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
Problem
Source: IGO 2024 Advanced Level - Problem 3
Tags: geometry
14.11.2024 23:01
14.11.2024 23:36
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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15.11.2024 01:14
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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15.11.2024 01:26
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. $\blacksquare$
15.11.2024 11:36
Another way for $AI=AP$ is in complex plane $P=a^2+b^2+\frac{b^2c}{a}+bc+ba$
16.11.2024 19:20
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.
22.11.2024 23:43
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: \[ \angle I_ADC = \angle IDC \implies \angle IDA = \angle ADQ \]
23.11.2024 06:23
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. .
13.12.2024 20:13
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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18.12.2024 15:28
We will show that $AP = AI = AQ$. The proof will be split into two parts. Part 1: $AQ = AI$. [asy][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][/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][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][/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.
20.12.2024 00:31
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.