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Source: Serbia 2024 MO Problem 4

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676 posts

#1 h Apr 4, 2024, 4:26 PM • 3 Y

Y by Rounak_iitr, Funcshun840, PikaPika999

Let $ABC$ be a triangle with incenter and $A$ -excenter $I, I_a$ , whose incircle touches $BC, CA, AB$ at $D, E, F$ . The line $EF$ meets $BC$ at $P$ and $X$ is the midpoint of $PD$ . Show that $XI \perp DI_a$ .

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#2 h Apr 4, 2024, 6:55 PM • 4 Y

Y by GeoKing, KevinYang2.71, Funcshun840, PikaPika999

This is Spoiler config, but here's a way to solve if you didn't know that.
Let $(PD) \cap (DEF)=X'$ , $H$ feet of altitude from $A$ to $BC$ , $DD'$ diameter in $(DEF)$ , $AD \cap (DEF)=J$ , $AI \cap BC=K$ and $N$ midpoint of $AH$ .
As $-1=(P, D; E, F)$ by McLaurin we have $XD^2=XB \cdot XC$ so $XX'$ is tangent to $(BX'C)$ in addition as $XX'=DX$ we must have $XX'$ tangent to $(DEF)$ as well, also note that $P,X',D'$ sre colinear due to the $90º$ angles so as $-1=(J, D; E, F)$ and the given colinearity we have that $-1=(J, D; X', D') \overset{D}{=} (A, H; DX' \cap AH, \infty_{AH})$ therefore $D,X',N$ are colinear and now we finish with $-1=(A, H; N, \infty_{AH}) \overset{D}{=} (A, K; X'D \cap AI, I)$ therefore $X',D,I_a$ are colinear and now this means that since $XI \perp DX'$ we in fact have $XI \perp DI_a$ as desired thus we are done

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Motivational Remarks

This post has been edited 1 time. Last edited by MathLuis, Apr 4, 2024, 6:58 PM

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PROF65

#3 h Apr 4, 2024, 10:47 PM • 2 Y

Y by Steff9, PikaPika999

Since $(DP,BC)=-1$ then $XB\cdot XC=XD^2$ hence $X$ in the radical axis of $(I)$ and $(ABC)$
but $X$ is also on the radical axis of $(ABC)$ and $(IBC)$
therefore $X$ is the radical center of $(ABC),(IBC),(I)$
hence $X$ is on the radical axis of $(IBC)$ and $(I)$ which is $ST$ where $S,T$ their intersections;
but $II_a$ is diameter then $I_aS\perp SI, I_aT\perp ST$ i.e. $X$ on the polar of $I_a$ so $I_a$ is on the polar of $X$
therfore $DI_a$ is the polar of $X$ which leads to the desired result.

Best regards.
RH HAS

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72 posts

#4 h Apr 5, 2024, 8:43 AM • 3 Y

Y by VicKmath7, Assassino9931, PikaPika999

Solution 1:

WLOG assume $AB>AC$ . Let $I_aK\perp BC$ , where $K\in BC$ and let $S=XI\cap I_aD.$ Since $\angle IDX=90^{\circ}$ , we will be done if we show that $\angle DIS=\angle SDX \iff \angle DIS=\angle KDI_a \iff \triangle KDI_a \sim \triangle DIX\iff \frac{KI_a}{KD}=\frac{DX}{DI}.$
Firstly, we will calculate $PC$ . Since $(B,C;D,P)=-1$ from the Ceva/Menelaus picture, we obtain that $\frac{DB}{DC}\div\frac{PB}{PC}=1\iff \frac{PC}{PC+BC}=\frac{DC}{DB}\iff PC=\frac{BC\cdot CD}{DB-DC}=\frac{a(p-c)}{c-b}.$ Hence, $PD=PC+CD=(p-c)\left(\frac{a}{c-b}+1\right)=\frac{(p-c)(a+c-b)}{c-b}.$ Therefore, $XD=\frac{PD}{2}=\frac{(p-c)(a+c-b)}{2(c-b)}=\frac{(p-c)(p-b)}{c-b}$
Also, note that $KB=CD=p-c$ , so $KD=BC-BK-CD=a-2(p-c)=c-b.$
Moreover, $KI_a\cdot DI=r_a\cdot r=\frac{S}{p-a}\cdot \frac{S}{p}=\frac{S^2}{p(p-a)}=(p-b)(p-c),$ where we used Heron's formula. Thus, $XD\cdot KD=(p-c)(p-b)=KI_a\cdot DI,$ as desired.

Solution 2: With help from VicKmath7

Let $Y=XI\cap(IBC)$ . We wish to show that $Y, D, I_a$ are collinear.

Since $(B,C;D,P)=-1$ and $X$ is the midpoint of $PD$ , we obtain that $XB\cdot XC=XD^2.$
However, from PoP we know that $XB\cdot XC=XY\cdot XI$ .

Hence, $XD^2=XY\cdot XI$ and since $\angle IDX = 90^{\circ}$ , we obtain that $DY\perp XI$ .
Note that $II_a$ is the diameter in $(IBC)$ , thus $\angle IYI_a = 90^{\circ}=\angle IYD$ , so $Y, D, I_a$ are collinear, as desired.

This post has been edited 2 times. Last edited by rstenetbg, Apr 5, 2024, 9:36 AM

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gvole

#5 h Apr 6, 2024, 5:35 PM • 1 Y

Y by PikaPika999

MathLuis wrote:

This is Spoiler config, but here's a way to solve if you didn't know that.
Let $(PD) \cap (DEF)=X'$ , $H$ feet of altitude from $A$ to $BC$ , $DD'$ diameter in $(DEF)$ , $AD \cap (DEF)=J$ , $AI \cap BC=K$ and $N$ midpoint of $AH$ .
As $-1=(P, D; E, F)$ by McLaurin we have $XD^2=XB \cdot XC$ so $XX'$ is tangent to $(BX'C)$ in addition as $XX'=DX$ we must have $XX'$ tangent to $(DEF)$ as well, also note that $P,X',D'$ sre colinear due to the $90º$ angles so as $-1=(J, D; E, F)$ and the given colinearity we have that $-1=(J, D; X', D') \overset{D}{=} (A, H; DX' \cap AH, \infty_{AH})$ therefore $D,X',N$ are colinear and now we finish with $-1=(A, H; N, \infty_{AH}) \overset{D}{=} (A, K; X'D \cap AI, I)$ therefore $X',D,I_a$ are colinear and now this means that since $XI \perp DX'$ we in fact have $XI \perp DI_a$ as desired thus we are done

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Motivational Remarks

How unexpected that yet another problem is well-known. Let's hope they came up with something original for the TST! :oops_sign:

:oops_sign:

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171 posts

#6 h Apr 8, 2024, 3:48 PM • 1 Y

Y by PikaPika999

Let ${N}=DI_A\cap (DEF)$ ,obviously $(P,D,B,C)=-1$ so well known X is the center of circle appolonius for points $S$ such that $\frac{SB}{SC}=\frac{DB}{DC}$ .By 2002 G7 $(BNC)$ is tangent to $(DFEN)$ and since $BC$ is a chord tangent to incircle, $ND$ is bisector of $\angle BNC$ so $\frac{NB}{NC}=\frac{DB}{DC}$ so $N$ is on the circle appolonius.
Finish: ND is the radical axis of the appolonius circle and incircle so $ND\perp XI$ (the line of centers). Since $ND$ is the same as $I_aD$ we are done

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#7 h Apr 30, 2024, 6:23 AM • 1 Y

Y by PikaPika999

Another Solution
Let $G$ be a point on incircle such that $XG$ tangent to incircle.Then $GD \perp XI$ . Since $(P,D;B,C)=-1$ and $X$ midpoint of $PD$ , from harmonic properties we get that $XD^2=XB \cdot XC$ , from $XD=XG$ we get that $XG^2=XB \cdot XC$ , so $XG$ tangent to the $(BGC)$ . From IMO 2002 G7 we get that $G, D, I_a$ collinear, so $I_aD \perp XI$

This post has been edited 2 times. Last edited by NuMBeRaToRiC, Apr 30, 2024, 6:25 AM

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#8 h Apr 30, 2024, 10:48 AM • 1 Y

Y by PikaPika999

WLOG, assume $AB<AC$ . Let the $A-$ excircle touches $BC$ at $Q$ .

Let $DI$ and $DI_a$ meet the incircle again at $R$ and $S$ ,
respectively. Note that $\triangle DRS\sim\triangle I_aDR$ .

By length chasing, we obtain that $PD=\dfrac{\left(a-b+c\right)\left(a+b-c\right)}{2\left(b-c\right)}$ and $DQ=b-c$ .

Hence, $PD\cdot DQ=\dfrac{\left(a-b+c\right)\left(a+b-c\right)}{2}=2rr_a=DR\cdot I_aQ=DS\cdot DI_a,$ i.e. $P,S,Q,I_a$ are concyclic. This implies $P,S,R$ are collinear and $PR\perp DI_a$ .

Since $XI\parallel PR$ , we are done.

This post has been edited 1 time. Last edited by jrpartty, Apr 30, 2024, 10:50 AM

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#9 h Jun 6, 2024, 4:17 PM • 5 Y

Y by GeoKing, ehuseyinyigit, zzSpartan, Steff9, PikaPika999

Let the excircle touch $BC$ at $G$ and $H$ be the antipode of $G$ .

Claim: $HD\perp IP$
By homthety $A$ , $D$ , and $H$ are collinear. Let $AD$ intersect the incircle again at $Q$ . By construction $EQFD$ is harmonic so $PQ$ is tangent to the incircle. So $DQ\perp IP$ as desired.

Claim: $XI\perp DI_a$
Notice triangles $IPD$ and $DHG$ are similar so triangles $IXD$ and $DI_aG$ are also similar finishing the problem.

Attachments:

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#10 h Nov 12, 2024, 3:09 AM • 1 Y

Y by PikaPika999

Just Bash!
Let $AB=c$ , $BC=a$ , $CA=b$ , $s=\frac{a+b+c}{2}$ , inradius $=r$ and $A$ -exradius $=R$
WLOG, assume that $b>c$
Step 1 Identify the position of $X$
It is well-known that $(P,D;B,C)=-1$
Since $X$ is midpoint of $BC$ , we obtain $XD^2=XB\cdot XC$ .
Let $XB=k \implies \left(k+\frac{a+c-b}{2}\right)^2=(k)(k+a)\implies k=\frac{(a+c-b)^2}{4(b-c)}$
$\implies XD= k+\frac{a+c-b}{2}=\frac{(a+c-b)(a+b-c)}{4(b-c)}$
Step 2 Spot similar triangles
Let $A’$ be a point which $A$ -excircle is tangent to $BC$ $\implies DA’=b-c$
Note that $\angle IDX=\angle DA’I_A$
Consider $XD\cdot DA’=\frac{(a+c-b)(a+b-c)}{4}=(s-b)(s-c)=\frac{s(s-a)(s-b)(s-c)}{s(s-a)}=\frac{4[ABC]^2}{(a+b+c)(b+c-a)}=\frac{2[ABC]}{a+b+c}\cdot \frac{2[ABC]}{b+c-a}=rR$ So, $\triangle IDX \sim \triangle DA’I_A \implies XI \perp DI_A \qquad \blacksquare$

This post has been edited 1 time. Last edited by WLOGQED1729, Nov 12, 2024, 3:09 AM
Reason: Minor mistake

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SomeonesPenguin

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SomeonesPenguin

#11 h Nov 26, 2024, 4:14 PM • 1 Y

Y by PikaPika999

Beautiful problem :10:

:10:

Setup

Since $II_a$ is a diameter in $(BIC)$ we get $\overline{I_a-T-R}$ and $I_aV\parallel BC$ . Notice that $-1=(P,D;X,\infty_{BC})\overset{I}{=}(Q,V;K,R)$ Now, it suffices to prove that $I_a$ , $D$ and $K$ are collinear. If we let $I_aD$ meet $(BIC)$ again at $K'$ , it suffices to then prove $-1=(Q,V;K',R)$ .

We also have $-1=(P,D;B,C)\overset{I}{=}(Q,V;B,C)\overset{I_a}{=}(I_aQ\cap BC,\infty_{BC};B,C)$ Hence $Q$ , $M$ and $I_a$ are collinear and since $BD=TC$ , by butterfly theorem we get that $R$ , $M$ and $U$ are also collinear. Therefore $(Q,V;K',R)\overset{D}{=}(U,I;I_a,S)\overset{R}{=}(M,\infty_{BC};D,T)=-1. \ \ \blacksquare$

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#12 h Nov 28, 2024, 2:02 PM • 2 Y

Y by Steff9, PikaPika999

:love:

Denote the incircle by $\omega$ .
Let $(BIC) \cap \omega = {J, K}$ . Since $(P, D; B, C)=-1$ and $M$ is the midpoint of $PD$ , from EGMO Lemma 9.17 we have $XB \cdot XC = XD^2$ . Therefore $X$ has equal power wrt $(BIC)$ and $\omega$ so $X-J-K$ collinear. But $\angle IJI_a = \angle IBI_a = 90^{\circ}$ so $JI_a$ is tangent to $\omega$ . Similarly, $KI_a$ is tangent to $\omega$ , hence $JK$ is the pole of $I_a$ wrt to $\omega$ . By La Hire’s Theorem, $I_a$ lies on the pole of $X$ wrt $\omega$ . Since $XD$ is tangent to $\omega$ , $D$ lies on the pole of $X$ too, so the conclusion follows $\blacksquare$

This post has been edited 1 time. Last edited by cosdealfa, Nov 28, 2024, 2:08 PM
Reason: Forgot to add the number of the lemma

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tugra_ozbey_eratli

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tugra_ozbey_eratli

#13 h Dec 13, 2024, 7:09 PM • 3 Y

Y by bin_sherlo, poirasss, PikaPika999

Let $K$ be the feet of the altitude from $A$ to $BC$ , $M$ be the midpoint of $AK$ , $D'$ be the antipode of $D$
in $(DEF)$ . We know that $M,D,I_A$ are colinear. Because of homothety from $D$ , we have $XI\parallel PD'$ .
So we want to show that $MD \perp PD'$
Now we will use complex bashing. WLOG
$|d|=|e|=|f|=1$ $d=1$ $d'=-1$ $PD$ tangent to $(DEF)\implies$ $p=2-\overline{p}$ $P;E,F$ are colinear $\implies$ $p=e+f-ef\overline{p}$ from two of this, $p=\frac{e+f-2ef}{1-ef}$
$a=\frac{2ef}{e+f}$
$k=\frac{a+2-\overline{a}}{2}=\frac{ef+e+f-1}{e+f}$
$m=\frac{a+k}{2}=\frac{3ef+e+f-1}{2(e+f)}$
$\frac{m-d}{p-d'}=\frac{ef-1}{2(e+f)}\in i\mathbb{R} \implies MD \perp PD'$ We are done.

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Bonime

#14 h Apr 6, 2025, 7:15 PM • 1 Y

Y by PikaPika999

Cute and simple geo.

Define $G=(BIC) \cap XI$ since $(B,C; D, P)=-1$ we get that $XD^2=XB\cdot XC=XG\cdot XI$ so $G$ is the feet of perpendicular from $D$ in $XI$ . Then, by IE-lemma, $II_a$ is diameter of $(BIC)$ , so the perpendicular from $G$ in $XI$ pass trough $I_a$ and then, we´re done! $\blacksquare$

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