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Source: 2019 Belarus Team Selection Test 6.1

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#1 h Sep 2, 2019, 8:20 AM • 1 Y

Y by Adventure10

Two circles $\Omega$ and $\Gamma$ are internally tangent at the point $B$ . The chord $AC$ of $\Gamma$ is tangent to $\Omega$ at the point $L$ , and the segments $AB$ and $BC$ intersect $\Omega$ at the points $M$ and $N$ . Let $M_1$ and $N_1$ be the reflections of $M$ and $N$ about the line $BL$ ; and let $M_2$ and $N_2$ be the reflections of $M$ and $N$ about the line $AC$ . The lines $M_1M_2$ and $N_1N_2$ intersect at the point $K$ .
Prove that the lines $BK$ and $AC$ are perpendicular.

(M. Karpuk)

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niksic

#2 h Apr 6, 2020, 6:00 PM

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Lemma 1: $BL$ bisects $\angle ABC$
Let $BL \cap \odot ABC \equiv L'$ . Consider an inversion through $B$ with radius $\sqrt{AB\cdot BC}$ with reflection through bisector of $\angle ABC$ , and for clarity denote the inversion as a function $\psi$ . Then $\psi (A)=C$ and reverse, $\psi (\Gamma) = \overline{AC}$ and reverse, and $\psi (\Omega)$ is a line through $\psi (L)$ parallel to $AC$ . But we see that $\psi (L) \equiv L'$ reflected through angle bisector of $\angle B$ and so $\psi (L)$ had to be the midpoint of the arc $AC$ of $\Gamma$ , and this implies that $BL$ is the actual angle bisector of $\angle B$ .

Lemma 2: $L$ is the center of $\odot MM_1M_2NN_1N_2$
Lemma 1 tells us $M_1 \in BC$ and $N_1 \in BA$ . Also, by the reflection conditions we get $LM = LM_1 = LM_2 (1)$ and $LN = LN_1 = LN_2 (2)$ . Also by lemma 1 we have $\angle NBL = \angle MBL \implies LN = LM$ . Combining this with $(1)$ and $(2)$ gives the result

Lemma 3: $B, N_1, M_1, K$ are concyclic
$NN_2 \perp AC$ and $MM_2 \perp AC$ so $NN_2 \parallel MM_2$ . Also, by the tangency condition we have $\angle NLC = \angle NBL = \angle MBL = \angle MLA$ , and this along with reflection gives $\angle NLN_2 = \angle MLM_2$ . Combining this with $NN_2 \parallel MM_2$ tells us $M, L , N_2$ are collinear, as well as $N, L, M_2$ . Because L is the center of the circle given in lemma 2, we have $MN_1 \perp N_1N_2 \implies \angle KN_1B = \pi /2$ and $NM_1 \perp M_1M_2 \implies \angle KM_1B = \pi /2$ , which give the result. For the next lemma keep in mind that $BK$ is the diameter of this circle

Lemma 4: $L\in \odot B, N_1, M_1, K$
$M_1NN_1M$ is and isosceles trapezoid, so we have $M_1N_1 = MN = M_2N_2$ , where the last equality follows from reflecting $MN$ through $L$ . Also $M_1N_2 \parallel N_1M_2$ because $M_1N_2 \perp M_1M$ and $N_1M_2 \perp NN_1$ , and obviously $MM_1 \parallel NN_1$ . All of this tells us $N_1M_1N_2M_2$ is an isosceles trapezoid, so $KL \perp M_1N_2$ and $N_1M_2$ so $KL\parallel MM_1$ . But we also have $BL\perp MM_1$ so $\angle KLB = \pi /2$ , and this gives the result

Final angle chase:
We have $\angle KBN_1 =\angle KM_1N_1 =\angle M_2M_1N_1 =\angle M_2MN_1 \implies BK \parallel MM_2$ , but $MM_2 \perp AC$ so $BK\perp AC$ $\blacksquare$

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#3 h May 1, 2021, 7:29 AM

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By Archimedes $BL$ bisects $\angle{ABC}$ . Then $LN_{2}=LN_{1}=LN=LM=LM_{1}=LM_{2}$ Therefore $NM_{1}M_{2}MN_{1}N_{2}$ is cyclic. By Pascal on it we get that $B=NM_{1} \cap MN_{1};\; K=M_{1}M_{2} \cap N_{1}N_{2}$ and $X=M_{2}M \cap N_{2}N$ are collinear. But $X$ is a point at infinity for altitude from $B$ . QED

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#4 h May 1, 2021, 8:22 AM

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Obviously, $L$ is the circumcenter of $\triangle MM_1M_2$ and $\triangle NN_1N_2$ .
By a well known homothety, we know that $BL$ is the angle bisector of $\angle ABC$ and since $BMLN$ is cyclic, we have $LM=LN$ , implying that $MM_1M_2NN_1N_2$ are concyclic with center $L$ . And by symmetry, we have $M,L,N_2$ and $N,L,M_2$ are collinear.
Since $MN_1 = NM_1$ , we have $\angle N_1N_2M = \angle KN_2L = \angle NM_2M_1 = \angle LM_2K \implies LKN_2M_2$ is cyclic.
We have $\angle LKM_1= 180 - \angle LKM_2 = 180 - \angle LN_2M_2 = 180 - \angle MN_2M_2 = 180 - \angle MNM_2 = 180 - \angle MNL = 180 - \angle LBN \implies LBM_1K$ is cyclic. Since $N_1 \in (LBM_1)$ , we have $LN_1BM_1K$ is cyclic.
Finally, let $\angle MBL = x$ , $\angle LBK = y$ .
We have $\angle M_2MB = 90 - x + 90 - y = 180 - x - y = 180 - \angle MBK \implies BK\parallel MM_2 \implies BK\perp AC$ . $\square$

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