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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

) . 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.**
Let's make a trip to the world of geometry 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:

### 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.