TST PERU - IMO 2007 [Day 2- Q02]

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TST PERU - IMO 2007 [Day 2- Q02]

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#1 h May 16, 2007, 4:11 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Let $ABC$ be a triangle such that $CA \neq CB$ ,
the points $A_{1}$ and $B_{1}$ are tangency points for the ex-circles relative to sides $CB$ and $CA$ ,
respectively, and $I$ the incircle.
The line $CI$ intersects the cincumcircle of the triangle $ABC$ in the point $P$ .
The line that trough $P$ that is perpendicular to $CP$ , intersects the line $AB$ in $Q$ .
Prove that the lines $QI$ and $A_{1}B_{1}$ are parallels.

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#2 h May 22, 2007, 8:57 PM • 1 Y

Y by Adventure10

WLOG, let a = BC < CA = b. Let $A_{1}B_{1}$ meet AB at $C_{1},$ let $K \in AB$ be foot of the angle bisector $CI \equiv CP,$ Let parallel to CK through $A_{1}$ meet AB at L and let M be the midpoint of c = AB. As usual, R is the triangle circumradius, s its semiperimeter, and $\triangle$ its area.

$\frac{IK}{CK}= \frac{c}{a+b+c},$ $\frac{A_{1}L}{CK}= \frac{s-c}{a},$ $\frac{A_{1}L}{IK}= \frac{2s(s-c)}{ca}.$

By Menelaus theorem for the $\triangle ABC$ cut by the line $A_{1}B_{1}C_{1},$ $\frac{C_{1}B}{C_{1}A}= \frac{s-a}{s-b}.$ In addition, $C_{1}B-C_{1}A = c,$ hence $C_{1}B = \frac{c(s-a)}{b-a}.$ Since $\frac{BL}{BK}= \frac{s-c}{a}$ and $BK = \frac{ca}{a+b},$ we have $BL = \frac{c(s-c)}{a+b}$ and $C_{1}L = C_{1}B-BL = \frac{bc^{2}}{b^{2}-a^{2}}.$

The right $\triangle PQK \sim \triangle MPK$ are similar, hence $\frac{PK}{QK}= \frac{MK}{PK},$ $QK = \frac{PK^{2}}{MK}= \frac{MK^{2}+MP^{2}}{MK}.$ $MK = BM-BK = \frac{c(b-a)}{2(a+b)}$ and

$MP = R(1-\cos C) = \frac{4R(s-a)(s-b)}{2ab}= \frac{c(s-a)(s-b)}{2 \triangle}$

$MP^{2}= \frac{c^{2}(s-a)(s-b)}{4s(s-c)}$

$QK = \left((b-a)^{2}+\frac{(s-a)(s-b)}{s(s-c)}(b+a)^{2}\right) \cdot \frac{c^{2}}{4(b+a)^{2}}\cdot \frac{2(b+a)}{c(b-a)}=$

$= \frac{s(s-c)(b-a)^{2}+(s-a)(s-b)(b+a)^{2}}{s(s-c)}\cdot \frac{c}{2(b^{2}-a^{2})}=$

$= \frac{[(b+a)^{2}-c^{2}](b-a)^{2}+[c^{2}-(b-a)^{2}](b+a)^{2}}{4s(s-c)}\cdot \frac{c}{2(b^{2}-a^{2})}=$

$= \frac{4abc^{2}}{4s(s-c)}\cdot \frac{c}{2(b^{2}-a^{2})}= \frac{abc^{3}}{2s(s-c) (b^{2}-a^{2})}$

$\frac{C_{1}L}{QK}= \frac{2s(s-c)}{ca}= \frac{A_{1}L}{IK}$

The $\triangle QKI \sim \triangle C_{1}LA_{1}$ are centrally similar by SAS, hence $QI \parallel C_{1}A_{1}\equiv B_{1}A_{1}.$

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#3 h May 23, 2007, 11:36 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

PP. Let $\triangle ABC$ so that $a\ne b$ with circumcircle $\alpha =C(O,R)$ and incircle $w=\mathbb C(I,r)$ .

Let $\left\{\begin{array}{c} A_{1}\in BC\cap w_a\\\\ B_{1}\in AC\cap w_b\\\\\ \{C,P\}=CI\cap\alpha\end{array}\right\|$ and $Q\in AB$ so that $PQ\perp PC$ . Prove that $IQ\parallel A_{1}B_{1}$ .

Proof. Suppose w.l.o.g. $a>b$ . Denote the diameters $[CC']$ , $[PP']$ of $\alpha$ and $\left\{\begin{array}{c}K\in CI\cap AB\ ,\ S\in C'P'\cap AB\\\\ U\in CB\cap IQ\ ,\ V\in CA\cap IQ\end{array}\right\|$ .

$1\blacktriangleright$ At first I"ll find the position of $Q\in AB$ , i.e. the ratio $\frac{QA}{QB}$ . Thus, $\left\{\begin{array}{c}P'C'\parallel CP\implies C'S\perp C'Q\\\\ P'A=P'B\implies \widehat{AC'S}\equiv\widehat{BC'S}\end{array}\right\|$ $\implies$

the division $(A,B,S,Q)$ is harmonically, i.e. $\frac{QA}{QB}=\frac{SA}{SB}=$ $\frac{C'A}{C'B}=\frac{2R\sin\widehat{ACC'}}{2R\sin\widehat{BCC'}}=$ $\frac{\sin(90^{\circ}-B)}{\sin(90^{\circ}-A)}=$ $\frac{\cos B}{\cos A}$ $\implies$

$\boxed{\frac{QA}{\cos B}=\frac{QB}{\cos A}}=\frac{c}{\cos B-\cos A}$ .

$2\blacktriangleright$ At second I"ll find the positions of $U\in CB$ , $V\in CA$ and I"ll value the ratio $\frac{CV}{CU}$ . Observe that $\boxed{\frac{CB_{1}}{CA_{1}}=\frac{p-a}{p-b}}$

and $QK=QB+BK=\frac{c\cdot\cos A}{\cos B-\cos A}+\frac{ca}{a+b}=$ $c\cdot\frac{(a+b)\cos A+a(\cos B-\cos A)}{(a+b)(\cos B-\cos A)}=$

$c\cdot\frac{b\cos A+a\cos B}{(a+b)(\cos B-\cos A)}=$ $\frac{(b^{2}+c^{2}-a^{2})+(a^{2}+c^{2}-b^{2})}{2(a+b)(\cos B-\cos A)}$ $\implies$ $\boxed{QK=\frac{c^{2}}{(a+b)(\cos B-\cos A)}}$ $\implies$

$\left\{\begin{array}{c}\frac{QA}{QK}=\frac{c\cos B}{\cos B-\cos A}\cdot\frac{(a+b)(\cos B-\cos A)}{c^{2}}\implies \frac{QA}{QK}=\frac{(a+b)\cos B}{c}\\\\ \frac{QB}{QK}=\frac{c\cos A}{\cos B-\cos A}\cdot\frac{(a+b)(\cos B-\cos A)}{c^{2}}\implies \frac{QB}{QK}=\frac{(a+b)\cos A}{c}\end{array}\right\|$ . Apply the Menelaus' theorem to the transversal $\overline{QUIV}$

in: $\left\{\begin{array}{cccccc} \triangle ACK\ : & \frac{QA}{QK}\cdot\frac{IK}{IC}\cdot\frac{VC}{VA}=1 & \implies & \frac{(a+b)\cos B}{c}\cdot \frac{c}{a+b}\cdot \frac{VC}{VA}=1 & \implies & \frac{VA}{\cos B}=\frac{VC}{1}=\frac{b}{1+\cos B}\\\\ \triangle BCK\ : & \frac{QB}{QK}\cdot\frac{IK}{IC}\cdot\frac{UC}{UB}=1 & \implies & \frac{(a+b)\cos A}{c}\cdot\frac{c}{a+b}\cdot\frac{UC}{UB}=1 & \implies & \frac{UB}{\cos A}=\frac{UC}{1}=\frac{a}{1+\cos A}\end{array}\right\|$ $\implies$

$\frac{CV}{CU}=\frac{b}{a}\cdot\frac{1+\cos A}{1+\cos B}=$ $\frac{b}{a}\cdot\frac{2\cos^{2}\frac{A}{2}}{2\cos^{2}\frac{B}{2}}=$ $\frac{b}{a}\cdot\frac{p(p-a)}{bc}\cdot\frac{ac}{p(p-b)}$ $\implies$ $\boxed{\frac{CV}{CU}=\frac{p-a}{p-b}}$ .

In conclusion, $\frac{CV}{CU}=\frac{CB_{1}}{CA_{1}}$ $\implies$ $\triangle CVU\sim\triangle CB_{1}A_{1}$ $\implies$ $UV\parallel A_{1}B_{1}$ $\implies$ $\boxed{IQ\parallel A_{1}B_{1}}$ .

This post has been edited 10 times. Last edited by Virgil Nicula, Feb 6, 2016, 9:16 AM

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#4 h May 24, 2007, 4:36 PM • 11 Y

Y by vi1lat, Adventure10, Mango247, and 8 other users

Let $(I)$ touch $BC,AC$ at $F,E$ . $PQ$ meets $(O)$ at $K$
Let the perpendicular line from $I$ to $BK,AK$ meet $BK,AK$ at $M,N$
We have $IN//AC$ and $IM//BC$ so $\angle NIM=\angle ACB$
Because $NAEI, MIFB$ are rectangle so $IN=AE=CB1$ and $IM=FB=CA1$
Thus triangle $IMN$ and $CA1B1$ are congruent.
So $\angle MNI=\angle A1B1C$ . But $IN//AC$ so $MN//A1B1$ $(1)$
we have $\angle PQB=\angle PAB-\angle KBA=\angle PBA-\angle KBA=\angle PBK=\angle PAK$
so triangle $PAK$ and $POA$ are similar.
Then $PK*PO=PA^{2}=PI^{2}$
Thus triangle $PKI$ and $PIQ$ are similar
then $\angle PKI=\angle PIQ$
we have $K,N,P,M,I$ are cyclic so
$\angle INM=\angle IKM=\angle IKP-\angle PKM=\angle PIQ-\angle BCP=$ $\angle PIQ-\angle ACP=\angle PIQ-\angle NIP=\angle NIQ$
Thus $MN//IQ$ $(2)$
From (1) and (2) $QI//A1B1$ $q.e.d$

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#5 h Jan 31, 2010, 3:54 PM • 2 Y

Y by tnoodee, Adventure10

I'm changing the notations according to my solution.

Quote:

Let $ABC$ be a triangle such that $AB \neq AC,$ the points $B_{1}$ and $C_{1}$ are tangency points for the ex-circles relative to sides $AC$ and $AB,$ respectively and $I$ the incircle. The line $AI$ intersects the cincumcircle of the triangle $ABC$ in the point $P.$ The line that goes trough $P$ perpendicular to $AP$ intersects the line $BC$ in $Q.$ Prove that $QI$ and $B_{1}C_{1}$ are parallel.

Let us use barycentric coordinates with respect to $\triangle ABC.$ Then

$B_1 (s - a: 0: s - c) \ , \ C_1 (s - a: s - b: 0)$

Line $B_1C_1$ has infinite point $T_{\infty} ((s - a)(s - c): b(s - b): - c(s - c))$

Since $P$ is the midpoint of the arc $BC$ of the circumcircle, then its coordinates are

$P ( - a^2: b(b + c): c(b + c))$

Line $\mathcal{L}_a$ passing through $P$ and orthogonal to $AP$ is parallel to the external bisector of $\angle BAC,$ hence they have the same infinite point $(c - b: b: - c).$ Its equation is then

$\mathcal{L}_a \equiv bc(b + c)x - cS_Cy - bS_Bz = 0 \Longrightarrow Q \equiv BC \cap \mathcal{L}_a \equiv (0: bS_B: - cS_C)$

$\Longrightarrow IQ \equiv abcx - cS_Cy - bS_Bz = 0.$ Coordinates of its infinite point $R_{\infty}$ are:

$R_{\infty} (cS_C - bS_B: b(S_B + ac): - c(S_C + ab))$

Now, using the identities

$cS_C - bS_B = 2s(s - a)(b - c) \ , \ S_B + ac = 2s(s - b) \ , \ S_C + ab = 2s(s - c).$

$R_{\infty} ((s - a)(b - c): b(s - b): - c(s - c)) \Longrightarrow T_{\infty} \equiv R_{\infty} \Longrightarrow IQ \parallel B_1C_1$

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#6 h Feb 1, 2010, 7:01 AM • 2 Y

Y by Adventure10, Mango247

let $\{L,C \}=CO \cap (O)$ and obviously we have $PQ \cap (O)= \{P,L \}$ let $\{B',B \}=BI \cap (O)$ and define similarly $A'$ .Let $S_1=LA' \cap BC$ and $S_2=LB' \cap AC$ .by pascal's theorem for $ACBB'LA'$ and $BCPLA'A$ we get the collinearity of $S_1,S_2,I,Q$ .it remains to prove that $S_1S_2 || A_1B_1$ .so we need to compute $CS_1,CS_2$ for proving $\frac{CS_1}{CS_2}=\frac{p-b}{p-a}$ .but we have $\frac{CS_1}{BC}=\frac{CL}{BL+LC}=\frac{1}{1+\cos \angle A}$ and similarly for $\frac{CS_2}{AC}=\frac{1}{1+\cos \angle B}$ .now we must prove $\frac{1+\cos \angle B}{1+\cos \angle A}=\frac{(p-b)b}{(p-a)a}$ which is obvious since $\cos \angle B= \frac{a^2+c^2-b^2}{2ac}$ and similarly for $\cos \angle A$ .

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#7 h Oct 22, 2014, 1:40 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

My solution:

Let $A_2, B_2$ be the projection of $I$ on $CB, CA$ .
Let $C'=PQ \cap \odot (ABC)$ and $A', B'$ be the projection of $I$ on $C'A, C'B$ , respectively .

Since $IA'=B_2A=CB_1, IB'=A_2B=CA_1$ and $IA' \parallel CA, IB' \parallel CB$ ,
so $\triangle IA'B'$ and $\triangle CB_1A_1$ are homothety (also congruent) .
Since $I, A', B', C', P$ are concyclic at $\odot (IC')$ ,
so $\angle PQA= \angle PAB- \angle APQ =\angle ACP -\angle ACC'=\angle C'AP$ ,
hence we get $\triangle PAC' \sim \triangle PQA$ and $PI^2=PA^2=PC' \cdot PQ$ .
i.e. $\triangle PIC' \sim \triangle PQI$
Since $\angle B'A'I=\angle B'C'I=\angle PC'I -\angle PC'B'=\angle QIP- \angle A'IP=\angle QIA'$ ,
so we get $QI \parallel A'B'$ . i.e. $QI \parallel A_1B_1$

Q.E.D

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jayme

#8 h Nov 25, 2014, 10:43 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
see
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=576791942&t=189585

Sincerely
Jean-Louis

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#9 h Jan 14, 2015, 9:31 AM • 10 Y

Y by Dukejukem, rassom, buzzychaoz, AlastorMoody, Ard_11103, Arefe, future_gold_in_IMO, Adventure10, Mango247, and 1 other user

I found another simpler proof by using Reim theorem

My solution:

Let $Z$ be the tangent point of $\odot (I)$ with $AB$ .
Let a line passing through $A$ and parallel to $A_1B_1$ intersect $CP, CB$ at $X, Y$ , respectively .

From $\angle IPQ=\angle IZQ=90^{\circ} \Longrightarrow I, P, Q, Z$ are concyclic . ... $(1)$

From $\frac{AX}{XY}=\frac{CA}{CY}=\frac{CB_1}{CA_1}=\frac{AZ}{ZB} \Longrightarrow XZ \parallel CB \Longrightarrow A, P, X, Z$ are concyclic (Reim theorem). ... $(2)$

Combine with $(1), (2) \Longrightarrow IQ \parallel XA$ (Reim theorem). i.e. $IQ \parallel A_1B_1$

Q.E.D

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jayme

#10 h Jan 14, 2015, 10:01 AM • 2 Y

Y by Adventure10, Mango247

Dear,
very nice proof with my favorite theorem....which make some proof very elegant in my mind.
Sincerely
Jean-Louis

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#11 h Feb 5, 2016, 2:59 PM • 2 Y

Y by Adventure10, Mango247

Let $QI$ meet $CB,CA$ at $A_2,B_2$ respectively, incircle touches $AB$ at $Z$ .

WLOG assume $CA>CB$ . Note $I,Z,P,Q$ concyclic, so $\angle APZ=\angle APC + \angle IPZ= \angle ABC + \angle IQZ=\angle CA_2B_2$ , similarly $\angle BPZ=\angle CB_2A_2$ .

Thus $\frac{CA_1}{CB_1}=\frac{BZ}{AZ}=\frac{\sin \angle BPZ}{\sin \angle APZ}=\frac{\sin \angle CB_2A_2}{\sin \angle CA_2B_2}=\frac{CA_2}{CB_2}$ , so $A_1B_1\parallel A_2B_2\equiv QI$ .

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#15 h Jul 3, 2016, 3:40 AM • 1 Y

Y by Adventure10

Let $R$ be the intersection point of $PQ$ with the circumcircle of $\triangle ABC$ . Let $M$ be the midpoint of $RI$ . We want to say that $QI$ and $A_{1}B_{1}$ are parallels. $PB^2=PI^2=PR.PQ$ , so using the inversion with the center $P$ and the radius $PB$ , we have to prove that $PM$ is perpendicular to $A_{1}B_{1}$ . Because the line $QI$ , goes to the circumcircle of $\triangle PRI$ and $\angle RPI=90$ , so $M$ is the center of the circumcircle of $\triangle PRI$ . Let $T$ be the midpoint of the arc $ACB$ of the circumcircle of $\triangle ABC$ . And let $O$ be its circumcenter. Let $L$ and $N$ be the reflection points of $I$ and $M$ trough $O$ . The reflection of $R$ and $P$ trough $O$ are $C$ and $T$ . So it is enough to prove that $TN$ is perpendicular to $A_{1}B_{1}$ . $N$ is the midpoint of $CL$ . $L$ is the circumcenter of the circumcircle of the excenters of $\triangle ABC$ . So $LA_{1}$ and $LB_{1}$ are perpendicular to $AC$ and $BC$ . So $LB_{1}CA_{1}$ is concyclic with the center $N$ . $AA_{1}=BB_{1}$ , so $TB_{1}=TA_{1}$ and $CTB_{1}A_{1}$ is concyclic. So $CTB_{1}LA_{1}$ is concyclic with the center $N$ . So $TN$ is the perpendicular bisector of $A_{1}B_{1}$ . So $TN$ is perpendicular to $A_{1}B_{1}$ .

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#16 h Nov 30, 2017, 2:43 PM • 2 Y

Y by Adventure10, Mango247

my solution (L.B.T) : X = CO cut (O)) , (I) contact AB = D , PD cut (O) = Y . X,I,Y collinear (easy) . YA/YB=AD/BD=CB1/CA1 ==> YBA=CA1B1
A1B1 cut AB = Z . A1ZA = B-CA1B1 = B-YBA=CBY=IPD=IQD

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