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Source: Iran second round 2016, day 2, problem 5

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164 posts

#1 h Apr 29, 2016, 12:30 PM • 2 Y

Y by Adventure10, Mango247

$ABCD$ is a quadrilateral such that $\angle ACB=\angle ACD$ . $T$ is inside of $ABCD$ such that $\angle ADC-\angle ATB=\angle BAC$ and $\angle ABC-\angle ATD=\angle CAD$ . Prove that $\angle BAT=\angle DAC$ .

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532 posts

K.N

#2 h Apr 29, 2016, 12:40 PM • 1 Y

Y by Adventure10

سلام من از راه حلم عكس دارم ولي چطور بايد بذارمش تو اين جا؟

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K.N

#3 h Apr 29, 2016, 12:42 PM • 1 Y

Y by Adventure10

See here too
Iran second round 2016 day 2-p6
http://artofproblemsolving.com/community/q1h1235174p6270026

This post has been edited 1 time. Last edited by K.N, Apr 29, 2016, 12:45 PM

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#4 h Apr 29, 2016, 1:31 PM • 2 Y

Y by seyyedmohammadamin_taheri, Adventure10

Finally i found
Here's my solution
Sorry!
A little in farsi

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doxuanlong15052000

269 posts

doxuanlong15052000

#5 h Apr 29, 2016, 3:21 PM • 2 Y

Y by Adventure10, Mango247

My solution:
Let $x,y$ be the tangent of $(ATD)$ and $(ATB)$ pass through $A$ . We have $\angle XAC=\angle ATD+\angle DAC=\angle ABC$ $\Longrightarrow$ $x$ is the tangent of $(ABC)$ . Similarly, we have $y$ is the tangent of $(ADC)$ . From $\angle BCA=\angle ACD$ $\Longrightarrow$ $\angle ATB=\angle ATD$ $\Longrightarrow$ $TA$ is the bisector of $\angle DTB$ .
Let $TD$ cut $CB$ at $L$ . We will define the problem again: Suppose $A$ is the incenter of $\triangle DAC$ . $B$ is a point on $LC$ and $T$ is a point on the ray $LD$ sucth that $TA$ is the bisector of $\angle DTB$ . $\Longrightarrow$ $\angle TAB=\angle DAC=90^o+\frac{\angle DLC}{2}$ $\Longrightarrow$ $\angle ABC=\angle DAC+\angle ATB$ and $\angle ADC=\angle BAC+\angle ATB$ $\Longrightarrow$ done.

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Gabriel_ZaslaVsky

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Gabriel_ZaslaVsky

#11 h Apr 30, 2016, 3:10 AM • 7 Y

Y by Ryoma.Echizen, YellowDog, Dadgarnia, Romarios, H.HAFEZI2000, Adventure10, Mango247

$\angle ABC$ + $\angle CAB$ = $\angle ADC$ + $\angle CAD$ $\Longrightarrow$ $\angle CAD$ + $\angle ATD$ + $\angle CAB$ = $\angle ATB$ + $\angle CAB$ + $\angle CAD$ $\Longrightarrow$ $\angle ATD$ = $\angle ATB$ it means $C$ & $T$ move on a hyperbola( $B$ & $D$ are focal points).
Let $C$ & $T$ go to $\infty$ you see their hyperbola is the same $\Longrightarrow$ $\angle CAD$ = $\angle TAB$ .
Good luck in your olympiad

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#13 h May 1, 2016, 7:16 AM • 2 Y

Y by Adventure10, Mango247

It' easy to prove <ATB=<ATD. K on CD that ABCK cyclic quadrilateral . Intersection BK AND AD is X . L on BC that CDAL cyclic quadrilateral . Intersection DL and AB is Y . It's easy to prove XTBA and YTDA cyclic quadrilateral .Intersection BK and DL is M . We have<BXT=<BAT=<YDT .===》》》XDMT cyclic quadrilateral .===》》》》<DTM=<MXD=<ATB=<ATD .===》》》 A ,T,M lies on line . it's easy to prove AMKD cyclic quadrilateral .===》》》<MAD=<MKC=<BAC . ===》》》<BAT=<DAC.

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Mahdi_Mashayekhi

689 posts

Mahdi_Mashayekhi

#14 h Apr 6, 2022, 1:45 PM

Y by

Awesome Geometry.
Let $\angle BCA = \angle ACD = a$ , $\angle BAC = y$ , $\angle DAC = x$ , $\angle BTA = b$ and $ATD = t$ . Let $M,N$ be points on extension of $CB,CD$ . $\angle 180 - a - x - b = y$ and $\angle 180 - a - y - t = x$ so $b = t$ . Let $S$ be on $AC$ such that $\angle SBC = x$ . Now we have $SBC$ and $DAC$ have spiral similarity and also $ABC$ and $DSC$ . we have $\angle ABS = \angle ABC - x = t$ and $ADS = \angle ADC - y = b$ . Let $AT$ meet $BS$ at $K$ . Note that for proving $\angle BAT = x$ we need to prove $BAK$ is tangent to $BC$ or $\angle BKA = \angle ABM$ . $\angle BKA = b - \angle TBK = t - \angle TBK = \angle ABT$ . Let $BT$ meet $DS$ at $Q$ . $\angle ADQ = b = \angle ATB \implies ATQD$ is cyclic so $\angle AQD = ATD = t = \angle ABS \implies ABSQ$ is cyclic. $\angle BKA = \angle ABT = \angle ABQ = \angle ASQ = \angle 180 - \angle DSC= \angle 180 - \angle ABC = \angle MBA$ so $\angle BKA = \angle ABM$ so we're Done.

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559 posts

#15 h Apr 8, 2022, 7:24 PM • 1 Y

Y by XX-math-XX

Let $\odot (ABT),\odot (ADT)$ intersect $AC$ again at $P,Q,$ and $AD,AB$ at $X,Y$ respectively. From $\measuredangle AXP=\measuredangle ATP=\measuredangle ATB+\measuredangle BAP=\measuredangle ADC$ we obtain $PX\parallel CD$ and analogously $PY\parallel CB.$ Therefore $\measuredangle YBX=\measuredangle ATX=\measuredangle ACD=\measuredangle BCA=\measuredangle YTA=\measuredangle YDX,$ which yields $AXYT\stackrel{-}{\sim} ABDC,$ and the conclusion follows.

This post has been edited 1 time. Last edited by JAnatolGT_00, Apr 8, 2022, 7:28 PM

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