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Source: IGO 2024 Advanced Level - Problem 3

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Shayan-TayefehIR

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Shayan-TayefehIR

#1 h Nov 14, 2024, 7:59 PM • 2 Y

Y by Rounak_iitr, cubres

In the triangle $\bigtriangleup ABC$ let $D$ be the foot of the altitude from $A$ to the side $BC$ and $I$ , $I_A$ , $I_C$ be the incenter, $A$ -excenter, and $C$ -excenter, respectively. Denote by $P\neq B$ and $Q\neq D$ the other intersection points of the circle $\bigtriangleup BDI_C$ with the lines $BI$ and $DI_A$ , respectively. Prove that $AP=AQ$ .

Proposed Michal Jan'ik - Czech Republic

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#2 h Nov 14, 2024, 8:01 PM • 2 Y

Y by ehuseyinyigit, cubres

Sketch

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1388 posts

#3 h Nov 14, 2024, 8:36 PM • 2 Y

Y by iamnotgentle, cubres

I didn't like it very much, because it's basically two not much related problems combined into one problem (or at least my solution makes it look like that). Anyway, we show that $AP=AI=AQ$ , refer to the first image for $AQ=AI$ and to the second for $AP=AI$ .

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#4 h Nov 14, 2024, 10:14 PM • 3 Y

Y by SomeonesPenguin, egxa, cubres

Part $I$ : $AI=AQ$ .
Proof of Part $I$ : We have $I_AI.I_AA=I_AB.I_AC=I_AD.I_AQ$ hence $Q,A,I,D$ are concyclic. Also since $-1=(DC,DA;DI_A,DI)=(DC,DA;DQ,DI)$ and $\measuredangle ADC=90$ , we observe that $\measuredangle QDA=\measuredangle ADI$ which implies $AQ=AI$ . $\square$

Part $II$ : $AI=AP$ .
Proof of Part $II$ : First we present a lemma.

Lemma: $ABC$ is a triangle with $AB=AC$ and altitude $AD$ . Let $P$ be an arbitrary point on $BC$ where $I_1,I_2$ are the incenters of $\triangle PAB$ and $\triangle PAC$ . Prove that $I_1,I_2,P,D$ are concyclic.
Proof: This is a special case of Serbia 2018 P1 and it can be proved by the method of moving points with rotations centered at $A$ and $D$ . $\square$

Let $K\in BI$ with $AI=AK$ We will show that $\measuredangle I_CDK=\measuredangle I_CBK=90$ . Let $C'$ be the reflection of $C$ over $D$ . $CI\cap AD=F,BI_C\cap C'F=S$ . By above lemma, since $S$ and $I$ are the incenters of triangles $\triangle AC'B$ and $\triangle ACB$ we see that $S,B,I,D$ are concyclic. Apply DDIT on quadrilateral $SI_CIK$ to get that $(\overline{DS},\overline{DI}),(\overline{DI_C},\overline{DK}),(\overline{DB},\overline{DF})$ is an involution. Note that $\measuredangle BDF=90=\measuredangle SDI$ hence this involution must be rotating $90$ degrees. Thus, $\measuredangle I_CDK=90$ as desired. $\blacksquare$

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Reason: typo

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SomeonesPenguin

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SomeonesPenguin

#5 h Nov 14, 2024, 10:26 PM • 2 Y

Y by zzSpartan, cubres

This is also very easy using barycentric coordinates.

Claim: $AQ=AI$ .

Proof: Notice that $AI_cBI$ is cyclic so by PoP we have $I_aA\cdot I_aI=I_aB\cdot I_aI_c=I_aD\cdot I_aQ$ So $DIAQ$ is cyclic. Now note that $-1=(I_a,I;AI\cap BC,A)$ and $\angle ADC=90^\circ$ , so $DA$ is the angle bisector of $\angle QDI$ so $AQ=AI$ .

Now we use barycentric coordinates to show that $AP=AI$ . Set $\triangle ABC$ as the reference triangle.

$\begin{align*} D&=(0:a^2+b^2-c^2:a^2+c^2-b^2)\\ I_c&=(a:b:-c) \end{align*}$
The equation of the circle $(I_cBD)$ is $-a^2yz-b^2zx-c^2xy+(x+y+c)(ux+vy+wz)=0$ Now since $B$ lies on this circle we get $v=0$ . Plugging $D$ and canceling a factor of $a^2(a^2+c^2-b^2)$ gives $w=(a^2+b^2-c^2)/2$ . Finally, plugging $I_c$ in the equation yields $ua-wc=-abc\implies ua+wc=2wc-abc=c(a^2+b^2-c^2-ab)$
Now let $P=(a:t:c)$ . Plugging $P$ in the equation and canceling a factor of $c(a-b+c)$ gives $t=\frac{a^2}{b+c}-c$ Therefore $P=\left(\frac{b+c}{a+b+c},\frac{a^2-bc-c^2}{a(a+b+c)},\frac{bc+c^2}{a(a+b+c)}\right)$ So $\overrightarrow{AP}$ has displacement vector $\left(\frac{-a}{a+b+c},\frac{a^2-bc-c^2}{a(a+b+c)},\frac{bc+c^2}{a(a+b+c)}\right)$ Finally, plugging this into the length formula gives

$\begin{align*} (a+b+c)^2\cdot AP^2&=-(a^2-bc-c^2)(bc+c^2)+b^2(bc+c^2)+c^2(a^2-bc-c^2)\\ &=-a^2bc+2b^2c^2+b^3c+c^3b \end{align*}$
The displacement vector of $\overrightarrow{AI}$ is $\left(\frac{-b-c}{a+b+c},\frac{b}{a+b+c},\frac{c}{a+b+c}\right)$ So by the distance formula $(a+b+c)^2\cdot AI^2=-a^2bc+2b^2c^2+b^3c+c^3b$ And this concludes the proof. $\blacksquare$

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egxa

#6 h Nov 15, 2024, 8:36 AM • 2 Y

Y by bin_sherlo, cubres

Another way for $AI=AP$ is in complex plane $P=a^2+b^2+\frac{b^2c}{a}+bc+ba$

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#7 h Nov 16, 2024, 4:20 PM • 2 Y

Y by AylyGayypow009, cubres

One can easily prove that $AIDQ$ is cyclic by PoP. Next, harmonic bundles imply that $AD$ bisects $QDI$ . Then, $AQ=AI$ . Then, $\angle QAI=\angle IDI_a=2\angle QDB=2\angle QPB$ . Hence, $A$ is the center of $(QPI)$ . So, we are done.

This post has been edited 3 times. Last edited by X.Allaberdiyev, Nov 17, 2024, 2:52 AM

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NumberzAndStuff

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NumberzAndStuff

#8 h Nov 22, 2024, 8:43 PM • 1 Y

Y by cubres

Alternative synthetic proof of $AQ = AI$ :
As said above, by $POP$ we have $AIDQ$ cyclic and want to show $\angle IDA = \angle ADQ$ .
Note that $\angle ADQ = \angle I_ADC$ .
Now consider the homothety centered at $A$ which sends the incircle to the A-excircle. This sends $BC$ to a parallel line $B'C'$ , also tangent to the excircle and $D \rightarrow D'$ on that line. Because the excircle is tangent to $BC$ and the new parallel line, $I_A$ mus lie on the perpendicular bisector of $DD'$ . Thus we have:
$\angle I_ADD' = \angle I_AD'D \implies \angle I_ADC = \angle I_AD'C'$ However we have $\angle I_AD'C' = \angle IDC$ because of the homothety. This now implies:
$\angle I_ADC = \angle IDC \implies \angle IDA = \angle ADQ$

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#9 h Nov 23, 2024, 3:23 AM • 2 Y

Y by GeoKing, cubres

Lemma: $AQ=AI$
Proof: We first claim that $QAID$ is cyclic. This follows because $AIBI_C$ is cyclic by Fact 5, and radical axis theorem on $(AIBI_C), (BDI_C)$ from point $I_A$ finishes.

Now, we claim that $AD$ bisects $QDI$ , which would suffice by Fact 5. By Appolonian circles, it suffices to show $(DQ\cap CI, I; DA\cap CI, C)=-1$ . Yet projecting through $D$ onto $AI$ , we see that this follows from $(I_A, I; A, AD\cap CI)=-1$ which is well-known (or follows from $IBI_A$ right and $BI$ bisecting $\angle ABC$ ).

Now, note that we wish to show $\angle AIP=\angle IPA$ , yet the left hand side is clearly $\angle AI_CB$ by Fact 5. Now, consider forced overlaid inversion at $A$ . If $D'$ is the antipode of $A$ in $(ABC)$ , and $(I_BCD')\cap (ACI_A)$ is $P'$ . Now forced overlaid inversion at $C$ . We note that the perpendicular at $C$ to $BC$ intersects $AB$ at $D''$ , and $D''I_A\cap BI_B=P''$ , it remains to show $BC$ is the bisector of $\angle P''CI_A$ . However, as $\angle BCD''$ is right, by Appolonian circles, it suffices to show that $(D'', BC\cap I_AD''; P'', I_A)=-1$ . Yet
$(D'', BC\cap I_AD''; P'', I_A)\overset{B}{=}(A, C; BI\cap AC, BI_A\cap AC)=-1$ from right angles and bisector.

Thus, we have $AP=AI=AQ$ , as desired.
.

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886 posts

#10 h Dec 13, 2024, 5:13 PM • 2 Y

Y by GeoKing, cubres

Claim: $AP = AI$ (The tricky part)

Proof: Let $BB'$ be a diameter of the circumcircle of $\triangle BDI_C$ . Notice $BPB'I_C$ is a rectangle. Let $AI$ meet $B'P$ at $I'$ . Then $AB'I'I_C$ is cyclic with diameter $I'I_C$ . Also, since $\angle B'DB = 90^\circ = \angle ADB$ , the points $B'$ , $A$ , and $D$ are collinear. As $AI_C \perp II'$ and
$\angle AI'I_C = \angle AB'I_C = \angle DB'I_C = 180^\circ - \angle CBI_C = \angle ABI_C = \angle AII_C,$ it follows that $AI = AI'$ . Since $\angle IPI' = 90^\circ$ , the claim follows.

Claim: $AQ = AI$ (The projective part)

Proof: Let $AI$ meet $BC$ at $K$ . As $\angle IBI_A = 90^\circ$ and $\angle ABI = \angle IBK$ , it follows that $(AK; II_A)$ is harmonic. But as $\angle ADK = 90^\circ$ , we must have $\angle IDK = \angle KDI_A$ , or equivalently $\angle IDA = \angle ADQ$ . Since $AIBI_C$ is cyclic,
$I_AA \cdot I_AI = I_AI_C \cdot I_AB = I_AQ \cdot I_AD,$ thus $AIDQ$ is cyclic. Since $\angle IDA = \angle ADQ$ , we have that $A$ is the midpoint of arc $\widehat{IQ}$ , finishing the claim.

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Batsuh

#11 h Dec 18, 2024, 12:28 PM • 2 Y

Y by sami1618, cubres

We will show that $AP = AI = AQ$ . The proof will be split into two parts.
Part 1: $AQ = AI$ .
$[asy] size(13cm); import geometry; draw(unitcircle, blue+white); pair A = dir(130); pair B = dir(210); pair C = dir(330); draw(A -- B -- C -- cycle); pair I = incenter(A,B,C); pair I_A = intersectionpoint(line(A,I),perpendicular(B,line(B,I))); pair I_C = intersectionpoint(line(C,I),line(B,I_A)); pair D = intersectionpoint(line(B,C),perpendicular(A,line(B,C))); transform reflect = reflect(perpendicular(circumcenter(I_C,B,D),line(I_A,D))); pair Q = reflect * D; dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I_A$", I_A, dir(I_A)); dot("$I_C$", I_C, dir(I_C)); dot("$D$", D, S); dot("$I$", I, dir(I)); dot("$Q$", Q, dir(Q)); draw(circumcircle(I_C,B,D), red); draw(A -- I_A -- I_C -- C); draw(I_A -- Q); draw(circumcircle(A,I,D), red); draw(I -- D -- A); markangle(I,D,A, grey); markangle(A,D,Q, grey); [/asy]$
By PoP, we have $I_AD \cdot I_AQ = I_AB \cdot I_AI_C = I_AI \cdot I_AA$ so $Q$ , $A$ , $I$ and $D$ are cyclic. On the other hand, we have $-1 = (A, AI \cap BC; I, I_A)$ so $DC$ bisects the angle $\angle I_ADI$ . Thus $DA$ bisects the angle $\angle IDQ$ , so $AI = AQ$ .

Part 2: $AP = AI$ .
Let $I_B$ be the $B$ -excenter. Let $K$ be the reflection of $I$ across $A$ , and redefine $P$ to be the foot from $K$ to $BI_B$ . Since $AI = AK$ , we have $AI = AP$ , so it suffices to show that $I_CPDB$ is cyclic
$[asy]size(13cm); import geometry; pair A = dir(130); pair B = dir(210); pair C = dir(330); pair I = incenter(A,B,C); pair I_A = intersectionpoint(line(A,I),perpendicular(B,line(B,I))); pair I_C = intersectionpoint(line(C,I),line(B,I_A)); pair D = intersectionpoint(line(B,C),perpendicular(A,line(B,C))); pair I_B = intersectionpoint(line(I_C,A),line(I_A,C)); pair K = reflect(line(I_C,A)) * I; pair P = intersectionpoint(line(B,I),perpendicular(K,line(B,I))); filldraw(I_C -- I_B -- P -- cycle, orange+white+white+white); filldraw(I_C -- C -- D -- cycle, orange+white+white+white); draw(unitcircle, blue+white); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I_A$", I_A, dir(I_A)); dot("$I_C$", I_C, dir(I_C)); dot("$D$", D, S); dot("$I$", I, dir(I)); dot("$I_B$", I_B, dir(I_B)); dot("$K$", K, dir(K)); dot("$P$", P, dir(P)); draw(A -- B -- C -- cycle); draw(K -- I_A -- I_C -- C); draw(D -- A); draw(circumcircle(I_A,I_B,I_C), blue); draw(I_C -- I_B -- I_A); draw(B -- I_B); draw(A -- P -- K); draw(I_B -- K); markrightangle(K,P,I); draw(circumcircle(I_C,B,D), red+dashed); [/asy]$
By a quick angle chase, we see that $\triangle KPI_B \sim ADC$ , so $\frac{PI_B}{DC} = \frac{KI_B}{AC} = \frac{I_CI_B}{I_CC}$ Combining this with the fact that $\angle I_CI_BP = \angle I_CCD$ , we see that $\triangle I_CI_BP \sim \triangle I_CCD$ . Thus, $\angle I_CDB = \angle IPB$ so the conclusion follows.

This post has been edited 2 times. Last edited by Batsuh, Dec 18, 2024, 12:30 PM

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ezpotd

#12 h Dec 19, 2024, 9:31 PM • 2 Y

Y by sami1618, cubres

We first show $AQ = AI$ .
Since $BI_C, DQ, AI$ concur at $I_A$ , and $(DQBI_C), (BI_CAI)$ are cyclic, we can conclude $(DQAI)$ is cyclic by radical axis. Then $AQ = DI\frac{AI_A}{DI_A}$ by $\triangle I_ADI \sim \triangle I_AAQ$ . Letting $EI_A$ be the reflection of $DI_A$ over the perpendicular from $I_A$ to $AD$ , it suffices to prove $EI_A \parallel DI$ , as this will give $\triangle EI_AA \sim \triangle DIA$ and then $\frac{DI}{DI_A} = \frac{DI}{EI_A} = \frac{AI}{AI_A}$ , as desired. Since $\angle AEI_A = \angle EDI_A$ , it suffices to prove $\angle EDI_A = \angle ADI$ , or equivalently $\angle IDC = \angle I_ADC$ , but this is obvious from the angle bisector and right angle harmonic lemma.

To show $AP = AI$ , we show that $P$ is the reflection of $I$ over the Iran point that is the intersection of the $A$ midline and $BI$ . We can identify this Iran point in barycentrics as $(\frac 12, \frac{a - c}{2a}, \frac{c}{2a})$ . The incenter is given as $(\frac{a}{a + b + c}, \frac{b}{a + b + c}, \frac{c}{a + b+ c})$ . The desired reflection is then given as $(\frac{b + c}{a + b + c}, \frac{a^2 - bc- c^2}{(a(a + b+ c)} , \frac{bc + c^2}{a(a + b + c)})$ , or $(b + c : \frac{a^2 - bc - c^2}{a} : \frac{bc + c^2}{a})$ . Next, we compute the coefficients of the circle $(BDI_C)$ , clearly $v = 0$ . Then plugging in the coordinates of $I_c$ gives $-a^2bc -b^2ac + c^2ab = (b + a - c)(ua - vc)$ , or $-abc = ua - wc$ . Plugging in the coordinates of $D$ (which are $(0:S_{AC}:S_{AB})$ ) gives $a^2S_AS_AS_BS_C = (a^2S_A)(wS_AS_B)$ , or $w = S_C$ . Going back, $u = \frac{cS_C - abc}{a}$ .

Plugging in the coordinates of the reflection into the circle, we desire to prove $(a^2 - bc - c^2)(bc + c^2) + b^2(b +c)\frac{c(b + c)}{a} + c^2 (b + c)(\frac{a^2 -bc - c^2}{a}) = (a + b + c)(\frac{cS_C -abc}{a}(b + c) + cS_C \frac{(b + c)}{a})$ . Combining terms and dividing by $c\frac{b + c}{a}$ , we desire $(a^2 -bc-c^2)a + b^2(b + c) + c(a^2 - bc -c^2) = (a + b + c)(a^2 + b^2 - c^2 - ab)$ . The left side can be expanded as $a^3 - abc - ac^2 + b^3 + b^2c + a^2c -bc^2 - c^3$ . The right side can be expanded as $a^3 + ab^2 - ac^2 -a^2b + a^2b + b^3 -bc^2 - ab^2 + a^2c +b^2c -c^3 -abc = a^3 -abc - ac^2 + b^3 + b^2c + a^2c -bc^2 - c^3$ , so we are done.

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Math_legendno12

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Math_legendno12

#13 h Feb 11, 2025, 2:59 AM • 2 Y

Y by cubres, sami1618

Another way to get AP=AI after you have AQ=AI is to notice A is meant to be the center of (PQI)
So our goal is equivalent to $<QAI=2<QPI$

But this is just by angle chasing as
$<QPI=<QPB=<QDB=<CDI_A$
$=(1/2) <IDI_A$ , which is from the fact $CD$ is the internal angle bisector of $<IDI_A$ (from the projective stuff in the first part of the other solutions)
$=(1/2) <QAI$ as $QAID$ cyclic
Which is as desired.

This post has been edited 2 times. Last edited by Math_legendno12, Feb 11, 2025, 3:01 AM

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#14 h Feb 12, 2025, 5:46 PM • 2 Y

Y by cubres, sami1618

Synthetic:
Claim 1. AP = AI
Redefine P as the point on BI satisfying AP = AI. Let Q, R, J be the orthogonal projection from P to AB, BC, I to AD. (Ic) touches BC at G, (I) touches BC at T.
With a simple angle chase: <PAR = <AIJ. Then, PA = AI, <ARP = <IJA = 90. Thus, triangles APR and IAJ are congruent, so AR = IJ = DT
Now, BQ = BR = BA - AR = BA - DT = BA - BT + BD = BA - (AB + BC - CA)/2 + BD = (AB - BC + CA)/2 + BD = BG + BD = DG.
Let S be the midpoint of PIc. It is well known that triangle QSG is isoceles at S. Thus by symmetry one can point out that SD = SB = IcP/2, thus IcDP = 90.
Then, P lies on (IBD), and we are done.
Claim 2. AQ = AI
Let AI intersect BC at L. Note that BIAIc is cyclic, so by power of a point: ID.IQ = IaB.IaIc = IaI.IaA. Thus quadrilateral QAID is cyclic.
Note that: LI/LIa = AI/AIa, but <ADL = 90 so we can conclude that DA is external bisector of <IDIa, thus DA is internal bisector of <IDIc
Then, A is midpoint of minor arc IQ in (AQDI). Thus AQ = AI and we are done.
From the above claim, we now have that AP = AI = AQ, so AP = AQ, as desired.

This post has been edited 1 time. Last edited by Retemoeg, Feb 12, 2025, 5:49 PM
Reason: typo

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mathuz

#15 h Mar 29, 2025, 10:03 PM

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Denote the circumcircle of $BDI_C$ by $w$ . Let $X$ be the second intersection point of $AD$ and $w$ . Then $BI_CXP$ forms a rectangle.
As $A$ is a given point inside this rectangle, we can determine its angles relative to $X$ , $I_C$ , $B$ ; however, we don't know yet the exact values of $\angle APX$ and $\angle APB$ .

One way to find these angles is by applying Ceva's trigonometric theorem.

However, we prefer a different approach: $A$ has the isogonal conjugate w.r.t. $BI_CXP$ . This follows from the fact that the projections of $A$ onto $BP$ , $PX$ , $I_CX$ , $BI_C$ lie on a circle (since $\angle AXP = \angle ABP$ ). From this, we get $\angle I_CAB+\angle XAP = 180^{\circ}$ and $\angle XAP = 90^{\circ} + \frac{\angle A}{2}$ . Thus, $\angle XPA = \frac{\angle C}{2}$ which shows that $\triangle AIP$ -- isosceles (namely, $AI=AP$ ).

$AI=AQ$ part can be done in the same way as above solution ( $AIDQ$ is cyclic and $DA$ is the bisector of $\angle QDI$ ).

Combining both, we obtain $AI=AQ=AP$ .

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mcmp

#16 h Mar 30, 2025, 1:34 AM • 1 Y

Y by ohiorizzler1434

Solved with ohiorizzler1434 in 13 minutes.

We construct $I_b$ as the $B$ -excentre of $\triangle ABC$ . I claim that $AQ=AI=AP$ .

Step 1: $AQ=AI$ .
First notice by PoP spamming that $AI_a\cdot II_a=BI_a\cdot I_cI_a=QI_a\cdot DI_a$ , thus $AIDQ$ cyclic. But then notice that if $T=\overline{AI}\cap\overline{BC}$ , then $(AT;II_a)=-1$ , so $(\overline{DA},\overline{DT};\overline{DI},\overline{DI_a})=-1$ . But then as $\overline{AD}\perp\overline{DT}$ , $\overline{DT}=\overline{BDC}$ is the angle bisector of $\angle IDI_a$ , therefore $\overline{AD}$ is the bisector of $\angle QDI$ . Since $ADQI$ cyclic this forces $AQ=AI$ .

Step 2: $A$ is the circumcentre of $\triangle QPI$
This time, recall that $QPDBI_c$ is cyclic. Therefore:
$\begin{align*} 2\measuredangle QPI&=2\measuredangle QPB\\ &=2\measuredangle QDB\\ &=2\measuredangle I_aDT\\ &=\measuredangle I_aDI\\ &=\measuredangle QDI\\ &=\measuredangle QAI \end{align*}$ where the second last line follows from $ADQI$ cyclic. Therefore by the inscribed angle theorem $A$ circumcentre of $QPI$ . This finishes everything.

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