SAMO Problem 5: Cyclic quadrilateral through points of two isosceles triangles

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SAMO Problem 5: Cyclic quadrilateral through points of two isosceles triangles

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Source: South African Mathematics Olympiad 2020, Problem 5

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#1 h Jan 17, 2021, 3:30 AM

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Let $ABC$ be a triangle, and let $T$ be a point on the extension of $AB$ beyond $B$ , and $U$ a point on the extension of $AC$ beyond $C$ , such that $BT = CU$ . Moreover, let $R$ and $S$ be points on the extensions of $AB$ and $AC$ beyond $A$ such that $AS = AT$ and $AR = AU$ . Prove that $R$ , $S$ , $T$ , $U$ lie on a circle whose centre lies on the circumcircle of $ABC$ .

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#3 h Jan 17, 2021, 7:45 AM

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$\triangle RAT\equiv\triangle SAU$ since are isosceles and $\angle RAT=\angle SAU$ then $RTUS$ are isosceles trapezoid. Take $D$ a circumcenter of $RTUS$ and we have by angle chasing:
$\angle TDU=2\angle TRU=2\frac{\angle BAC}{2}=\angle BAC$ Suppose that $G=TD\cap\odot(ABC)$ and $F=UD\cap\odot(ABC)$
Then if $D\in \odot(ABC)$ we must have:
$\angle ABD=\angle ACD$ Clearly $\triangle DRU\simeq DST$ since $ST=RU$
Since $BT=CU$ by LAL $\triangle BTF\simeq\triangle CDU$ then $\angle DBT=\angle DCU$ then $\angle ABD=\angle ACD$ then $ABCD$ are cyclic and $D\in \odot(ABC)$ $\blacksquare$

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This post has been edited 2 times. Last edited by MiltonMath12, Jan 17, 2021, 7:53 AM

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#4 h Jan 20, 2021, 10:18 AM

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Easy, but nice.

By Power of a Point,
$AU\cdot AS=AR\cdot AT\implies RSTU\text{ is cyclic.}$
Now we claim that the midpoint of arc $BAC$ is the centre of $(RSTU)$ .
Let that midpoint be $F$ .

We have $\triangle FCU\cong \triangle FBT$ , since
$\measuredangle UCF=\measuredangle ACF=\measuredangle ABF=\measuredangle TBF$ and $FB=FC \text{ and } CU=BT.$
Also, $\triangle SFC\cong \triangle RFB$ , since
$\measuredangle SCF=\measuredangle ACF=\measuredangle ABF=\measuredangle RBF$ and $FB=FC \text{ and } CS=AC+AS=AU+AB=BR.$
Thus, $F$ lies on the perpendicular bisectors of $UT$ and $SR$ , meaning it is indeed the centre of $(RSTU)$ .

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