Russia-Sharygin-geometry olympiad 2016 correspondence round

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Russia-Sharygin-geometry olympiad 2016 correspondence round

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Source: http://geometry.ru/

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237 posts

#1 h Apr 13, 2016, 4:15 PM • 1 Y

Y by Adventure10

Hi every one, here is problems of the first (correspondence) round of Sharygin geometry olympiad.
Let's solve them in this topic Good Luck

Attachments:

first round 2016.pdf (160kb)

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695 posts

#2 h May 1, 2016, 1:48 PM • 3 Y

Y by anantmudgal09, Adventure10, Mango247

Attached are my solutions for problems 14-20, 22

. 3D geo (23, 24) is hard

EDIT: Attachment removed for now. Are we allowed to discuss? I know the submission deadline was April 1, but I don't see it posted anywhere that we're allowed to discuss after this deadline.

EDIT: Attachment restored.

Attachments:

2016 Sharygin Correspondence Round.pdf (228kb)

This post has been edited 2 times. Last edited by Dukejukem, May 2, 2016, 11:00 AM

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#3 h May 1, 2016, 2:49 PM • 2 Y

Y by Adventure10, Mango247

I submitted problems 9-20 and 22. I am in grade 9 as per Russian schools. Is that good enough?

I will post mine here as soon as it is sure that there is no problem in doing so.

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237 posts

#4 h May 1, 2016, 5:49 PM • 2 Y

Y by Adventure10, Mango247

Dukejukem wrote:

Attached are my solutions for problems 14-20, 22

. 3D geo (23, 24) is hard

EDIT: Attachment removed for now. Are we allowed to discuss? I know the submission deadline was April 1, but I don't see it posted anywhere that we're allowed to discuss after this deadline.

Hi,I think there is no problem for discussion on problems and you can post your solutions ,Because now the solutions are available in main site of Sharygin geometry olympiad.

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1246 posts

#5 h May 1, 2016, 9:18 PM • 2 Y

Y by Adventure10, Mango247

14 caught my eye right away hence the solution:
Lemma :Let the A-mixtlinear excircle touch $\odot ABC$ in $A_{2}$ and let the touch point of the incircle with $BC$ be $A_{2}$ $\implies$ $AA_{1}$ , $AA_{2}$ are isogonals.
Proof:
Consider the inversion around $A$ with arbitrary radius. $A_{1}$ is taken to $A_{2}$ hence the result.
Let the desired A-circle cut $AB,AC$ in $B^{*},C^{*}$
$\triangle AB^{*}C^{*}$ and $\triangle ABC$ are homothetic so but by lemma $AA_{1},AA_{2}$ are isogonals in $AB^{*}C^{*}$ so also in $\triangle ABC$

The proof can also go along the line of Mongue-De Alambert's but then you would have to explain that the centar of negative homothety taking the incentre to $\odot ABC$ is isogonal conjugate of $G_{e}$

Notice the part of the configuration of USAMO 2007 6.

This post has been edited 1 time. Last edited by nikolapavlovic, May 1, 2016, 10:46 PM

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479 posts

#6 h May 2, 2016, 2:54 AM • 2 Y

Y by Adventure10, Mango247

Too bad i am in grade 11

14. a) We claim that the point of concurrence is $X(56)$ , the exsimilicenter of incircle and circumcircle. For simplicity, call the circle touching its circumcircle at A and touching externally its incircle be $\omega_a$ . Let $AA_1$ intersect the circumcircle again at $A'$ , consider the homothety $\phi$ centered $A$ that maps $\omega_a$ to the circumcircle $\Gamma$ , then it maps $A_1$ to $A'$ .

Consider the tangent line of $A_1$ to $\omega_a$ , then this tangent line is also tangent to the incircle at $A_1$ . But since $\phi(\omega_a)=\Gamma$ and $\phi(A_1)=A'$ , the tangent line of $A_1$ to $\omega_a$ , which is the tangent line of $A_1$ to the incircle, is parallel to the tangent line of $A'$ to the circumcircle. So the negative homothety that maps the incircle to the circumcircle must map $A_1$ to $A'$ , so $X(56)$ lies on $A_1A'$ , which is line $AA_1$ . Similarly, $X(56)$ lies on lines $BB_1$ and $CC_1$ .

b) We prove the well known fact that the isogonal conjugate of $X(56)$ is the Gergonne point $G$ , which also solves the problem.

Consider a circle $\omega'_a$ such that it is tangent to $AB, AC$ and the circumcircle $\Gamma$ externally. Consider the transformation $\tau$ which is an inversion centered $A$ with radius $\sqrt{AB \cdot AC}$ , then a reflection about the $A$ -inner angle bisector. Then $\tau(\omega'_a)$ , the image of $\omega_a$ under this transformation is tangent to $AB, AC$ and $\tau(\Gamma)=BC$ . So its image is the incircle.

Consider the point $A'$ as in part (a), then note that $\phi(\omega_a)$ is tangent to $AB, AC$ and $\Gamma$ , so $\phi(\omega_a)=\omega'_a$ . So the contact point of $\omega'_a$ and $\Gamma$ is $A'$ , so $\tau(A')=A_2$ . This implies that the lines $AA'=AA_1$ and $AA_2$ are isogonal, which solves the problem.

To prove the claim, just note the same properties for vertices $B$ and $C$ , consequently $X(56)$ and $G$ are isogonal conjugates.

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479 posts

#7 h May 2, 2016, 2:55 AM • 2 Y

Y by Adventure10, Mango247

15. We will use two well known lemmas. (I will include proof if requested)

Lemma 1. $IO=IH$ $\iff$ one of the angles $\angle A, \angle B, \angle C$ is $60^{\circ}$ .

Lemma 2. If $I$ is the incenter of $\triangle ABC$ , then $MN=2MI$ , and $I, M, N$ are colinear in this order.

Now, denote $H, O_9$ be the orthocenter, nine-point center of $\triangle ABC$ respectively. Then since $MN=2MI$ by Lemma 2 and it is well known that $MO=2MO_9$ , so $\angle MON=\angle MO_9I$ , and since $O, O_9, H, G$ are colinear with $OO_9=HO_9$ , then $\angle MON=90^{\circ}\iff \angle MO_9I=90^{\circ} \iff IO_9\perp OH \iff IO=IH \iff \text{one of the angles $\angle A, \angle B, \angle C$ is $60^{\circ}$.}$ where the last part is due to the Lemma 1.

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#8 h May 2, 2016, 3:19 AM • 1 Y

Y by Adventure10

17. Consider the transformation $\tau$ which is an inversion centered $A$ with radius $\sqrt{AB \cdot AC}$ , then a reflection about the $A$ -inner angle bisector. We denote $\tau(T)$ to be the image of object $T$ under $\tau$ . To start, note that $D$ lies on $BC$ , so $\tau(D)$ lies on $\tau(BC)=(ABC)$ .

Consider the circle $\omega_1$ , then $\tau(\omega_1)$ is a line $\ell_1$ , passes $\tau(D)$ , and parallel to $A\tau(B)=AC$ . Similarly, $\tau(\omega_2)$ is a line $\ell_2$ , passes through $\tau(D)$ , and parallel to $AB$ . Now consider the line $BX$ , then $\tau(BX)$ is a circle passes through $A, \tau(B)=C$ and tangent to $\tau(\omega_1)=\ell_1$ . Conseuqently the contact point of $\ell_1$ and $\tau(BX)$ is $\tau(X)$ . Similarly define $\tau(Y)$ , and $\tau(CY)$ is tangent to $\ell_2$ at $\tau(Y)$ .

To prove $(XDY)$ is tangent to $BC$ , it suffices to show that $\tau((XDY))=(\tau(X)\tau(D)\tau(Y))$ is tangent to $\tau(BC)=(ABC)$ . Let $\tau(D)\tau(X)$ intersect $\Gamma$ again at $X'$ , and $\tau(D)\tau(Y)$ intersect $\Gamma$ again at $Y'$ , then $AB\tau(D)X'$ is an isosceles trapezoid, so the circle passes though $A$ and $B$ and tangent to the segment $\tau(D)X'$ must be tangent to it at its midpoint. So $\tau(X)$ is the midpoint of $\tau(D)X'$ , and similarly $\tau(Y)$ is the midpoint of $\tau(D)Y'$ . So $X'Y'\parallel \tau(X)\tau(Y)$ , implying the circles $(\tau(X)\tau(D)\tau(Y))$ , $(ABC)$ are homothetic at $\tau(D)$ . Therefore they are tangent to each other at $\tau(D)$ .

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136 posts

#9 h May 2, 2016, 2:27 PM • 2 Y

Y by Adventure10, Mango247

Any solution to problem 20??

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479 posts

#10 h May 2, 2016, 2:50 PM • 2 Y

Y by aopser123, Adventure10

20. Observe $P$ and $Q$ lies on $ABA_0$ and $ACA_0$ respectively. This is because $AB\neq BA_0$ , $PB$ bisects $ABA_0$ , and $PA=PA_0$ , and likewise for $Q$ .

Next, observe that triangles $\triangle A_0PQ$ and $\triangle A_0B_0C_0$ are similar, because $\angle QPA_0=\frac{1}{2}\angle APA_0=90-\frac{\angle B}{2}=\angle C_0B_0A_0$ , and likewise $\angle PQA_0=\angle B_0C_0A_0$ .

So consider the spiral center $Z$ mapping $\triangle A_0PQ$ to $\triangle A_0B_0C_0$ , it lies on circle $(A_0B_0C_0)$ , and the lines $PB_0, QC_0$ , i.e. $PB_0, QC_0$ meet at $Z\in (A_0B_0C_0)$ , as desired.

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136 posts

#11 h May 2, 2016, 8:10 PM • 1 Y

Y by Adventure10

Thanks navi_09220114.

Also is there any solution to problem 16?

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82 posts

#12 h May 3, 2016, 11:57 AM • 1 Y

Y by Adventure10

Here is a solution for problem 16:
Let $A',B',C'$ be midpoints of $BC,AC,AB$ respectively. We immediately have $A',B_1,N$ and $A',C_1,M$ are colinear. Now because $A'C'||AM$ we have $\triangle C_1A'C' \sim \triangle C_1MA \sim B_1MC_1$ (last similarity is because $MC_1$ is tangent to circumcircle of $\triangle AB_1C_1$ ). From this similarity we get $C_1C' \cdot B_1M= A'C_1 \cdot B_1C_1$ . Doing the exact same way on the other side we get $B_1B' \cdot C_1N=A'B_1 \cdot B_1C_1$ so we conclude $\frac{C1C'}{C_1N}=\frac{B_1B'}{B_1M}$ . Now because $C_1,B_1$ are perpendiculars from $H$ and $C',B'$ perpendiculars from $O$ on $AB,AC$ we conclude that perpendiculars at $M,N$ to $AC,AB$ intersect on $OH$ at some point $X$ . Note that $AMXN$ is cyclic. Now if $Y$ is perpendicular from $A$ to $OH$ we have $\angle AYH=90=\angle AYX$ so $Y$ is the second intersection of $AC_1B_1H$ and $AMNX$ so we are finished.

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1059 posts

#13 h May 3, 2016, 12:28 PM • 3 Y

Y by buzzychaoz, Adventure10, Mango247

Everyone, could we please discuss these problems one by one in different topics? It would be nice to add the 2016 Sharygin Olympiad to the Contests page as well. Has anyone posted some of these problems yet separately? If yes, please PM me the link.

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