Let $ABC$ be an acute-angled triangle with circumcenter $O$ and incenter $I$. The midpoint of arc $BC$ of the circumcircle of $ABC$ not containing $A$ is denoted $S$. Points $E, F$ were chosen on line $OI$ for which $BE$ and $CF$ are both perpendicular to $OI$. Point $X$ was chosen so that $XE\perp AC$ and $XF\perp AB$. Point $Y$ was chosen so that $YE\perp SC$ and $YF\perp SB$. $D$ was chosen on $BC$ so that $DI\perp BC$. Prove that $X$, $Y$, and $D$ are collinear.
Problem
Source: 2023 Israel TST Test 7 P3
Tags: TST, geometry, triangle centers, circumcircle, incenter
ckliao914
10.05.2023 13:51
There is a generalized problem: Let $ABCD$ be a cyclic quadrilateral inscribed in $(O)$, $X$ is a point on $AC$, then the orthopole of $XO$ wrt $ABD$, $CBD$, and the foot of $X$ on $BD$ are collinear. Let the line connecting the midpoint of $AB,AD$ be $\ell_A$, the line connecting the midpoint of $CB,CD$ be $\ell_C$. Let the foot of $X$ on $AB,AD,CB,CD,BD$ be $F_1,F_2,F_3,F_4,H$. Fontene's first theorem says that $\ell_A\cap F_1F_2$, $H$, and the orthopole of $OX$ wrt $ABD$ are collinear. Therefore it suffices to prove that $\ell_A\cap F_1F_2=P$, $\ell_C\cap F_3F_4=Q$, and $H$ are collinear. We let $X$ be a variable point on $AC$. It is clear that $X\to F_1F_2\to P$ is projective since the direction of $F_1F_2$ is fixed. Therefore $X,P,Q,H$ are all projective. When $X=\infty_{AC}$, $P=Q=H=\infty_{BD}$, so by Steiner conic $PH,QH$ both passes through a fixed point, so it suffices to check for two more points. Take $X=A$, $PQH$ becomes the Simson line of $C$ wrt $ABD$. $X=C$ is similar, so we are done
hakN
03.01.2024 23:09
Let $P=BE\cap AC$ and $Q$ be the second intersection of $SB$ with $(ABP)$. We have $\angle EXF = \angle PAB$ and $\angle XFE = \angle ABP$, so $\triangle ABP \stackrel{\rm +}{\sim} \triangle XFE$. Also, $\angle YFE = \angle QBP$ and $\angle EYF = \angle PQB$, so $\triangle QBP \stackrel{\rm +}{\sim} \triangle YFE$. Since $\angle EXF = \angle EYF = \angle CAB$, we have $APBQ \stackrel{\rm +}{\sim} XEFY$. Let the spiral similarity taking $XEFY$ to $APBQ$ take $D$ to $D'$. To show $D\in XY$, we need $D'\in AQ$. Let $BD'$ and $PD'$ intersect $(ABP)$ at $M,N$ respectively. Note that since $\triangle DEF \stackrel{\rm +}{\sim} \triangle D'PB$, we have $\angle D'BP = \angle BCI$ and $\angle BPD' = \angle IBC$. Let $K=BI\cap (ABC)$, $L=CI\cap (ABC)$, $U=BM\cap (ABC)$ and $V=BN\cap (ABC)$. Denote the reflections of $B,C$ across $IO$ as $B',C'$. Since $\angle PAB = \angle C'CB + \angle CBC' = \angle CBP + \angle CBC' = \angle PBC'$, we obtain $BC'$ is tangent to $(ABP)$. Now an easy angle chase shows that $BB' \parallel KV$ and $CC'\parallel LU$. Thus, $B'V$ and $C'U$ pass through $I$, so we obtain $(A,S),(B',V),(C',U)$ are pairs of involution. Projecting them from $B$ to $(APB)$, we get $(A,Q),(P,N),(B,M)$ are pairs of involution, which implies that $AQ,PN,BM$ are concurrent, so we are done. $\square$
starchan
10.03.2024 15:28
We denote line $OI {}$ as $\ell$.
Let $L$ denote the intersection $\ell \cap BC$. Without losing generality, we assume that $L$ lies on ray $CB$ past $B$.
From the perpendicularities, we note the following: \[\measuredangle EXF = \measuredangle CAB, \measuredangle XFY = \measuredangle ABS, \measuredangle FYE = \measuredangle BAC, \measuredangle YEX = \measuredangle SCA\]In particular, since \[\measuredangle EXF = \measuredangle CAB = -\measuredangle BAC = -\measuredangle FYE = \measuredangle EYF\]points $E, X, F, Y$ are concylic.
Let $\Gamma$ denote the circle $(ABC)$ and $\Gamma'$ its reflection over $BC$. Consider $\tau$, the spiral similarity centred at $L$ which maps segment $EF$ to $BC$.
Claim: Under $\tau$, circle $(XEYF)$ maps to $\Gamma'$.
Proof: We see that for a point $P$ on $(XEYF)$ we have \[- \measuredangle BAC = \measuredangle EPF = \measuredangle B\tau(P)C\]and thus $\tau(P)$ lies on $\Gamma'$.
Claim: Under $\tau$, point $D$ maps to $I'$, the reflection of $I$ over $BC$.
Proof: We note that $\triangle LEB$ and $\triangle LDI$ are inversely similar (due to the right angles), implying that $\triangle LEB$ is directly similar to $\triangle LDI'$. This implies the claim.
Let $U, V \in \Gamma'$ denote the images of points $X, Y$. We note that $U, V$ can be now redefined as
$U$ is the point on $\Gamma'$ such that $\measuredangle UCB = 90^{\circ} + \measuredangle (\overline{AB}, \ell)$
$V$ is the point on $\Gamma'$ such that $\measuredangle VCB = 90^{\circ} + \measuredangle (\overline{SB}, \ell)$.
We wish to show that $U, V, I'$ are collinear. Let us reflect points $U, V$ over line $BC$ to obtain points $M, N$ respectively. Then we see that $M, N \in \Gamma$ with the following: \[\measuredangle MCB = 90^{\circ} - \measuredangle (\overline{AB}, \ell) ~;~ \measuredangle NCB = 90^{\circ}-\measuredangle (\overline{SB}, \ell) \]We now wish to show that $M, N, I$ are collinear.
Claim: Lines $AM, SN$ are both perpendicular to $\ell$.
Proof: We have \[\measuredangle (\overline{MA}, \ell) = \measuredangle MAB + \measuredangle (\overline{AB}, \ell) = \measuredangle MCB + \measuredangle (\overline{AB}, \ell) = 90^{\circ}\]and \[\measuredangle (\overline{NS}, \ell) = \measuredangle NSB + \measuredangle (\overline{SB}, \ell) = \measuredangle NCB + \measuredangle (\overline{SB}, \ell) = 90^{\circ}\]
Now, since $OA = OM = ON = OS$, we actually obtain that $\ell$ is the perpendicular common bisector of segments $AM$ and $SN$.
Thus line $MN$ is the reflection of $AS$ over $\ell$. It follows that $MN, AS, \ell$ concur at $AS \cap \ell = I$. (note that these are collinear since $S$ is the midpoint of arc $BC$ in $\Gamma$ not containing $A$)
In particular, points $M, N, I$ are collinear. Tracing back the transformations above (the reflection, then the spiral map), we obtain that points $X, Y, D$ are collinear; as desired. $\square$
math_comb01
12.06.2024 23:33
here's a different synthetic solution (sketch) : It is easy to see by angle chase that $XEYF$ is cyclic, now perform incircle inversion; notice $E,F$ gets sent to intersection of euler line with sides of contact triangle, we claim that $X$ lies on the incircle, to prove so, redefine $X$ as the antisteiner point of the euler line in the contact triangle, and now the property follows from well known properties of antisteiner of euler line. Let $M_{BC}$ denote the arc midpoint of arc $BC$ containing $A$, the idea is that, $BM_{BC}$ is parallel to line through $F$ perpendicular to $SB$, so after inversion these turn into tangent circles, and now the problem reduces to simple angle chase.
YaoAOPS
08.10.2024 06:33
Feuerbach point property spam Note that $\measuredangle EXF = \measuredangle EYF = \measuredangle CAB$ so $EXYF$ is cyclic. Let $\triangle MNP, \triangle DE_1F_1$ be the medial triangle of $\triangle ABC$. Let $X'$ be the Feuerbach point, which is known to be the anti-Steiner line of $OI$ wrt $\triangle MNP, \triangle DE_1F_1$, and that $AOI \sim^+ FeMD$. Then if $E'$ is the reflection of $X'$ over $MP$, then $E' \in (BO)$ and $E' \in OI$ so $E' = E$. Since $X'E \perp MP \parallel AC$, it follows that $X \in X'E$, so by symmetry $X = X'$. As such, rename $X$ to $Fe$. As such, it follows that \[ \measuredangle EFeD = \perp AC - \measuredangle FeD = \perp AC - ((\measuredangle AI - \measuredangle IO) + \measuredangle BC) = \perp AC - \measuredangle AI + \measuredangle IO - \measuredangle BC \]and that \[ \measuredangle EFeS = \measuredangle EBS = \perp IO - \measuredangle BS = \perp IO + \measuredangle AC - \measuredangle AI - \measuredangle BC \]which implies the result where we use directed lines.