ST passes through midpoint of arc ABC

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ST passes through midpoint of arc ABC

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Source: saint petersburg 2015,grade 11,P7

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#1 h Mar 14, 2016, 3:58 PM • 4 Y

Y by buratinogigle, anantmudgal09, Adventure10, Mango247

Let $BL$ be angle bisector of acute triangle $ABC$ .Point $K$ choosen on $BL$ such that $\measuredangle AKC-\measuredangle ABC=90º$ .point $S$ lies on the extention of $BL$ from $L$ such that $\measuredangle ASC=90º$ .Point $T$ is diametrically opposite the point $K$ on the circumcircle of $\triangle AKC$ .Prove that $ST$ passes through midpoint of arc $ABC$ .(S. Berlov)
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This post has been edited 1 time. Last edited by MRF2017, Mar 14, 2016, 4:02 PM

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#2 h Mar 15, 2016, 3:13 AM • 2 Y

Y by Adventure10, Mango247

Taking our point of reference at $T$ and extending $AT,CT$ to intersect $(ABC)$ gives us the following equivalence:

https://lh3.googleusercontent.com/GJCu4jGL-vSZFi_C38KcuX9zh4YPzgddXlup_I3fyL_4iq1mYHVIqeJR5rRPfKKMnQY2TLn58wIeAdJrqxAbgVWbbbvDTvxuhfDfP-q8fvhTZPdXjxN6UQPKyP-SzEuMm_dK4vnjoopBcIwtwB1bm_69-l4GvEcgX2pHOS9j2gR7kUePxa-cmlbkXOdcZXNVyzE6N-jjcZ_HPEKzmndQDDLnDbTZclAmTYP8qzp3e9dWeFrwvbqvVRm5eZmZQyqGvMxyx4k0M5Eqib6JV3fmFl2Zq1T9x7AvB5ECHSvXX-vahdEaKE7rX3DNv-XdWtWDgtbT0CXDq0UDMSa2ePEPFugme12TnUBClIIkhLUw7sxzlozr2ILxGYk-D6VhnXIXmgAVBi_jU0SJLDIvkGuK6V1Rv9mznd3FOICr_r1lAleVft0VO3iah0I3SWtZa7ljtT8T7WsZtUzB-jw8AGbH2xu83X2jA_M4v2WRbqCfMkIorR9KzqzVyVL-wnO-PdxnIckXL647YYCD-wcVHbUAqQFnqs_5IfAhnNok9CwsnTlWrh30rnPMYv1PZic2xI3KirQ=w532-h626-no

Let the altitudes $BE,CF$ of $ABC$ meet at $H$ . Let $M,N$ denote the midpoints of arc $FE$ such that $M$ lies on the same side of $EF$ as $A$ . Suppose $K$ lies on $MH$ such that $\angle FKE=90^{\circ}$ and $K$ lies on the same side of $EF$ as $A$ . Prove that $A,N,K$ are collinear.

Proof: Since $AFE\sim ABC$ , construct $M'$ in $AFE$ such that it corresponds to $M$ in $ABC$ . This implies $AM,AM'$ are isogonal wrt $\angle A$ . Since $\angle ECM=\angle FBM$ because $M$ is the midpoint of arc $EF$ , and $M,N$ are symmetric about the midpoint of $BC$ , therefore it is well known that $AM,AN$ are isogonal. Thus $M'\in AN$ . Once we show that $K\equiv M'$ , the equivalence follows.

Since $\angle FM'E=\angle BMC=90^{\circ}$ , it suffices show $M'\in HM$ . Let $BM\cap HF=X, CM\cap HE=Y$ . Note that $\angle AEM'=\angle ABM=\angle ACM\implies CM||EM'$ , and $BM||FM'$ similarly. Since $BX,CY$ are angle bisectors in similar triangles $BHF, CHE$ respectively, therefore
$\frac {FX}{XH}=\frac {YE}{YH}\implies XY||EF$ It follows that $EFM',YXM$ are homothetic about $H$ , so $H,M,M'$ are collinear and we are done.

This post has been edited 1 time. Last edited by XmL, Mar 15, 2016, 3:18 AM
Reason: Added diagram

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#3 h Jun 5, 2016, 10:19 PM • 3 Y

Y by tenplusten, Adventure10, Mango247

Let bisector $BM$ (external) meet $BC$ at point $N$ and let $NK$ meet $MS$ at point $J$ . We claim that $NK \perp MS$ . This follows since an inversion about $B$ of radius $\sqrt{BA.BC}$ maps $S$ to $K$ and $M$ to $N$ giving that $MS$ and $NK$ are anti-parallel in angle $ABC$ and since $\angle MBL=90$ we have that $NK \perp MS$ . Now, by radical axis theorem, we get that $L$ lies on $(AKC)$ . Indeed, $NA.NC=NB.NM=NK.NJ$ and we get the claim. Now, clearly $\angle KJS=90$ implies that $JS$ passes through the anti-pose of $K$ in $AKC$ . This proves that $M,T,S,J$ are collinear.

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#4 h Jan 1, 2018, 9:05 PM • 2 Y

Y by Adventure10, Mango247

Can anyone post non-inversive solution please?
Thanks!!!
One thing İ proved is that:
The center of $(AKTC)$ is the intersection of $AA$ and $CC$ .

This post has been edited 1 time. Last edited by tenplusten, Jan 1, 2018, 9:06 PM

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