove that $Y'$ lies on circle $AXC$ i.e. $Y'$ coincides with $Y.$
Here, $\angle XAY'=\angle XAB=\angle BAF=\angle BDF=\angle BDY'=\angle BCY'=\angle XCY' \Rightarrow XACY'$ is cyclic as desired.

**I'll post another problem within a while if none other posts here till then.**

Re: Geometry Marathon : Season 3

Posted: Fri Jan 06, 2017 7:14 pm
by Raiyan Jamil
$\text{Problem 3:}$

In Acute angled triangle $ABC$, let $D$ be the point where $A$ angle bisector meets $BC$. The perpendicular from $B$ to $AD$ meets the circumcircle of $ABD$ at $E$. If $O$ is the circumcentre of triangle $ABC$ then prove that $A,E$ and $O$ are collinear.

Re: Geometry Marathon : Season 3

Posted: Fri Jan 06, 2017 7:16 pm
by nahin munkar
tanmoy wrote:Problem 2
In $\triangle ABC$, $\angle ABC=90^{\circ}$. Let $D$ be any point on side $AC$, $D \neq A,C$. The circumcircle of $\triangle BDC$ and the circle with center $C$ and radius $CD$ intersect at $D,E$. Let $F$ be a point on side $BC$ so that $AF \parallel DE$. $X$ is another point on $BC$(Different from $F$) so that $XB=BF$. The circumcircle of $\triangle BDC$ and the circumcircle of $\triangle AXC$ intersect at $C,Y$.
Prove that $Y,F,D$ are collinear.
A very nice problem indeed.

Solution of problem 2:

We denote the center of $\odot BCD$ by $O$ & $CY \cap ED = L$.

$\spadesuit$ Claim 1 : $A,Y,B$ collinear

proof :
By property of radical axis of $\odot BCD$ & $\odot (C,CD)$, we can see $CY \perp ED$ & $EL=DL$. It's easy to see that, CY is a diameter of $\odot BCD$ . $B,E,C,D,Y$ are con-cyclic by statement. So, $\angle YBC = 90^0$.
Given, $\angle ABC = 90^0$.
So, from here,
$\angle ABC = 90^0$ = $\angle YBC = 90^0$.
we get , $A,Y,B$ collinear.$\spadesuit$


$\clubsuit$ Claim 2 : $CY \perp AF$

proof :
We extend $CY$ and $CY \cap AF = C'$ .
As, $DE||AF$ & $CC' \perp ED \Longrightarrow CC' \perp AF \Longrightarrow CY \perp AF$.$\clubsuit$


$\bigstar$ So,By claim $1$ & $2$, $Y$ is the orthocenter of $\triangle AFC$ .

As, $\angle YDC = 90^0$ & $Y$ is orthocenter of $\triangle AFC$ , $FYD$ must be a line.
So, $F ,Y, D$ are collinear . Done ! $\blacksquare$

Re: Geometry Marathon : Season 3

Posted: Fri Jan 06, 2017 8:19 pm
by nahin munkar
Raiyan Jamil wrote:$\text{Problem 3:}$

In Acute angled triangle $ABC$, let $D$ be the point where $A$ angle bisector meets $BC$. The perpendicular from $B$ to $AD$ meets the circumcircle of $ABD$ at $E$. If $O$ is the circumcentre of triangle $ABC$ then prove that $A,E$ and $O$ are collinear.
Solution of problem 3 :

Let, $H$ be orthocenter of $\triangle ABC$.

So,We know, $O$ & $H$ are isogonal conjugates .
Let, $A'$ be the projection of $A$ on $BC$.
& $X$ be the projection of $A$ on $BE$.

$\bigstar$ In $\triangle AA'D$,

$\angle AA'D =90^0$
$\angle ADA'= \theta$

$\bigstar$ In $\triangle AXE$,

$\angle AXE= 90^0$
& $\angle AEX = \theta $

So, $\triangle AA'D \sim \triangle AXE $

$\Longrightarrow$ $\angle A'AD= \angle EAD $

For this , $AH$ & $AE$ also isogonal conjugate line. So, $A,O,E$ lie on a line . Thus, $A,O,E$ are collinear. $\blacksquare$

Re: Geometry Marathon : Season 3

Posted: Fri Jan 06, 2017 8:59 pm
by nahin munkar
Problem 4:

Let $ABC$ be a triangle and $m$ a line which intersects the sides $AB$ and $AC$ at interior points $D$ and $F$, respectively, and intersects the line $BC$ at a point $E$ such that $C$ lies between $B$ and $E$. The parallel lines from the points $A$, $B$, $C$ to the line $m$ intersect the circumcircle of triangle $ABC$ at the points $A_1$, $B_1$ and $C_1$, respectively (apart from $A$, $B$, $C$). Prove that the lines $A_1E$ , $B_1F$ and $C_1D$ pass through the same point.

Re: Geometry Marathon : Season 3

Posted: Fri Jan 06, 2017 9:42 pm
by joydip
Solution of problem 4 :

Let $B_1F$ meet (ABC) again at $K$, $KC_1\cap AB = D_1$.Applying pascal's theorem on hexagon $BACC_1KB_1$ we get $ CC_1 \| BB_1 \| FD_1$ . So , $D=D_1$. So $K,D,C_1$ are collinear. Similerly $ K,E,A_1$ are collinear.