## Geometry Marathon : Season 3

For discussing Olympiad level Geometry Problems
nahin munkar
Posts: 81
Joined: Mon Aug 17, 2015 6:51 pm
Location: banasree,dhaka

### Geometry Marathon : Season 3

$\Re$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
Last edited by nahin munkar on Fri Jan 06, 2017 1:03 pm, edited 14 times in total.
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss

Raiyan Jamil
Posts: 138
Joined: Fri Mar 29, 2013 3:49 pm

### Re: Geometry Marathon : Season 3

$\text {Solution of Problem 1:}$

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.
Last edited by Raiyan Jamil on Fri Jan 06, 2017 3:24 pm, edited 1 time in total.
A smile is the best way to get through a tough situation, even if it's a fake smile.

tanmoy
Posts: 302
Joined: Fri Oct 18, 2013 11:56 pm

### Re: Geometry Marathon : Season 3

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:
Last edited by tanmoy on Sat Jan 07, 2017 11:19 am, edited 1 time in total.
"Questions we can't answer are far better than answers we can't question"

tanmoy
Posts: 302
Joined: Fri Oct 18, 2013 11:56 pm

### Re: Geometry Marathon : Season 3

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.
"Questions we can't answer are far better than answers we can't question"

Raiyan Jamil
Posts: 138
Joined: Fri Mar 29, 2013 3:49 pm

### Re: Geometry Marathon : Season 3

$\text{Solution of Problem 2:}$

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.**
A smile is the best way to get through a tough situation, even if it's a fake smile.

Raiyan Jamil
Posts: 138
Joined: Fri Mar 29, 2013 3:49 pm

### Re: Geometry Marathon : Season 3

$\text{Problem 3:}$

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.
A smile is the best way to get through a tough situation, even if it's a fake smile.

nahin munkar
Posts: 81
Joined: Mon Aug 17, 2015 6:51 pm
Location: banasree,dhaka

### Re: Geometry Marathon : Season 3

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 ! $\blacksquare$
Last edited by nahin munkar on Fri Jan 06, 2017 9:04 pm, edited 2 times in total.
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss

nahin munkar
Posts: 81
Joined: Mon Aug 17, 2015 6:51 pm
Location: banasree,dhaka

### Re: Geometry Marathon : Season 3

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. $\blacksquare$
Last edited by nahin munkar on Fri Jan 06, 2017 9:01 pm, edited 1 time in total.
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss

nahin munkar
Posts: 81
Joined: Mon Aug 17, 2015 6:51 pm
Location: banasree,dhaka

### Re: Geometry Marathon : Season 3

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.
# Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss

joydip
Posts: 48
Joined: Tue May 17, 2016 11:52 am

### Re: Geometry Marathon : Season 3

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.
Last edited by joydip on Fri Jan 06, 2017 10:00 pm, edited 1 time in total.
The first principle is that you must not fool yourself and you are the easiest person to fool.