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Source: Turkey Tst 2017 p3

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160 posts

#1 h Mar 30, 2017, 9:11 AM • 4 Y

Y by den_thewhitelion, aopser123, Adventure10, Mango247

At the $ABC$ triangle the midpoints of $BC, AC, AB$ are respectively $D, E, F$ and the triangle tangent to the incircle at $G$ , $H$ and $I$ in the same order.The midpoint of $AD$ is $J$ . $BJ$ and $AG$ intersect at point $K$ . The $C-$ centered circle passing through $A$ cuts the $[CB$ ray at point $X$ . The line passing through $K$ and parallel to the $BC$ and $AX$ meet at $U$ . $IU$ and $BC$ intersect at the $P$ point. There is $Y$ point chosen at incircle. $PY$ is tangent to incircle at point $Y$ . Prove that $D, E, F, Y$ are cyclic.

This post has been edited 3 times. Last edited by KereMath, Sep 15, 2017, 10:43 AM

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778 posts

#2 h Mar 30, 2017, 4:06 PM • 1 Y

Y by Adventure10

any solution

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160 posts

#3 h Mar 30, 2017, 8:19 PM • 1 Y

Y by Adventure10

Hint: we can use feuerbach's theorem:the nine point circle is tangent to incircle.

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54 posts

#4 h Mar 30, 2017, 8:24 PM • 2 Y

Y by Adventure10, Mango247

Yep, but I think you should correct the statement as it is false.

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374 posts

#5 h Mar 30, 2017, 8:59 PM • 3 Y

Y by Garfield, Adventure10, Mango247

Let me restate problem first to nicer formulation.

Problem: Let $\triangle ABC$ be given and it's incircle $(I)$ . Let $(I)$ touches sides $AB$ , $BC$ and $AC$ at $F$ , $D$ and $E$ , respectively and denote by $P$ , $M$ and $N$ midpoints of same sides, respectively. Let $AM\cap PN=\{ J\}$ and $AD\cap BJ=\{ K\}$ . If circle centered at $C$ with radius $AC$ cuts $BC$ at $X$ , line parallel to $BC$ through $K$ meets $AX$ at $U$ and $FU\cap BC=\{ R\}$ , then prove that $Q$ lies on tangent from Feuerbach point $F_e$ onto $(I)$ .

Notation:
Let $A_H$ be foot of altitude from $A$ on $BC$ , let $R$ be intersection point of tangent from $F_e$ on $(I)$ with $BC$ and denote with $a$ , $b$ and $c$ sides $BC$ , $AC$ and $AB$ .

Proof:
Assume WLOG that $b>a$
Let $R'$ be harmonic conjugate of $R$ WRT $A_HM$ i.e. $-1=(R,\ R';\ A_H,\ M)\Longleftrightarrow \frac{RA_H}{RM}=\frac{R'A_H}{R'M}=\frac{F_eA_H^2}{F_eM^2}$ But from $\text{Casey's}$ we have
$\begin{align*} \left(\frac{F_eA_H}{F_eM}=\frac{A_HD}{MD}\right)^2& \Longrightarrow \frac{RA_H}{RM}=\frac{A_HD^2}{MD^2}\\ & \Longleftrightarrow \frac{MA_H}{RM}=\frac{MD^2-A_HD^2}{MD^2}\\ & \Longleftrightarrow RM=\frac{MD^2\cdot MA_H}{(MD-A_HD)\cdot MA_H}\\ & \qquad \qquad \ \ = \frac{MD^2}{MD-A_HD}\\ & \qquad \qquad \ \ = \frac{MD^2}{2MD-A_HM}\\ & \qquad \qquad \ \ = \frac{\left(\frac{a-(a+c-b)}{2}\right)^2}{(a-(a+c-b))-A_HM}\\ & \qquad \qquad \ \ = \frac{\frac{(b-c)^2}{4}}{b-c-\frac{b^2-c^2}{2a}}\\ & \qquad \qquad \ \ = \frac{a(b-c)}{2(2a-b-c)} \end{align*}$
Now let's finish it!
$\begin{align*} \begin{rcases} RB=RM- BM=\frac{a(b-a)}{2a-b-c}\\ RX=RB-(b-a)=\frac{(a-b)(a-b-c)}{2a-b-c}\\ BF=\frac{a+c-b}{2}\\ AF=\frac{b+c-a}{2}\\ \frac{AU}{UX}\stackrel{UK||XD}{=}\frac{AK}{KD}\stackrel{\text{Menelaus on }\triangle ADM}{===}\frac{BM}{BD}\cdot \frac{AJ}{JM}\stackrel{AJ=JM}{=}\frac{BM}{BD} =\frac{a}{a+c-b} \end{rcases}\Longrightarrow \frac{RX}{RB}\cdot \frac{BF}{FA}\cdot \frac{AU}{UX}&=\frac{\frac{(a-b)(a-b-c)}{2a-b-c}}{\frac{a(b-a)}{2a-b-c}}\cdot \frac{\frac{a+c-b}{2}}{\frac{b+c-a}{2}}\cdot \frac{a}{a+c-b}\\ & =1 \end{align*}$ And by Menelaus we are done!

This post has been edited 1 time. Last edited by mihajlon, Mar 30, 2017, 9:06 PM
Reason: forgot to write WLOG condition...

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370 posts

#6 h Jan 27, 2018, 5:19 PM • 2 Y

Y by Adventure10, Mango247

Sorry for revive, but is there any proof without calculations?

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#7 h Mar 2, 2020, 11:20 AM

Y by

KereMath wrote:

Hint: we can use feuerbach's theorem:the nine point circle is tangent to incircle.

Hortlattım sorry

Post solution without calc reco

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6 posts

#8 h Mar 2, 2020, 11:37 AM

Y by

İnmemories you want too much things bro. It`s Fehmi Kadan`s question. What dou you expect?

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#9 h Mar 3, 2020, 6:49 AM

Y by

Kyolcu22 wrote:

İnmemories you want too much things bro. It`s Fehmi Kadan`s question. What dou you expect?

Something synthetic

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61 posts

#10 h Feb 14, 2021, 4:36 AM

Y by

Once we see that all points are actually well-described points, what can stop us from bary bashing??

First, let's see that our point $P$ must satisfy $PG^2=PD\cdot PH$ , where $H$ is the foot of altitude from $A$ . Also, a point satisfying this condition is unique. Remember that some of the coordinates in our solution are non-normalized:
$D=(0:1/2:1/2)$ , $J=(1:1/2:1/2)$ and $G=\left( 0: \frac{s-c}{a} : \frac{s-b}{a}\right)$ so writing the equation of $AG: \frac{s-c}{a}\cdot z - \frac{s-b}{a}\cdot y = 0 \text{ and } BJ : x-2z=0$ We deduce that $K=(2\cdot(s-b) : s-c: s-b)$ and the equation of the line parallel to $BC$ is $ax -2(s-b)(y+z)=0$ .
Note that $X=(0:b:a-b)$ and the equation of line $AX: bz -(a-b)\cdot y=0$
From these, we get $U=(2(s-b): b: a-b)$ and we know that $I= \left( \frac{s-b}{c} : \frac{s-a}{c} : 0\right)$ , so $UI : -(s-a)(a-b)x + (s-b)(a-b)y + (s-b)(c-a)z = 0$ deducing $P=(0 : a-c : a-b)$ , so its normalized coordinates are $P= \left(0 , \frac{a-c}{2a-b-c}, \frac{a-b}{2a-b-c}\right)$ . Also note that normalized coordinates of $H = \left(0 , \frac{a^2+b^2-c^2}{2a^2}, \frac{a^2-b^2+c^2}{2a^2}\right)$
With these coordinates, we can find the vectors $\overrightarrow{PG}$ , $\overrightarrow{PD}$ and $\overrightarrow{PH}$ . All these vectors lie on $BC$ so the lenghts are basically $-a^2yz$ for any $\overrightarrow{PQ} = (x,y,z)$ , $Q \in \{G,D,H\}$ , because of the fact that $x=0$ . The rest is just some annoying computations.

If I didn't make a typo, desired equality holds for these coordinates

This post has been edited 2 times. Last edited by badumtsss, Apr 5, 2021, 4:03 PM

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#11 h Jul 29, 2021, 6:41 PM

Y by

BUMPing for a synthetic solution

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#12 h Dec 26, 2022, 5:53 AM • 1 Y

Y by Mango247

inmemories wrote:

Kyolcu22 wrote:

It`s Fehmi Kadan`s question. What dou you expect?

Something synthetic

I think Fehmi Kadan does not always propose extremely difficult or bashy problems, but of course sometimes he did ask harder ones. In fact, geometry solutions in the booklets of Turkish National Olympiad (the official source) never used advanced/ultimate concepts, for example, barycentric coordinates or complex numbers or just plain trigonometry.

This post has been edited 1 time. Last edited by charme7269, Dec 26, 2022, 5:55 AM

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