Is it correct ?

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skytin wrote: Is it correct ? skytin is right. I draw the picture using geogebra and I got the same result. part a of this problem is wrong. I don't know whats the benefit of giving wrong problems in exams such as TST. EDIT: see the next post!

Thanks to MahammadP that informed me of a typo in the problems statement. $K'$ and $H'$ are on angle bisectors of $\angle BCO$ and $\angle CBO$ respectively. now the problem is correct. (I used geogebra again!)

Solution for (a) : Let K' is on C'H we will prove that H' is on KB' Let C'K' intersect circle w at point S Use Reim's theorem , so HSH'X is cyclic (C'B || H'H) Let KK' and HH' intersect BC at points K_1 and H_1 angle HSC = HH_1C = 90 , so HSH_1C is cyclic angle CSH_1 = CHH_1 = CBB' = CSB' , so H_1 is on SB' Let circles (C'B'C) and (K_1H_1S) are intersected at points F and S Use Reim's theorem , so K_1 is on C'F K_1K'CS is cyclic and K_1 is on YS from Reim's Theorem for lines K'K_1 and Tangent line to w thru point Y Easy to see that KBK_1F is Cyclic , so Use Reim's theorem , so H_1 is on FX Use Reim's theorem , so FBH'H_1 is Cyclic So angle H'FB = H'H_1B = 90 , so H is on FB' Angle BFK = 90 , so F is on KB' . Done

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(b) : Easy to see after Homotety with center S that one of this circles is tangent to w at S like the same another is tangent to w at F That they are tangent to each other i prove later Easy to see that lines KJ and BS are intersected on first circle and JKBS is cyclic we only need to prove that point I is on circle KBS angle K'IB = 2*YXB = K'SB , so K'IBS is Cyclic , we only need to prove that ISis on circle (BK'K) Use Reim's Theorem . Done

To Prove that they tangent we only need to prove that circles (KBS) and (HSC) are intersected on second circle Let they intersects at points I and J' Easy to see that angle FJ'K = IJ'K - IJ'F = ... = FKB . Done

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Sorry for this mistake but it wasn't my fault actually. the problem has written wrong even in the exam paper of tst !

This problem also is very nice! (a). Replace, $H'$ and $K'$ by $H_1$ and $K_1$ and Let $H_2\equiv(H_1B_1)\cap(\omega)$ , $K_2\equiv(C_1K_1)\cap(\omega)$, $H_3$ and $K_3$ the inetersections of $BC$ with $B_1H_1$ and $C_1K_1$ respectively and denote $C_2$ and $B_2$ midpoint of arcs $BC_1$ and $CB_1$ respectively. Let $HH_1$ and $KK_1$ cut $BC$ at $H_4$ and $K_4$ respectively. Easy to see that quadrilateral $K_1K_4K_2C$ is cyclic, hence $\angle{K_4K_2K_1}=\angle{K_4CK_1}=\angle{C_2K_2K_1}$, so $C_2$ , $K_4$ and $K_2$ are collinear. Similarly, $B_2$ , $H_4$ and $H_2$ are collinear. Therefore, $\angle{C_2K_2K_3}=\angle{B_2H_2H_3}$. From another hand, observe that $\angle{H_2K_2K_4}=\angle{H_2CC_2}=\angle{H_2H_4K_4}$ which implies that points $K_4$ , $H_2$ , $K_2$ and $H_4$ are concyclic; Thus $\angle{K_4H_2H_4}=\angle{K_4K_2H_4}$ or again $\angle{K_3K_2H_4}=\angle{H_3H_2K_4}$, And since both of $KH_2K_4B$ and $H_4K_2HC$ are concyclic, we deduce that $K$ , $H_1$ and $B_1$ are collinear if and only if $C_1$ , $K_1$ and $H$ are collinear. (b). Let $OH_2$ meet the perpendicular to $TB$ trough $K$ at $O_k$. Not hard to see that $O_kH_2=O_kK$ and then $(O_k)$ is the circle tangent to $TB$ at $K$ and tangent to $(\omega)$, Similarly, let $OK_2$ meet the perpendicular to $TC$ trough $H$ at $O_h$, then $(O_h)$ is the circle tangent to $(\omega)$, and tangent to $TC$ at $H$. If those two circles are tangent at $J$. It's easy to see that $O_k$ , $O_h$ and $J$ are collinear. The rest now is only an easy angle chasing.

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for the second part, Our teacher used inversion wrt $G$ (the tangency point of two smaller circles). try it!!

[asy][asy] size(10cm); defaultpen(fontsize(9pt)); pen pri=deepblue; pen sec=royalblue; pen tri=blue; pen qua=rgb(41, 207, 255); pen qui=deepcyan; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pen qfil=invisible; pen qifil=invisible; pair O, T, B, C, Bp, Cp, I, K, Kp, H, Hp, M, NN, X, Y, P, Q, O1, O2, J, Kq, Hq; O=(0, 0); T=(1.7, 0); B=intersectionpoints(circle(O, 1), circle(T/2, length(T)/2))[0]; C=intersectionpoints(circle(O, 1), circle(T/2, length(T)/2))[1]; Bp=-B; Cp=-C; I=incenter(O, B, C); K=(2B+7T)/9; Kp=extension(C, I, K, foot(K, B, C)); H=extension(Kp, Cp, T, C); Hp=extension(B, I, H, foot(H, B, C)); M=dir(270); NN=dir(90); X=foot(K, B, C); Y=foot(H, B, C); P=foot(B, Bp, K); Q=foot(C, Cp, H); O1=extension(O, P, K, K+O-B); O2=extension(O, Q, H, H+O-C); J=intersectionpoint(O1 -- O2, circle(O1, length(K-O1))); Kq=extension(K, J, H, Hp); Hq=extension(H, J, K, Kp); filldraw(circle(O, 1), fil, pri); filldraw(circumcircle(B, X, P), sfil, sec); filldraw(circumcircle(C, Y, Q), sfil, sec); filldraw(circumcircle(P, X, Y), sfil, sec); filldraw(circumcircle(B, Y, P), tfil, tri); filldraw(circumcircle(C, X, Q), tfil, tri); filldraw(circle(O1, length(K-O1)), qfil, qua); filldraw(circle(O2, length(H-O2)), qfil, qua); filldraw(circumcircle(B, K, Q), qifil, qui); filldraw(circumcircle(C, H, P), qifil, qui); draw(K -- Bp, tri); draw(H -- Cp, tri); draw(B -- Bp, pri); draw(C -- Cp, pri); draw(B -- T -- C -- B, pri); draw(K -- Kp, pri); draw(H -- Hp, pri); draw(B -- M, sec); draw(C -- NN, sec); draw(M -- P -- Cp, qua); draw(NN -- Q -- Bp, qua); draw(B -- Kq -- K, qui); draw(C -- Hq -- H, qui); dot("$O$", O, W); dot("$T$", T, E); dot("$B$", B, N); dot("$C$", C, S); dot("$B'$", Bp, Bp); dot("$C'$", Cp, Cp); dot("$I$", I, W); dot("$K$", K, dir(10)); dot("$H$", H, dir(-10)); dot("$K'$", Kp, dir(160)); dot("$H'$", Hp, dir(120)); dot("$M$", M, M); dot("$N$", NN, NN); dot("$X$", X, NE); dot("$Y$", Y, SE); dot("$P$", P, SE); dot("$Q$", Q, dir(12)); dot("$J$", J, unit(O1-O2)); dot("$K^+$", Kq, dir(260)); dot("$H^+$", Hq, N); [/asy][/asy] Solution to part (a). Ignore the collinearities for now. Let $\overline{B'K}$ and $\overline{C'H}$ intersect $\omega$ again at $P$ and $Q$, and let $\overline{BC}$ intersect $\overline{KK'}$ at $X$ and $\overline{HH'}$ at $Y$. Denote by $M$ and $N$ the midpoints of arcs $CB'$ and $BC'$, so that $H'\in\overline{BM}$ and $K'\in\overline{CN}$. Claim. Points $Q$, $Y$, $B'$ are collinear; points $P$, $X$, $C'$ are collinear; and $XPQY$ is cyclic. Proof. Since $\measuredangle CQY=\measuredangle CHY=90^\circ-\measuredangle YCH=\measuredangle OCB=\measuredangle CBB'=\measuredangle CQB'$, points $Q$, $Y$, $B'$ are collinear. Analogously $P$, $X$, $C'$ are collinear. Finally the concyclicity follows from the converse of Reim's Theorem. To spell it out, $\measuredangle PXY=\measuredangle PC'B'=\measuredangle PQB'=\measuredangle PQY$. $\blacksquare$ Claim. Points $H$, $K'$, $C'$ are collinear if and only if points $Q$, $X$, $N$ are collinear. Symmetrically, points $K$, $H'$, $B'$ are collinear if and only if points $P$, $Y$, $M$ are collinear. Proof. Check that $\measuredangle K'QX=\measuredangle K'CX=\measuredangle NCB=\measuredangle C'QN$, so \[\overline{HK'C'}\iff\measuredangle C'QX=\measuredangle K'QX\iff\measuredangle C'QX=\measuredangle C'QN\iff\overline{QXN},\]where $\overline{UVW}$ is the assertion that $U$, $V$, $W$ are collinear. $\blacksquare$ Claim. Points $Q$, $X$, $N$ are collinear if and only if points $P$, $Y$, $M$ are collinear. Proof. Assume that $Q$, $X$, $N$ are collinear. It follows that $\measuredangle C'PY=\measuredangle XPY=\measuredangle XQY=\measuredangle NQB'=\measuredangle C'QN$, proving the claim. $\blacksquare$ Putting these claims together yields the desired conclusion. Solution to part (b). First we present a well-known lemma. Lemma (Folklore). Circles $\Gamma_1$ and $\Gamma_2$ are externally tangent at $P$, circles $\Gamma_2$ and $\Gamma_3$ are externally tangent at $Q$, and circles $\Gamma_3$ and $\Gamma_1$ are externally tangent at $R$. Let $A$ be an arbitrary point on $\Gamma_1$. Line $AP$ intersects $\Gamma_2$ again at $B$, line $BQ$ intersects $\Gamma_3$ again at $C$, and line $\overline{CR}$ intersects $\Gamma_1$ again at $D$. Then $\overline{AD}$ is a diameter of $\Gamma_1$. Proof. Let $O_1$, $O_2$, $O_3$ be the centers of $\Gamma_1$, $\Gamma_2$, $\Gamma_3$, respectively. Since $D$ must be the insimilicenter of $\Gamma_1$ and $\Gamma_2$, rays $O_1A$ and $O_2B$ are parallel but in opposite directions. Similarly rays $O_2B$ and $O_3C$ are opposite and parallel, and so are rays $O_3C$ and $O_1D$. Combining these, $\overline{O_1A}$ and $\overline{O_1D}$ are parallel but in opposite directions; in other words, $\overline{AD}$ must be a diameter of $\Gamma_1$, as claimed. $\blacksquare$ For now, ignore the two extra circles. We prove another concyclicity. Claim. $BKQIK'$ and $CHPIH'$ are cyclic. Proof. Since $\measuredangle BKK'=90^\circ-\measuredangle XBK=90^\circ-\measuredangle CC'B=\measuredangle BCC'=\measuredangle BQK'$, quadrilateral $BKQK'$ is cyclic. Furthermore $\measuredangle BIK'=\measuredangle BIN=\measuredangle BON=\measuredangle BQC'=\measuredangle BQK'$, as desired. $\blacksquare$ Claim. $(JPK)$ and $\omega$ are tangent at $P$. Analogously $(JQH)$ and $\omega$ are tangent at $Q$. Proof. The tangent to $(JPK)$ at $K$ is $\overline{BT}$, which is parallel to the tangent to $\omega$ at $B'$. Thus by homothety, the tangency point between $(JPK)$ and $\omega$ lies on $\overline{KB'}$ and therefore must be $P$. The other case follows in a similar fashion. $\blacksquare$ Let $(JPK)$ intersect $\overline{KK'}$ again at $H^+$ and let $(JQH)$ intersect $\overline{HH'}$ again at $K^+$. Claim. $J$ lies on $\overline{KK^+}$ and $\overline{HH^+}$. Proof. Angle chasing, $\measuredangle KJH^+=\measuredangle BKH^+=90^\circ-\measuredangle CBT$, and similarly $\measuredangle K^+JH=90^\circ-\measuredangle TCB$. These are equal, so $\measuredangle KJH^+=\measuredangle K^+JH$. Since $\overline{KH^+}\parallel\overline{HK^+}$, we must have that $J=\overline{KK^+}\cap\overline{HH^+}$. $\blacksquare$ By the lemma applied to $(JPK)$, $(JQH)$, $\omega$, we find that the second intersection of $\overline{QK^+}$ and $\omega$ must be the antipode of $B'$; id est it is $B$. It follows that $B$, $Q$, $K^+$ are collinear, and by symmetry so are $C$, $P$, $H^+$. To finish, remark that $\measuredangle BQJ=\measuredangle K^+QJ=\measuredangle JPK=\measuredangle BKJ$. Symmetry implies that $BKJQIK'$ and $CHJPKH'$ are cyclic, so we are done.

CK',BH',KB',HC' cuts w at D,E,F,A, KU,HG perp BC. Notice {A,G,B'},{F,U,C'} collinear, wlog K,H',B' collinear. ATGF cyclic by Reim hence A,R,E collienar, AC' cuts CE at K'' then UACK'' cylic so CUK''=90 so K'=K'' so done. By tangency, we have DEA=DHE+ECB, DFC=DKF+CBF, FDE=FKD+DHE so DEF+DFE+FEA+CFE=DHE+DKF+ECB+CBF or 180-2FDE+1/2(FB+CE)=1/2(EB+CF) or 180=2FDE+DFO hence EB,FC cuts FK,EH at L,J then DEJLF cylic, from DEB=DFK then DEKB cyclic, same for DFHC. Fix K, reconstruct H as a), we have KDF=IDK-ICF=180-IBK-ICF=ABF so tangency, by a) similar for DEH we have the conclusion. Else, D'<>D, wlog lies on smaller arc DK then H' lies outside CH while E' lies inside CH, absurd hence done

Here's a solution to part (a) by moving points. Let $M$ and $N$ be the midpoints of arcs $CB'$ and $BC'$ respectively. Choose a point $K$ on $\overline{BT}$; we construct $K'$ on $\overline{CN}$ with $\overline{KK'}\perp\overline{BC}$, let $H=\overline{CT}\cap\overline{C'K'}$, construct $H'$ on $\overline{BM}$ with $\overline{HH'}\perp\overline{BC}$, and let $K_0=\overline{BT}\cap\overline{B'H'}$. Our task is to prove that $K=K_0$. Move $K$ along line $BT$ at a linear rate, and let $\infty_{\perp BC}$ be the point at infinity perpendicular to $\overline{BC}$. By projection through $\infty_{\perp BC}$, $K'$ moves along $\overline{CN}$ at a linear rate. By projection through $C'$, $H$ moves along $\overline{CT}$ at a linear rate. By projection through $\infty_{\perp BC}$, $H'$ moves along $\overline{BM}$ at a linear rate. By projection through $B'$, $K_0$ moves along $\overline{BT}$ at a linear rate. Thus it suffices to verify the hypothesis for three values of $K$. We verify the following special cases: $K=B$: Then $K'$ lies on $\overline{BC}$, to $H$ is the reflection of $C$ over $T$. Then $H'=B$, so $K_0=B$. $K$ is the reflection of $B$ over $T$: Then $K'=C$, so $H=C$ and $H'$ is a point on $\overline{CB'}$. Thus $K_0=K$. $K$ is the point at infinity along $\overline{TB}$: Then $K'$ is the point at infinity along $\overline{CN}$, so $H=\overline{TC}\cap\overline{MC'}$. Then since $H'$ lies on $\overline{BM}$, it is the reflection of $H$ over $\overline{MN}$, so $\overline{B'H'}$ is tangent to $\omega$. It follows that $K_0=K$. This completes the proof. (Fortunately) the intended solution to (b) seems unavoidable.

Part (a) Move $H$ along $\overline{CT}$, and suppose that $B,H',K$ are collinear. Then $H\mapsto H'\mapsto K\mapsto K'$ is projective, so we just need to check that $C'$, $K'$, $H$ are collinear for three values of $H$. $\bullet$ When $H=C$, $K$ becomes $\overline{BT}\cap\overline{CB'}\Rightarrow K'=C$, and the collinearity is obvious. $\bullet$ When $H=T$, $H'$ becomes the incenter of $\triangle OBC$, so by Menelaus theorem, \[\frac{BC'}{KK'}=\frac{CB'}{KK'}=\frac{B'H'}{H'K}=\frac{BH}{HK}\]and hence $C'$, $H$, $K'$ are collinear too. $\bullet$ When $H$ is the reflection of $C$ over $T$, $K=B$ and hence $\overline{C'K'H}=\overline{C'B}$. Part (b) Invert at $C$ with arbitary radius. Then the problem becomes Inverted version wrote: Let $\triangle BTC$ be an isosceles triangle, and let $I$ be a point such that $BI=BC=BT$, and $\angle IBC$ is obtuse. Let $\Gamma_3$ be the tangent to $(CBT)$ at $B$. Let $\Gamma_1$ be the circle tangent to $(BTC)$ and $\Gamma_1$ at $K$ and $X$, and let $\Gamma_2$ be the circle tangent to $\overline{CT}$ and $\Gamma_3$ at $H$ and $Y$. Show that $I,H,J$ are collinear. To prove this, we will need to show the following easy lemma. Lemma: Let $\omega_1$ and $\omega_2$ be two circles externally tangent at $A$, and let $\overline{BC}$ be their common external tangent with $B\in\omega_1$ and $C\in\omega_2$. Then $\angle BAC=90^\circ$. Proof. Let $M$ be the midpoint of $BC$. Then $M$ lies on the radical axis of $\omega_1$ and $\omega_2$, so $MA$ is the common tangent. But this means that $MA=MB=MC$ so $\angle BAC=90^\circ$. $\square$ Applying the lemma, we see that $\angle YJX=90^\circ$ so $H,J,X$ are collinear. Let $X'$ and $B'$ be the antipodes of $X$ and $B$ in the respective circles. Then $\overline{YJX'}$, $\overline{BKX'}$ and $\overline{XKB'}$ by the lemma as well. Since $YH\parallel BI$, in order to show the collinearity, we just need to show that $YX:XB=YH:BI$. Let $M$ be the midpoint of $CT$. Then since $BMHY$ is a rectangle, \[\frac{YH^2}{BI^2}=\frac{BM^2}{BI^2}=\frac{BM^2}{BM\cdot BB'}=\frac{BM}{BB'}=\frac{YH}{BB'}\]This can be written as \[\frac{YH}{BB'}=\frac{YH}{X'X}\cdot\frac{X'X}{BB'}=\frac{YJ}{JX'}\cdot\frac{X'K}{KB}=\frac{YX^2}{XX'^2}\cdot\frac{XX'^2}{XB^2}=\frac{YX^2}{XB^2}\]and hence we are done. $\blacksquare$

Reims spam! Let $K^{*} = KB' \cap \omega$, similarly define $L^{*}$, Let $X,Y \in BC$ so that $KX \perp BC$ and $HY \perp BC$, let $(JK^{*}K) \cap KK' = U$ and similarly define $ V$, firstly notice that $\measuredangle CK^{*}Y =90^{\circ} - \measuredangle TCB = \measuredangle CQB' $ implying $K^{*},Y,B'$ are collinear, next $K^{*}L^{*}XY$ cyclic follows by reims, and let $BI,CI \cap \omega = M,N$, clearly $BXK^{*}K$ and $CYL^{*}L$ are cyclic. (a) If $K,H,B'$ are collinear then $K^{*},Y,M$ are collinear by reims, and now by reims, $L^{*},X,N$ are collinear implying the other collinearity, the other direction follows similarly by applying reims twice. (b) $CIH'K^{*}$ is cyclic by reims, while clearly $CIHH'$ is cyclic implying $CIHH'K^{*}$ is cyclic, similarly by reims and $BIKK'$ cyclic, we get $BKK'IL^{*}$ cyclic, now by homothety we have that the point of tangency are $P,Q$, so it is easy to see that $J = KU \cap HV$, and now use the following lemma: Lemma: Circles $\Gamma_1$ and $\Gamma_2$ are externally tangent at $P$, circles $\Gamma_2$ and $\Gamma_3$ are externally tangent at $Q$, and circles $\Gamma_3$ and $\Gamma_1$ are externally tangent at $R$. Let $A$ be an arbitrary point on $\Gamma_1$. Line $AP$ intersects $\Gamma_2$ again at $B$, line $BQ$ intersects $\Gamma_3$ again at $C$, and line $\overline{CR}$ intersects $\Gamma_1$ again at $D$. Then $\overline{AD}$ is a diameter of $\Gamma_1$.

Proof

to get that $B,L^{*},U$ are collinear and now $\measuredangle BL^{*}J = \measuredangle BKJ$ so we get $BKJI,CHIJ$ cyclic so done.