Turkey NMO 2009 Q2

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Turkey NMO 2009 Q2

G H J

Reveal topic tags

geometric transformation

geometry proposed

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1345 posts

#1 h Aug 31, 2010, 12:53 PM • 3 Y

Y by Omeredip, Adventure10, Mango247

Let $\Gamma$ be the circumcircle of a triangle $ABC,$ and let $D$ and $E$ be two points different from the vertices on the sides $AB$ and $AC,$ respectively. Let $A'$ be the second point where $\Gamma$ intersects the bisector of the angle $BAC,$ and let $P$ and $Q$ be the second points where $\Gamma$ intersects the lines $A'D$ and $A'E,$ respectively. Let $R$ and $S$ be the second points of intersection of the lines $AA'$ and the circumcircles of the triangles $APD$ and $AQE,$ respectively.
Show that the lines $DS, \: ER$ and the tangent line to $\Gamma$ through $A$ are concurrent.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4145 posts

#2 h Aug 31, 2010, 7:00 PM • 4 Y

Y by yunustuncbilek, darkeagle, Adventure10, and 1 other user

We assume that $D,E$ lie on the side-segments $AB,AC.$ Denote $\omega_1 \equiv \odot(APD)$ and $\omega_2 \equiv \odot(AQE).$ Because of $\angle SQE=\angle SAE=\angle SAB,$ it follows that $B,S,Q$ are collinear. Likewise, $C,R,P$ are collinear. Thus, $SE \parallel RD \parallel BC$ $\Longrightarrow$ arcs $ES$ and $RD$ of $\omega_2$ and $\omega_1$ are homologous $\Longrightarrow$ $ER,DS$ intersect at the exsimilicenter of $\omega_1,\omega_2.$ On the other hand, since the tangent line $\tau$ to $\Gamma$ through $A,$ $AA'$ and $BC$ bound an isosceles triangle, then $\tau$ cuts $\omega_2$ again at $F,$ such that $AFES$ is an isosceles trapezoid, i.e. $EF \parallel RA$ $\Longrightarrow$ $F,A$ are homologous points under the positive homothety that takes $\omega_1$ into $\omega_2.$ Hence, $DS,ER,\tau$ concur at the exsimilicenter of $\omega_1,\omega_2.$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

276 posts

#3 h Nov 8, 2011, 3:09 PM • 2 Y

Y by Adventure10, Mango247

Quote:

Arcs $ES$ and $RD$ of $\omega_2$ and $\omega_1$ are homologous $\Longrightarrow$ $ER,DS$ intersect at the exsimilicenter of $\omega_1,\omega_2.$

Luis, what is homologous and why $ER,DS$ intersect at the exsimilicenter?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

74 posts

Jeje

#4 h Oct 9, 2014, 12:57 PM • 2 Y

Y by Adventure10, Mango247

What ia homogolous???

Is there another sinplere and more elemntary solution????

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2312 posts

#5 h Oct 9, 2014, 2:11 PM • 6 Y

Y by buratinogigle, enhanced, Adventure10, This_deserves_a_like, and 2 other users

My solution:

Let $T$ be the intersection of $DS$ and $RE$
Let $X$ be the intersection of $TA$ and $(AQE)$

Easy to see $DR \parallel SE \parallel BC$ (Reim theorem).

Since arc $DR$ = arc $SE$ ,
so $DR$ and $SE$ are corresponding chord in $(APD)$ and $(AQE)$ ,
hence we get $T$ is the exsimilicenter of $(APD) \sim (AQE)$ .
Since $\triangle ADR$ and $\triangle XSE$ are homothetic with center $T$ ,
so $\angle TAB =\angle TXS =180^{\circ}-\angle AES =180^{\circ}-\angle ACB$ .
ie. $TA$ is tangent to $(ABC)$

Q.E.D

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

179 posts

#6 h Jul 8, 2016, 5:31 AM • 4 Y

Y by Omeredip, Mathasocean, Adventure10, ehuseyinyigit

Note that by Reim's theorem we have $DR \parallel SE \parallel BC$ . Let $PR\cap\Gamma=X$ , Then note that by Reim's theorem again we have $DR \parallel BX \implies X=C$ and $P,R,C$ collinear, similarly $Q,S,B$ .
Let the tangent at A intersect the line $PQ$ at $T$ . Then by pascal on $AABQPA'$ we have $T, D,S$ collinear, as are $T,R,E$ by pascal on $AACPQA'$ , so the three lines are concurrent.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

32 posts

#7 h Mar 3, 2019, 2:00 PM • 2 Y

Y by Adventure10, Mango247

Through elementary angle-chasing, we can see that $\measuredangle BQA' = \measuredangle BAA' = \measuredangle SAE = \measuredangle SQE = \measuredangle SQA'$ . Thus, points $B,S,Q$ must be collinear ( $\measuredangle BQA' = \measuredangle SQA'$ implies that both $B$ and $S$ lie on the same line passing through $Q$ ). Similarly, we can see that points $P,R,C$ are collinear.

Thus, $\measuredangle BCA = \measuredangle BQA = \measuredangle SQA = \measuredangle SEA$ , implying that $BC || SE$ . Similarly, we can use the collinearity described above between points $P,R,C$ to show that $DR || BC$ . This means that $DR || SE$ , implying that the point $DS \cap RE$ is the center of a homothety taking $DR$ to $SE$ .

Let $Y$ be the second intersection of the tangent to $A$ with $\odot SEQA$ . Now, we can again use angle chasing to show that $\measuredangle BAP = \measuredangle BCA = \measuredangle SEA = \measuredangle SYA$ , and therefore $DA || SY$ . This means that the homothety described above not only takes $DR$ to $SE$ , but in fact $\triangle DAR$ to $\triangle SYE$ !! Another way of saying this is that the center of this homothety lies at $AY \cap DS \cap ER$ , which implies that these three lines are concurrent. Q.E.D

This post has been edited 2 times. Last edited by Luis González, Mar 3, 2019, 10:55 PM
Reason: Unhiding solution

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

121 posts

#8 h Nov 30, 2021, 5:06 PM

Y by

Here is my solution:
Let's say $D,E$ lie on the side-segments $AB,AC$ respectively and denote $\omega_1 \equiv \odot(APD)$ and $\omega_2 \equiv \odot(AQE).$ By simple angle chasing we can observe that $B,S,Q$ are collinear likewise, $C,R,P$ are collinear. Thus, $SE \parallel RD \parallel BC$ . Let's define $K$ as $SK \parallel DA$ and $EK \parallel AR$ . Since all $3$ sides are parallel to each other triangle $ADR$ and triangle $KSE$ are similar $\Longrightarrow$ $\angle SKE=\angle DAR=\angle SAE$ it follows that $K$ lies on $\omega_1$ . Now we can say $\angle EAK=\angle KSE=\angle ABC$ $\Longrightarrow$ $AK$ is tangent to $\Gamma$ at $A$ . We are ready to finish by applying desargues (or homothety) to triangle $ADR$ and $KSE$ . ( $AK$ , $DS$ and $RE$ needs to concur)

This post has been edited 1 time. Last edited by GuvercinciHoca, Nov 30, 2021, 5:07 PM
Reason: punctuation error

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

672 posts

#9 h Dec 3, 2024, 5:31 PM • 1 Y

Y by tiny_brain123

Let $A'D\cap BC=D_1,A'E\cap BC=E_1,AA'\cap BC=K,RE\cap SD=T$ . Let $W\in (ABC)$ with $AW\parallel BC$ . Since $A'D.A'P=A'R.A'A$ and $A'D_1.A'P=A'K.A'A$ we get $DR\parallel BC$ . Similarily $ES\parallel BC$ . Apply DDIT on quadrilateral $RDSE$ to get $(\overline{AT},\overline{A(BC)_{\infty}}),(\overline{AD},\overline{AE}),(\overline{AA'},\overline{AA'})$ is an involution. Project this onto $(ABC)$ . We see that $A'A',BC,(AT\cap (ABC))W$ are concurrent. Since $A'A'\cap BC=BC_{\infty}$ , we observe that $AT\cap (ABC)=A$ so $AT$ is tangent to $(ABC)$ as desired. $\blacksquare$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a