Prove excircle is tangent to circumcircle

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Prove excircle is tangent to circumcircle

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Source: Philippine Mathematical Olympiad 2025 P4

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#1 h Mar 9, 2025, 3:22 PM • 4 Y

Y by MathLuis, mpcnotnpc, JollyEggsBanana, Rounak_iitr

Let $ABC$ be a triangle with incenter $I$ , and let $D$ be a point on side $BC$ . Points $X$ and $Y$ are chosen on lines $BI$ and $CI$ respectively such that $DXIY$ is a parallelogram. Points $E$ and $F$ are chosen on side $BC$ such that $AX$ and $AY$ are the angle bisectors of angles $\angle BAE$ and $\angle CAF$ respectively. Let $\omega$ be the circle tangent to segment $EF$ , the extension of $AE$ past $E$ , and the extension of $AF$ past $F$ . Prove that $\omega$ is tangent to the circumcircle of triangle $ABC$ .

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#2 h Mar 12, 2025, 8:16 AM

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bump bump this problem is so nice

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#3 h Mar 13, 2025, 5:53 AM

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ilovemath0402 wrote:

bump bump this problem is so nice

Thanks, I proposed this problem

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SimplisticFormulas

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#4 h Mar 13, 2025, 6:02 PM

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what’s the solution? I am completely stuck

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#5 h Mar 13, 2025, 8:15 PM • 4 Y

Y by drago.7437, sarjinius, Mysteriouxxx, radian_51

Well this geo is really amazing I have to say...solved in around 30 mins but I think this could even be around 30-35 MOHS because the way to find things on this problem requires deep intuition.
Let $BI \cap (ABC)=M_B$ and $CI \cap (ABC)=M_C$ , also let $N_A$ be midpoint of arc $BAC$ on $(ABC)$ , now let reflections of $D$ over $EX, FY, BI, CI, Y \infty_{\perp CI}, X \infty_{\perp BI}$ be $D_B, D_C, L', K', K, L$ respectively now let reflection of $D_B$ over $AX$ be $L_1$ and reflection of $D_C$ over $AY$ be $K_1$ .
Using the paralelogram we can easly see from the direction of the reflections that $KK'$ and $LL'$ are diameters on $(Y, YD), (X, XD)$ respectively, now let $I_B, I_C$ be the $B,C$ excenters of $\triangle ABC$ then notice we have $\measuredangle II_CA=\measuredangle CBI=\measuredangle CDY=\measuredangle YK'C$ which implies $I_CAK'Y$ cyclic and similarily $I_BAL'X$ is cyclic however since $\measuredangle CDY=\measuredangle YD_CF$ we also get that $I_CAK'YD_C$ is cyclic and similarily $I_BAL'XD_B$ is cyclic, however it doesn't end here...
Now notice that $YK'=YD_C$ so $Y$ is midpoint of arc $K'D_C$ on $(I_CAK')$ however $D, K'$ are symetric in $CI$ which means both $I_CD, I_CD'$ are reflections of $I_CK'$ over $CI$ and thus $I_C, D, D_C$ are colinear, and similarily $I_B, D, D_B$ are colinear.
Now $\measuredangle CDY=\measuredangle YD_CA=\measuredangle AK_1Y$ which means $CK_1YD$ is cyclic and similarily we have $L_1BXD$ cyclic, but also note that $\measuredangle L_1DL=\measuredangle L_1L'L=\measuredangle AI_BX=\measuredangle ACI=\measuredangle K_1DY$ which means that $L_1, D, K_1$ are colinear.
Now from here notice that $\measuredangle DL_1A=\measuredangle DXI=\measuredangle IYD=\measuredangle AK_1D$ which does in fact show that $\triangle L_1AK_1$ is isosceles and therefore $AK_1=AL_1$ , and from reflections this gives $AD_B=AD_C$ , but notice from other reflections we have $D_BG=DG=D_CG$ where $EX \cap FY=G$ (clearly then $G$ is A-excenter of $\triangle EAF$ ), but now also note that we have $\measuredangle AD_BG=\measuredangle GDE=\measuredangle AD_CG$ which means that $AD_BGD_C$ is cyclic but by summing arcs we end up realising $AG$ is diameter and in fact now this means $(D_BDD_C)$ is $\omega$ from the tangencies.
To finish let $J$ be the miquelpoint of $L_1BCK_1$ then $J$ lies on $(ABC)$ but also from Reim's we get $N_A, D, J$ colinear and then Reim's twice gives $M_CX \cap M_BY=J$ and from double Reim's once again we have that $(AL'X) \cap (AK'Y)=J$ and this is excellent news because now we can note that $\measuredangle D_CJD_B=\measuredangle D_CJA+\measuredangle AJD_B=\measuredangle D_CI_CA+\measuredangle AI_BD_B=\measuredangle D_CDD_B$ which shows that $J$ lies on $\omega$ as well, but since $N_A, D, J$ are colinear from the converse of Archiemedes Lemma (or just shooting Lemma/homothety) we have that $\omega, (ABC)$ are tangent at $J$ as desired thus we are done :cool:

:cool:

This post has been edited 1 time. Last edited by MathLuis, Mar 13, 2025, 8:16 PM

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#6 h Mar 14, 2025, 10:39 AM • 1 Y

Y by Lyzstudent

Notice that $X$ and $Y$ are the incenters of $\triangle ABE$ and $\triangle ACF$ . Let $X'$ and $Y'$ be the projections of $X$ and $Y$ onto $BC$ . Let $T$ and $S$ be the projections of $X$ onto $AE$ and $AB$ respectively, and let $K$ be the tangency point of $\omega$ with $AE$ .

Claim 1. The $A$ -excircle of $\triangle AEF$ is tangent to $EF$ in $D$ .
Proof: Since $IXDY$ is a parallelogram, by projecting onto $BC$ we get that $BX'+BY'=BT+BD$ . This is equivalent to $BE+AB-AE+2BF+FC+AF-AC=BA+BC-AC+2BD.$ Simplifying yields $AE+ED=AF+FD$ , which is equivalent to $D$ being the tangency point of the excircle.

Claim 2. Circle $\omega$ is tangent to $(ABC)$ .
Proof: By Casey's Theorem, we need to prove that $b\cdot BD + c\cdot CD = a\cdot AK.$ But $AK=AT+TK=AT+X'D=AT-BX'+BD=AS-BS+BD=c-2BS+BD.$ With Thales' Theorem, $\frac{BS}{p-b}=\frac{BX}{BI}=\frac{BD}{a},$ so $BS=\frac{BD(p-b)}{a},$ thus getting $AK=\frac{ac+BD(b-c)}{a},$ and the conclusion follows.

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SimplisticFormulas

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#7 h Mar 14, 2025, 10:45 AM

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I found that $X,Y$ are in centres, $XE$ meets $YF$ in $Z=$ $A$ - excentre of $AEF$ and that $A$ appears to be Miquel point of $IXYZ$ .

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#10 h Apr 21, 2025, 4:16 PM

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sarjinius, what solution did you have in mind at first, when you proposed this problem?

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#11 h Apr 24, 2025, 2:30 PM

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AndreiVila wrote:

Notice that $X$ and $Y$ are the incenters of $\triangle ABE$ and $\triangle ACF$ . Let $X'$ and $Y'$ be the projections of $X$ and $Y$ onto $BC$ . Let $T$ and $S$ be the projections of $X$ onto $AE$ and $AB$ respectively, and let $K$ be the tangency point of $\omega$ with $AE$ .

Claim 1. The $A$ -excircle of $\triangle AEF$ is tangent to $EF$ in $D$ .
Proof: Since $IXDY$ is a parallelogram, by projecting onto $BC$ we get that $BX'+BY'=BT+BD$ . This is equivalent to $BE+AB-AE+2BF+FC+AF-AC=BA+BC-AC+2BD.$ Simplifying yields $AE+ED=AF+FD$ , which is equivalent to $D$ being the tangency point of the excircle.

Claim 2. Circle $\omega$ is tangent to $(ABC)$ .
Proof: By Casey's Theorem, we need to prove that $b\cdot BD + c\cdot CD = a\cdot AK.$ But $AK=AT+TK=AT+X'D=AT-BX'+BD=AS-BS+BD=c-2BS+BD.$ With Thales' Theorem, $\frac{BS}{p-b}=\frac{BX}{BI}=\frac{BD}{a},$ so $BS=\frac{BD(p-b)}{a},$ thus getting $AK=\frac{ac+BD(b-c)}{a},$ and the conclusion follows.

Excellent!!!Much better than the solution above.

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