What the isogonal conjugate on IO

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Source: buratinogigle

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#1 h Jun 5, 2025, 2:11 PM • 1 Y

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Given a triangle $ABC$ with incircle $(I)$ tangent to $BC, CA, AB$ at points $D, E, F$ , respectively. Let $P$ be a point such that its isogonal conjugate lies on the line $OI$ (where $O$ is the circumcenter and $I$ the incenter of $ABC$ ). The line $PA$ intersects segments $DE$ and $DF$ at points $M_a$ and $N_a$ , respectively, such that the circle with diameter $M_a N_a$ meets $BC$ at points $P_a$ and $Q_a$ .

1) Prove that the circle $(AP_a Q_a)$ is tangent to the incircle $(I)$ at some point $X$ .

2) Similarly define points $Y, Z$ corresponding to vertices $B, C$ . Prove that the lines $AX, BY, CZ$ are concurrent.

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#2 h Jun 5, 2025, 4:23 PM

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waow elmo 2016/6 really leveled up

(a) Note that $P$ is obsolete for part (a). Now let $R$ be the center of $(M_A N_A)$ and let $AM_A$ meet $BC$ at $Q$ . Since $-1=(D, AD \cap (I); E, F) = (Q, A; M_A, N_A)$ and $\measuredangle M_A P_A N_A = \measuredangle M_A Q_A N_A = 90$ degrees, we have that $M_A$ , $N_A$ are the in/A-excenter of $AP_A Q_A$ and so $R$ is the south pole of $(AP_A Q_A)$ . Not let $RD$ meet $(I)$ at $X$ , by the shooting lemma it suffices to prove that $X$ lies on the incircle, i.e. $(M_A N_A)$ is orthogonal to $(I)$ . However is true as if we let $A M_A$ meet $(I)$ at $S$ , $T$ , we have $-1=(S, T; E, F)=(S, T; M_A, N_A)$ . and $M_A N_A$ is a diameter.

(b) Let the line through $D$ parallel to $AP$ meet $(I)$ at $(X_A)$ . Then $-1=(L, P_EF{\infty}; M_A, N_A)=(X, X_A; E, F)$ , so $A$ , $X_A$ , $X$ are collinear. By the Steinbart theorem, it suffices to prove $DX_A$ , $EY_B$ , $F Z_C$ are concurrent (where $Y_B$ , $Z_C$ are defined like $X_A$ ).

Note the “isogonal conjugate on $OI$ ” condition means $P$ is on the Feuerbach hyperbola. Now project $P$ through $I$ from the Feuerbach hyperbola to the Jerabek hyperbola of $DEF$ to get $Q$ . We claim this is the desired concurrency point. It suffices to prove that $DQ$ and $AP$ are parallel.

Note by a known lemma that these two hyperbolas intersect at four points: $I$ , the Gergonne point $G$ and their two points at infinity $U$ , $V$ . Now we have by projection at $I$ that $(AP, AG; AU, AV) = (P,G;U,V)=(Q,G;U, V)= (DQ,DG;DU,DV)$ . However since three pairs of these lines are parallel, the fourth are too! Hence we are done

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cursed_tangent1434

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cursed_tangent1434

#3 h Jun 9, 2025, 10:53 AM

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The attack on the problem is setup by discovering the configuration while it is launched with some MMP towards the end.

The configuration in the first part is pretty well-explored, having appeared in several contests such as the ELMO in the past. Let $O_a$ denote the midpoint of segment $M_aN_a$ . The following is the key claim.

Claim : Points $A$ , $P_a$ , $O_a$ and $Q_a$ are concyclic.

Proof : Let $T= \overline{EF} \cap \overline{BC}$ and $R = \overline{AP} \cap \overline{BC}$ . Note that,
$-1=(TD;BC)\overset{A}{=}(T,AD \cap EF ; F , E)\overset{D}{=}(RA;N_aM_a)$ which implies that
$RA \cdot RO_a=RN_a\cdot RM_a = RP_a \cdot RQ_a$ and hence the claim follows.

We claim that the tangency point $X$ is the second intersection of the line $\overline{DO_a}$ with the incircle. To confirm this, it suffices to see that
$\begin{align*} \measuredangle XFD + \measuredangle P_aO_aX &= \measuredangle FXD + \measuredangle XDF + \measuredangle P_aO_aX \\ &= \measuredangle FEB + \measuredangle O_aDB + \measuredangle BDF + \measuredangle P_aO_aX\\ &= \measuredangle O_aP_aD\\ &= \measuredangle O_aAQ_a \\ &= \measuredangle P_aXO_a \end{align*}$ since $\overline{AP}$ is the internal $\angle P_aAQ_a-$ bisector as $O_a$ is the minor $P_aQ_a$ -arc midpoint in $(AP_aQ_a)$ . We now proceed with the more difficult second part.

We first make sense of the weird condition on $P$ .

Lemma : Let $P^*$ be a point on the line $\overline{IO}$ . Let $P$ be the isogonal conjugate of $P^*$ and define $A_1 = \overline{AP} \cap \overline{DI}$ , $B_1= \overline{AP} \cap \overline{EI}$ and $C_1 = \overline{AP} \cap \overline{FI}$ . Then,
$IA_1=IB_1=IC_1$
Proof : We proceed via the method of tethered moving points. Let $\ell = \overline{IO}$ . Consider the maps,
$\ell \to \overline{DI} \to \overline{FI} \text{ by } P^* \to A_1 \to C_1'$ which is a composition of projections and reflection across the fixed lines $\overline{AI}$ and $\overline{BI}$ and
$\ell \to \overline{FI} \text{ by } P^* \to C_1$ which is a composition of projections and reflection across the fixed line $\overline{CI}$ . These maps are thus projective. Hence it suffices to check that these maps coincide for three choices of $P^*$ .

$\bullet$ When $P^* = I$ then $C_1\equiv C_1' \equiv I$ .
$\bullet$ When $P^* = O$ then $AP \parallel DI$ and $CP \parallel FI$ so $\overline{CP}$ is the image of $\overline{AP}$ under reflection across the $\angle B-$ bisector and thus $C_1=C_1'$ .
$\bullet$ When $P^* = IO \cap AC$ , $\overline{AP}$ is $\overline{AB}$ and $\overline{CP}$ is $\overline{BC}$ and since these lines are isogonal with respect to $\angle B$ we have $C_1=C_1'$ .

Hence, it follows that $\overline{BA_1}$ and $\overline{BC_1}$ are isogonal at $\angle B$ so $\triangle BIA_1 \cong \triangle BIC_1$ and $IA_1=IC_1$ . Repeating the argument cyclically proves the lemma.

Now, consider the (negative) homothety centered at $I$ mapping $\triangle A_1B_1C_1$ to $\triangle DEF$ . Let $A_2$ denote the second intersection of ray $\overrightarrow{A_1P}$ and $(A_1B_1C_1)$ . Define $B_2$ and $C_2$ similarly. Then, let $\triangle D'E'F'$ denote the image of $\triangle A_2B_2C_2$ under this homothety. Since $\triangle A_1B_1C_1$ and $\triangle A_2B_2C_2$ are perspective with center $P$ , it follows that so are $\triangle DEF$ and $\triangle D'E'F'$ .

To finish simply note that,
$-1=(M_aN_a;O_a\infty) \overset{D}{=}(EF;XD')$ and hence we must have points $A$ , $D'$ and $X$ being collinear. Repeating the argument on the other sides we have that by Steinbart's Theorem the result follows.

This post has been edited 1 time. Last edited by cursed_tangent1434, Jun 9, 2025, 10:54 AM
Reason: typoes

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