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Source: MODSMO 2021 July Contest P7

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#1 h Oct 26, 2021, 7:28 PM • 1 Y

Y by ehuseyinyigit

Consider a triangle $ABC$ with incircle $\omega$ . Let $S$ be the point on $\omega$ such that the circumcircle of $BSC$ is tangent to $\omega$ and let the $A$ -excircle be tangent to $BC$ at $A_1$ . Prove that the tangent from $S$ to $\omega$ and the tangent from $A_1$ to $\omega$ (distinct from $BC$ ) meet on the line parallel to $BC$ and passing through $A$ .

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#2 h Oct 26, 2021, 9:35 PM

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Let $DEF$ be the intouch triangle, $I$ be the incenter, $D'$ be the reflection of $D$ over $I$ , $R = \overline{AD'A_1} \cap \omega$ , $K = \overline{AD} \cap \omega$ , $T$ be where the second tangent from $A_1$ touches $\omega$ , $A_2$ be the foot of the $A$ -altitude, $N$ be the midpoint of $\overline{AA_2}$ , and $X = \overline{DI} \cap \overline{EF}$ .

Since $-1 = (DK;EF) = (RD';EF)$ , and $X = \overline{DD'} \cap \overline{EF}$ , we must have $K$ , $X$ , $R$ collinear.

By ISL 2002 G7, we know that $D$ , $S$ , $N$ are collinear. This means $-1 = (AA_2;N\infty_{AA_2}) \overset{D}= (KD;SD').$ By construction we also have $(TD;RD') = -1$ . Thus projecting through $X$ means $S$ , $X$ , $T$ collinear.

By La Hire $A$ lies on the polar of $X$ , so since $\overline{IX} \perp \overline{BC}$ , the polar of $X$ is the line through $A$ parallel to $\overline{BC}$ . But by La Hire $\overline{SS} \cap \overline{TT}$ lies on the polar of $X$ , so we are done.

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#3 h Apr 1, 2025, 5:53 PM • 1 Y

Y by ehuseyinyigit

We will define some new points.

$\bullet$ Let $\triangle DEF$ denote the intouch triangle.
$\bullet$ Let $D'$ and $D_1$ be $D$ -antipode and reflection of $D$ over $\perp$ bisector of $\overline{EF}$ respectively.
$\bullet$ Let $K$ be point on $\omega$ such that $\overline{KK} \cap \overline{BC}=A_1$ .
$\bullet$ Let $T$ be pole of $\overline{SK}$ .
$\bullet$ Let $X$ be foot of $D$ onto $\overline{EF}$ and $H$ be orthocenter of $\triangle DEF$ .

We need to prove $\overline{AT} \parallel \overline{BC}$ but by La Hire. this means pole of $\overline{AT}$ is $\overline{SK} \cap \overline{EF}$ must lie on perpendicular line of $\overline{BC}$ at $D$ . In other words we want to prove that $\overline{DD'}$ , $\overline{EF}$ , $\overline{SK}$ concur. Assume $AC \ge AB$ .

Now comes the pain part. We will prove this by Ratio lemma on $\overline{EF}$ . If we let $f(\bullet)=\frac{\bullet E}{\bullet F}$ . So we want to prove $f(D')f(D)=f(S)f(K)$ .

See that $f(D)=\frac{DE}{DF} \text{ }f(D')=\frac{D'E}{D'F}=\frac{HF}{HE}=\frac{\sin \angle HEF}{\sin \angle HFE}=\frac{\cos \angle F}{\cos \angle E}$ By IMO Shortlist 2002/G7, we have $S$ is $D$ -Why point of $\triangle DEF$ so $S=\overline{D_1X} \cap (DEF)$ . $f(S)=\frac{f(X)}{f(D_1)}=f(X)f(D)=\left(\frac{DE}{DF} \right)^2 \frac{\cos E}{\cos F}$ By Cosine rule and all, see that $EF=(b+c-a) \sin (\angle A/2) \text{ and } \cos \angle D=\sin (\angle A/2)$ And analogous.

We have all the stuff to find out $f(K)$ now. $\begin{align*} f(K) &=\frac{KE}{KF}=\frac{\sin \angle KFE}{\sin \angle KEF}=\frac{\sin \angle EKA_1}{\sin \angle FKA_1}=\frac{\sin(\angle EKD+\angle DKA_1)}{\sin(\angle FKD+\angle DKA_1)} \\ &= \frac{\sin \angle F \cos \angle DIA_1-\sin \angle DIA_1 \cos \angle F}{\sin \angle E \cos \angle DIA_1+\sin \angle DIA_1 \cos \angle E} \\ &= \frac{DE-4DM \cos \angle F}{DF+4DM \cos \angle E} \end{align*}$ So we want to prove that $(DE-4DNM \cos \angle F) DE (\cos \angle E)^2=(DF+4DM \cos \angle E) DF (\cos \angle F)^2$ Now see that $\begin{align*} & (f-2(b-c) \cos \angle F) f (\cos \angle E)^2 \\ =& \left((a+b-c) \sin (\angle C/2)+2(c-b) \sin (\angle C/2) \right) (a+b-c) \sin (\angle C/2) (\sin \angle B/2)^2 \\ =& (a+c-b)(a+b-c) (\sin \angle B/2 \cdot \sin \angle C/2)^2 \end{align*}$ And see the other expression gives the same thing and done.

This post has been edited 1 time. Last edited by ihategeo_1969, Apr 1, 2025, 5:53 PM

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#4 h Apr 1, 2025, 8:08 PM

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Let the tangent at $S$ meet $BC$ at $V$ . Let $D,E,F$ be touchpoints of incircle, $EF\cap BC=G$ . Let $M$ be the midpoint of $BC$ . Let $W$ be the intersection of tangents to incircle at $S$ and from $A_1$ .
Claim: $(M,G),(B,C),(D,BC_{\infty})$ is an involution.
Proof: Let $S_A$ be $A-$ sharky devil, $P,N$ be the midpoints of arcs $BC,BAC$ . Since $(S_AG,S_AP;S_AB,S_AC)\overset{S_A}{=}(G,D;B,C)=-1=(N,P;B,C)$ We see that $N,S_A,G$ are collinear. Thus, $DB.DC=DS_A.DP=DG.DM$ which proves the claim.
Since $DG.DM=DV.DA_1$ , we see that $(B,C),(D,BC_{\infty}),(V,A_1)$ is an involution. Also DDIT at $WVDA_1$ gives $(\overline{AW},\overline{AD}),(\overline{AV},\overline{AA_1}),(\overline{AB},\overline{AC})$ is an involution. Project this onto $BC$ which implies $(AW\cap BC,D),(V,A_1),(B,C)$ is an involution. By the previous observation, we get $AW\cap BC=BC_{\infty}$ so $AW\parallel BC$ as desired. $\blacksquare$

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