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Problem 3 of Finals

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Source: X International Festival of Young Mathematicians Sozopol 2019, Theme for 10-12 grade

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Pinko

#1 h Oct 3, 2019, 8:55 AM • 2 Y

Y by Adventure10, Mango247

$\Delta ABC$ is isosceles with a circumscribed circle $\omega (O)$ . Let $H$ be the foot of the altitude from $C$ to $AB$ and let $M$ be the middle point of $AB$ . We define a point $X$ as the second intersection point of the circle with diameter $CM$ and $\omega$ and let $XH$ intersect $\omega$ for a second time in $Y$ . If $CO\cap AB=D$ , then prove that the circumscribed circle of $\Delta YHD$ is tangent to $\omega$ .

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#2 h Nov 12, 2019, 1:58 PM • 3 Y

Y by Pinko, Adventure10, Mango247

let $Z$ intersection of altitudes. So $\angle CXM = 90^\circ$ let $XM\cap \omega=R$ and $CR$ diameter. So $\angle CBR =\angle CAR = 90^\circ$ . So $AZ \parallel RB$ and
$AR \parallel BZ$ . So $AZBR$ parallelogram and $ZR$ passes from $M$ so $ZR\cap \omega=X$ . Let $AB\cap XC=T$ . Because of $MZ,CZ$ are altitudes in $CTM$ , $TZ$ perpendicular to $CY$ . Let $TZ\cap CY=S$ . Power in $\omega$ we have $TA.TB=TX.TC$ and power in $(CXZ)$ we have $TZ.TS=TX.TC$ so $TZ.TS=TA.TB=TX.TC$ . So $ABSZ$ cyclic. Because of $CTHS$ cyclic we have $\angle TCH = \angle HST$ . Also $AZSB$ cyclic so $\angle ZSA = \angle ZBA = \angle ACZ$ . So $\angle HYA = \angle CXA = \angle TCH - \angle ACZ = \angle TSH - \angle ZSA = \angle ASH$ . Because of $HXCM,HTCS$ cyclic we have $\angle YHM = \angle XHT = \angle SCX = \angle SHM$ . So because of $\angle HYA = \angle ASH$ and $\angle YHM = \angle SHM$ . So we have $S$ and $Y$ symetric respect to $AB$ .
power lines of $CXZ$ , $ABC$ , $ABPQ$ intersect one point. Where $BZ\cap AC=Q$ . And $AZ\cap BC=P$ . So $P,Q,T$ linear. So we have because of $AP,BQ,CH$ intersect at $Z$ $T,A,H,B$ harmonic. So $HM.MT=MA^2$ . So $HM.MT=MS.MC=MA^2$ . Because of $CSHT$ cyclic $HM.MT=MS.MC$ . Because of $MS.MC=MA^2$ we have $\angle SAB = \angle ACM$ and $\angle SBA = \angle MCB$ . Because of $Y,S$ symetric $\angle YAB = \angle ACM$ and $\angle YBA = \angle BCM$ .
Because of $\angle YAB = \angle ACM$ , $\angle YBA = \angle BCM$ and $AYBC$ cyclic we have $\angle YCH = \angle ACY - \angle ACH = \angle ABY - \angle DCB = \angle BCM - \angle DCB = \angle DCM$ . Because of $CR$ diameter $D$ is on $CR$ .
So we have $\angle YCD = \angle YCM + \angle DCM = \angle YCM + \angle YCH = \angle HCS$ . Because of $S,Y$ symetric and $THSC$ cyclic we have $\angle YTD = \angle STM = \angle SCH = \angle YCD$ so $TDYC$ cyclic. So we had $\angle XCY = \angle HDY$ .
If we take $l=LY$ tangent to $(HYD)$ we have $\angle LYX = \angle HDY$ . Let $L$ closer to $H$ . So we have $\angle LYX = \angle XCY$ . So we have $l$ tangent to $(ABC)$ but
we didn't use isosceles. So we have result.

This post has been edited 1 time. Last edited by kocaman_math10, Nov 12, 2019, 2:00 PM

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ABCCBA

#3 h Nov 12, 2019, 2:54 PM • 2 Y

Y by Pinko, Adventure10

It's well-known that $CY$ is $C-$ symmedian of triangle $CAB,$ combine with $(CHD)$ is tangent to $(O):$ call $T$ the intersection of tangent at $Y$ of $(O)$ and common tangent at $A$ of $(CHD)$ and $(O)$ with $BC$
Then $TY^2=TA^2=TH.TD,$ so $(YHD)$ is tangent to $(O)$

This post has been edited 1 time. Last edited by ABCCBA, Nov 12, 2019, 2:55 PM

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