Let ABC be a triangle, let E,F be the tangency points of the incircle Γ(I) to the sides AC, respectively AB, and let M be the midpoint of the side BC. Let N=AM∩EF, let γ(M) be the circle of diameter BC, and let X,Y be the other (than B,C) intersection points of BI, respectively CI, with γ. Prove that NXNY=ACAB. Cosmin Pohoata
Problem
Source: Romanian TST 2007, Day 6, Problem 2
Tags: geometry, trigonometry, AMC, USA(J)MO, USAJMO, perpendicular bisector, geometry proposed
08.06.2007 23:30
pohoatza wrote: Let ABC be a triangle such that b>c and denote by D,E,F the intersections of the its incircle with the sides BC, AC, AB respectively. Consider the circle w=C(M) with the diameter [BC]. Denote the intersections {X∈w∩(NF Y∈w∩(NE N∈AM∩EF. Prove that NXNY=ABAC. Proof I. The well-known properties ‖IN⊥BCX∈CI , Y∈BI‖ ⟹ {m(^YXI)=B2 , m(^XYI)=C2m(^NIX)=90∘−C2 , m(^NIY)=90∘−B2‖ ⟹ NXNY=IXIY⋅sin^NIXsin^NIY= sin^XYIsin^YXI⋅sin^NIXsin^NIY= sinC2sinB2⋅cosC2cosB2 ⟹ NXNY=sinCsinB ⟹ NXNY=ABAC . Remark. NFNE=ACAB. Proof II. Denote the intersection Z∈BX∩CY. Observe that Z∈ID and the point I is the orthocenter of the triangle BZC, i.e. the triangle XDY is the orthic triangle of △BZC. From the well-known property "the ray [DI is the bisector of the angle ^XDY" obtain NXNY=DXDY. Prove easily that {m(^DXY)=B m(^DYX)=C‖ ⟹ △XDY∼△BAC ⟹ DXDY=ABAC. In conclusion, NXNY=ABAC. A general remark. If you know the above mentioned remarkable properties, then this very nice Pohoatza's problem is easily ! Virgil Nicula wrote: A similar problem. Let C(I) be the incircle of △ABC. Denote the second intersections M, N, P of the circumcircle C(O) of △ABC with the lines AI, BI, CI respectively. Prove that AM⊥NP and R∈OM∩NP⟹ RNRP=ABAC.
19.07.2007 15:58
Virgil Nicula wrote: X∈CI , Y∈BI Please, Virgil Nicula, I have no idea how this comes!
19.07.2007 21:22
cancer wrote: Virgil Nicula wrote: X∈CI , Y∈BI Please, Virgil Nicula, I have no idea how this comes! See http://www.mathlinks.ro/Forum/viewtopic.php?t=155800
20.07.2007 05:19
What? I can't see how it's related... are there any tangent circles?
20.07.2007 07:53
See the definitions of the points X, Y from the above enunciation !!.
20.07.2007 10:27
Virgil Nicula wrote: See the definitions of the points X, Y from the above enunciation !!. I already see! Unless the incircle and w are tangent??
20.07.2007 10:39
NO ! 1. N∈AM∩EF. 2. (NE and (NF are rays (semiline) ! 3. The circle w has the center in the middle M of the side [BC] and the length of the its radius is equal to MB ! Sorry, this is the all ! Now please, ask you Pohoatza to explain away ...
05.10.2008 01:39
pohoatza wrote: Let ABC be a triangle, let E,F be the tangency points of the incircle Γ(I) to the sides AC, respectively AB, and let M be the midpoint of the side BC. Let N=AM∩EF, let γ(M) be the circle of diameter BC, and let X,Y be the other (than B,C) intersection points of BI, respectively CI, with γ. Prove that NXNY=ACAB. Author: Cosmin Pohoata We have some well-known results in this configuration: 1. X, Y lie on line EF. 2. N lies on ID, where D is the tangency point of Γ(I) to the side BC. Now, denote by P the intersection of AI with BC. It's easily to see that ∠NIX=∠PIC, therefore triangles NIX and PIC are similar. It implies NXNI=PCPI. Similarly, we conclude that NINY=PIPB. Hence, we have NXNY=NXNI⋅NINY=PCPI⋅PIPB=PCPB=ACAB. This completes our solutions.
05.10.2008 13:00
Dear April, Your proof is Cosmin's. It was introduced in an aritcile of his.
18.11.2009 15:36
Let D be the tangency point of BC and (I) and K≡EF∩BC. Also, let X′,Y′ respectively be the intersections of BI,CI with EF. It is well known that (KDBC)=−1 ⟹ (X′K,X′D,X′B,X′C)=−1. In the other hand, since X′ lies on the perpendicular bisector of DF, therefore, X′B is the bisector of ∠KX′D, thus, BX′⊥CX′, which implies X′≡X. With the same argument, we can also conclude that Y′≡Y. As the consequence, E,F,X,Y are collinear. Let A′ be the intersection of ID and the line △ passes through A parallel to BC. Let N′≡ID∩EF and K′ be the intersection of EF with △. Since ∠AA′I=90∘ ⟹ A′∈(AEF). Therefore, A′N′ is the bisector of ∠EA′F, but A′N′⊥A′K′ ⟹ (A′E,A′F,A′N′,A′K′)=−1, which implies (EFN′K′)=−1 ⟹ A(CBN′K′)=−1. But AK′‖BC, therefore, AN′ passes through M, which implies N′≡N ⟹ I,D,N are collinear. Now, it is easy to prove that △YDX ≅ △BAC (a.a) and DI is the bisector of ∠YDX, then NXNY=DXDY=ACAB. ◻
21.05.2014 17:50
Well, the same trick ( MR, BI, FE and γ intersect in X) appears 7 years later at USAJMO. (http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3478583&sid=5997de18ba862fa0672276d599ecf7cb#p3478583). I first saw this trick in one of Cosmin's book. (R is midpoint of AC)
22.05.2014 08:38
USAJMO #6 is my problem too, so no copyright issues I hope . By the way, the problem didn't mean to have a part (a) initially, but we figured the problem might be too difficult without isolating the known lemma (which unfortunately turned out not to be so well known, as there weren't more than 30 students who solved the first part).
23.05.2014 16:03
Dear mathlinkers, any ideas for the similar problem of Virgil Nicula? Sincerely Jean-Louis
23.05.2014 17:42
Dear Virgil and Mathlinkers, are you sure of your result (similar problem )? Toying with the figure in question, I have a doubt? Sincerely Jean-Louis
11.03.2016 23:51
Lemma: NI⊥BC (This is lemma 7 here) Proof: Let the line passing through A parallel to BC be l. Let AN∩Γ(I)≡P,Q such that AP<AQ. Let FP∩QE≡S and FQ∩PE≡T. We have that PEQF is a harmonic quadrilateral. Applying Brokard to PEQF we have NI⊥TS, so it is enough to show that T,A,S,Z are collinear. By Brokard, ST is the polar of N, and by tangency N is on the polar of A, thus La Hire yields A is on the polar of N. It follows that T,A,S collinear. Also, PEQF is harmonic so (Z,N;E,F)=−1. Hence, N is on the polar of Z, and La Hire again yields the desired. Also, by the Right Angles on Intouch Chord Lemma X,Y∈γ(M). Now we length chase a little. By the Ratio Lemma, NXNY=XIYIsin∠XINsin∠YIN. By PoP or just similar triangles, this is CIBIsin∠BIDsin∠CID. It's easy to see with lengths around the incircle and LoS that CIBI=sinB2sinC2 and sin∠BIDsin∠CID=cosB2cosC2. Then by the double angle formula for sine we have NXNY=sinBsinC=ACAB as requested.
12.03.2016 01:02
It is well-known that X,Y lie on FE. It is well-known that NI⊥BC. We have ∠XIN=90∘−∠B2 and ∠NIY=90∘−∠C2. From Ratio Lemma, we have NXNF=XIIY⋅sin(90∘−∠B2)sin(90∘−∠C2)=XIIY⋅cos∠B2cos∠C2 We have BI=rsin∠B2 and CI=rsin∠C2. So, NXNF=sin∠B2cos∠B2sin∠C2cos∠C2=sin∠Bsin∠C=bc=ACABas desired.
12.03.2016 02:17
Let the midline MI cut EF at X′ since AB∥MX′ then ˆB=^MX′B+^X′BM,^X′BA=^BX′M so MX′B is isoceles which implies ^BX′C=π2 so X=X′ likewise we prove Y∈EF∩MJ . NXNY=sin^XMN⋅MXsin^YMN⋅MY=sin^MABsin^MAC=ACAB. R HAS
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26.08.2019 20:15
15.01.2022 13:40
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15.01.2022 15:05
Here we go again. By Iran Lemma X,Y∈EF also note that △IXY∼△ICB and bcs of meh config stuff N=DI∩¯EF where D is the other in touch point. NXNY=IXsin∠XINIYsin∠YIN=ICsin∠BIDIBsin∠CID=ICIDcosB/2IBIDcosC/2=sinB/2cosB/2sinC/2cosC/2=sinBsinC=ACAB
12.09.2022 06:56
Let P=AI∩EF, Q=AI∩BC and D be the tangency point on BC. From a well known lemma from EGMO we know DI passes through N. Now delete AM and consider N as the intersection of DI and EF. Notice that P,D are the corresponding point of the similar triangles △DIY and △CIB. So G,N are also corresponding points. Hence NX/NY=CQ/QB=AC/AB.
19.04.2023 02:27
Let DEF be the intouch triangle. We have X=BI∩EF, Y=CI∩EF, and N=DI∩EF. Since XYBC is cyclic, we get △IXY∼△ICB so if the isogonal of ID with respect to IBC intersects BC at Z, then NXNY=ZCZB. However, by Incenter-Excenter, the isogonal of ID is AI, so ZCZB=ACAB.
19.04.2023 10:33
new favorite geo problem