We are given an isosceles triangle ABC such that BC=a and AB=BC=b. The variable points M∈(AC) and N∈(AB) satisfy a2⋅AM⋅AN=b2⋅BN⋅CM. The straight lines BM and CN intersect in P. Find the locus of the variable point P. Dan Branzei
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Tags: geometry, trapezoid, trigonometry, incenter, modular arithmetic, ratio, parallelogram
18.08.2010 17:01
Lemma: Let ABC be a triangle; M,N∈[BC]. Then ^BAM=^CAN⇔AB2AC2=MBMC.NBNC Proof: Draw CE‖ Then \Delta AFB\sim \Delta AEC so \frac{AB^2}{AC^2} =\frac{AB}{AC}.\frac{AF}{AE}=\frac{AF}{AC}.\frac{AB}{AE}=\frac{MB}{MC}.\frac{NB}{NC} Let Q\in AC, \widehat{QBC}=\widehat{MBA} so \frac{BC^2}{BA^2}=\frac{MA}{MC}.\frac{QA}{QC} but \frac{BC^2}{BA^2}=\frac{MA}{MC}.\frac{NA}{NB} \Rightarrow \frac{QA}{QC}=\frac{NA}{NB}\Rightarrow NQ \|BC\Rightarrow BNQC is Isosceles trapezium We have \widehat{BPC}=\widehat{PMC}+\widehat{PCM}=\widehat{BMC}+\widehat{NBQ}=\widehat{BMC}+\widehat{MBC}=180^{0}-\widehat{ACB} Let I be incircle center of \Delta ABC so \widehat{BIC}=180^{0}-\widehat{ACB}=\widehat{BPC} Thus P is on arc AIC
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18.08.2010 19:05
Nice tuanh208's proof with the Steiner's theorem ! Here is and a trigonometrical proof. Quote: Let ABC be a triangle with AB=AC=b and BC=a . Let M\in (AC) , N\in (AB) be two mobile points such that \boxed{a^2\cdot AM\cdot AN=b^2\cdot BN\cdot CM}\ (*) and denote P\in BM\cap CN . Find the locus of the point P . Proof 1 (trigonometric). Denote AN=x , AM=y . Thus, BN=b-x , CM=b-y . The relation (*) becomes \underline{a^2xy=b^2(b-x)(b-y)}\ (1) . Denote m\left(\widehat{CBM}\right)=\alpha , m\left(\widehat{BCN}\right)=\beta . Thus, m\left(\widehat{ABM}\right)=B-\alpha , m\left(\widehat{ACN}\right)=B-\beta . Apply theorem of Sinus in triangles : \left\|\begin{array}{cccc} \triangle\ ABM\ : & \frac {y}{\sin (B-\alpha )}=\frac {b}{\sin (B+\alpha )} & \implies & y=\frac {b\cdot\sin (B-\alpha )}{\sin (B+\alpha )}\\\\ \triangle\ ACN\ : & \frac {x}{\sin (B-\beta )}=\frac {b}{\sin (B+\beta )} & \implies & x=\frac {b\cdot\sin (B-\beta )}{\sin (B+\beta )}\\\\ \triangle\ BCM\ : & \frac {b-y}{\sin\alpha }=\frac {a}{\sin (B+\alpha )} & \implies & b-y=\frac {a\cdot\sin\alpha }{\sin (B+\alpha )}\\\\ \triangle\ BCN\ : & \frac {b-x}{\sin \beta }=\frac {a}{\sin (B+\beta )} & \implies & b-x=\frac {a\cdot\sin \beta }{\sin (B+\beta )}\end{array}\right\| . The relation (1) becomes \frac {a^2b^2\sin (B-\alpha )\sin (B-\beta )}{\sin (B+\alpha )\sin (B+\beta )}=\frac {b^2a^2\sin \alpha \sin\beta }{\sin (B+\alpha )\sin (B+\beta )} \iff \sin (B-\alpha )\sin (B-\beta )=\sin\alpha\sin\beta \iff \cos (\alpha -\beta )-\cos [2B-(\alpha +\beta )]=\cos (\alpha -\beta )-\cos (\alpha +\beta ) \iff \cos [2B-(\alpha +\beta )]=\cos (\alpha +\beta ) \iff \boxed {\alpha +\beta =B} . Thus, m\left(\widehat{BPC}\right)=A+B (constant). Construct the rombus ABDC . Observe that D is A-exincenter and m\left(\widehat{BDC}\right)=B . In conclusion, the point P belongs to the circle with diameter [ID] (where I is incenter) so that the sideline BC separates P , D.
18.08.2010 21:12
Isn't there any solution by Menelaus Theorem?
19.08.2010 03:15
Proof 2 (metric). Denote AN=x , AM=y . Thus, BN=b-x , CM=b-y . The relation (*) becomes \underline{a^2xy=b^2(b-x)(b-y)}\ (1) . Denote m\left(\widehat{CBM}\right)=\alpha , m\left(\widehat{BCN}\right)=\beta . Thus, m\left(\widehat{ABM}\right)=B-\alpha , m\left(\widehat{ACN}\right)=B-\beta . Apply an well-known area relation [PBN]\cdot [PCM]=[PMN]\cdot [BPC] . Thus PB\cdot (b-x)\cdot \sin (B-\alpha )\cdot PC\cdot (b-y)\cdot\sin (C-\beta )= =\frac {a^2\sin \alpha\sin\beta}{\sin (\alpha +\beta )}\cdot PN\cdot PM\cdot \sin (\alpha +\beta ) \iff \frac {\sin (B-\alpha )\sin (B-\beta )}{\sin\alpha\sin \beta}=\frac {a^2}{(b-x)(b-y)}\cdot\frac {PM\cdot PN}{PB\cdot PC}\ (2) . Apply the Menelaus' theorem to : \left\|\begin{array}{cc} \overline {CPN}/ \triangle ABM \ : & \frac {b-y}{b}\cdot \frac {x}{b-x}\cdot \frac {PB}{PM}=1\\\\ \overline{BPM}/ \triangle ACN\ : & \frac {b-x}{b}\cdot\frac {y}{b-y}\cdot\frac {PC}{PN}=1\end{array}\right\| \bigodot\implies \frac {xy}{b^2}\cdot\frac {PB\cdot PC}{PM\cdot PN}=1\ (3) . Therefore, \left\|\begin{array}{cc} (1) & a^2xy=b^2(b-x)(b-y)\\\\ (2) & \frac {\sin (B-\alpha )\sin (B-\beta )}{\sin\alpha\sin \beta}=\frac {a^2}{(b-x)(b-y)}\cdot\frac {PM\cdot PN}{PB\cdot PC}\\\\ (3) & 1=\frac {xy}{b^2}\cdot\frac {PB\cdot PC}{PM\cdot PN}\end{array}\right\|\ \bigodot \implies \sin (B-\alpha )\sin (B-\beta)=\sin \alpha\sin\beta \iff \cos (\alpha -\beta )-\cos [2B-(\alpha +\beta )]=\cos (\alpha -\beta )-\cos (\alpha +\beta ) \iff \cos [2B-(\alpha +\beta )]=\cos (\alpha +\beta ) \iff \boxed {\alpha +\beta =B} . Thus, m\left(\widehat{BPC}\right)=A+B (constant). Construct the rombus ABDC . Observe that D is A-exincenter and m\left(\widehat{BDC}\right)=B . In conclusion, the point P belongs to the circle with diameter [ID] (where I is incenter) so that the sideline BC separates P , D.
19.08.2010 04:12
I is the incenter of \triangle ABC. The inversion with pole A and radius AB=AC=L transforms \odot(IBC) into itself. Let Q,M',N' be the inverses of P,M,N. Q \equiv \odot(ABM') \cap \odot(ACN'). Then we have \frac{BN'}{BN}=\frac{L}{AN} \ , \ \frac{CM'}{CM}=\frac{L}{AM} \Longrightarrow BN' \cdot CM'=L^2 \cdot \frac{BN \cdot CM}{AN \cdot AM} By the hypotesis, we have then BN' \cdot CM'=BC^2 \Longrightarrow \frac{BN'}{BC}=\frac{BC}{CM'} Thus, \triangle N'BC \sim \triangle BCM' by SAS \Longrightarrow \angle BN'C=\angle CBM'=\angle AQC and \angle BCN'= \angle CM'B=\angle AQB. Hence, \angle BQC=\angle BN'C+\angle BCN'=\angle ABC \pmod\pi \Longrightarrow Q \in \odot(IBC). If circles \odot(ABM') and \odot(ACN') meet on \odot(IBC), their inverse lines BM,CN meet on the double circle \odot(IBC), i.e. P \in \odot(IBC).
19.08.2010 04:26
Quote: Let ABC be a triangle with AB=AC=b and BC=a . Let M\in (AC) , N\in (AB) be two mobile points such that \boxed{a^2\cdot AM\cdot AN=b^2\cdot BN\cdot CM}\ (*) and denote P\in BM\cap CN . Find the locus of the point P . Proof 3 (metric). Denote AN=x , AM=y . Thus, BN=b-x , CM=b-y . The relation (*) becomes \boxed{a^2xy=b^2(b-x)(b-y)}\ (1) . Denote m\left(\widehat{CBM}\right)=\alpha , m\left(\widehat{BCN}\right)=\beta . Thus, m\left(\widehat{ABM}\right)=B-\alpha , m\left(\widehat{ACN}\right)=B-\beta . Apply an well-known relation to assess a ratio : \left\|\begin{array}{c} \frac {x}{b-x}=\frac {NA}{NB}=\frac {CA}{CB}\cdot\frac {\sin\widehat{NCA}}{\sin\widehat{NCB}}=\frac ba\cdot\frac {\sin (B-\beta )}{\sin \beta}\\\\ \frac {y}{b-y}=\frac {MA}{MB}=\frac {BA}{BC}\cdot\frac {\sin\widehat{MBA}}{\sin\widehat{MBC}}=\frac ba\cdot\frac {\sin (B-\alpha )}{\sin \alpha}\end{array}\right\| \bigodot\ \stackrel{(1)}{\implies} \sin(B-\alpha )\sin (B-\beta )=\sin \alpha\sin\beta \iff \cos (\alpha -\beta )-\cos [2B-(\alpha +\beta )]= \cos (\alpha -\beta )-\cos (\alpha +\beta ) \iff \cos [2B-(\alpha +\beta )]= \cos (\alpha +\beta ) \iff \boxed {\alpha +\beta =B} . Thus, m\left(\widehat{BPC}\right)=A+B (constant). Construct the rombus ABDC . Observe that D is A-exincenter I_a and m\left(\widehat{BDC}\right)=B . In conclusion, the point P belongs to the circle with diameter [II_a] (where I is incenter) so that the sideline BC separates P , I_a.
19.08.2010 13:56
Very nice proofs above! Assuming Steiner formula is unknown, let's see another way of construction of the points M, N. Take a random point M on (AC) and \{B, Q\}\in BM \cap \odot (ABC), then draw CR||AQ, R on \odot (ABC). We claim that \{N\}\equiv AB\cap CR and \{P\} \equiv BM \cap CN are our required points, i.e. N verifies the given relation. As \widehat{CBQ}=\widehat{CAQ}=\widehat{ACR} and \widehat{ABQ}=\widehat{ACQ}, we get \widehat{QCR}=\widehat{ABC}\;(\;1\;) and, with \widehat{QPC}=\widehat{AQB}=\widehat{ACB}\;(\;2\;), hence \triangle QPC is isosceles, ans so is \triangle RBP, both of them being similar to \triangle ABC \;(\;*\;). ARCQ is isosceles trapezoid and, PQ=CQ makes ARPQ parallelogram, i.e. AR||PQ, consequently \frac{BN}{AN}=\frac{BP}{PQ} and \frac{CM}{AM}=\frac{CP}{PR} or, multiplying side by side the above two equalities: $\frac{{BN}{AN}\cdot \frac{CM}{AM}=\frac{BP}{PR}\cdot \frac{CP}{PQ}; with (*), each fraction from right side of the last equality is \frac{BC}{AB}, so our claim has been proved. From (2) we have \widehat{QPC}=\angle B, i.e. \widehat{BPC}=\widehat{BIC} and P lies on the arc BIC of \odot (BIC)$, done. Best regards, sunken rock
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19.08.2010 16:16
Virgil Nicula wrote: \sin (B-\alpha )\sin (B-\beta)=\sin \alpha\sin\beta \iff \cos (\alpha -\beta )-\cos [2B-(\alpha +\beta )]=\cos (\alpha -\beta )-\cos (\alpha +\beta ) Why is it true?
19.08.2010 21:28
\left\|\begin{array}{c} \cos (x+y)=\cos x\cos y-\sin x\sin y\\\\ \cos (x-y)=\cos x\cos y+\sin x\sin y\end{array}\right\| \begin{array}{ccc} \ + & \implies & 2\cos x\cos y=\cos (x+y)+\cos (x-y)\ .\\\\ \ - & \implies & \boxed {2\sin x\sin y=\cos (x-y)-\sin (x+y)}\ .\end{array}
19.08.2010 23:18
Can establish somebody directly or at least using geogebra what is the geometrical locus of the interior point P w.r.t. \triangle ABC so that \boxed{\frac {AM\cdot AN}{BN\cdot CM}=\frac {bc}{a^2}} , where M\in AC\cap BP , N\in AB\cap CP ?!
20.08.2010 01:15
Virgil Nicula wrote: \left\|\begin{array}{c} \cos (x+y)=\cos x\cos y-\sin x\sin y\\\\ \cos (x-y)=\cos x\cos y+\sin x\sin y\end{array}\right\| \begin{array}{ccc} \ + & \implies & 2\cos x\cos y=\cos (x+y)+\cos (x-y)\ .\\\\ \ - & \implies & \boxed {2\sin x\sin y=\cos (x-y)-\sin (x+y)}\ .\end{array} Thanks very much! It was an easy equation! I don't know why i couldn't realize it!
20.08.2010 02:46
Virgil Nicula wrote: Can establish somebody directly or at least using geogebra what is the geometrical locus of the interior point P w.r.t. \triangle ABC so that \frac {AM\cdot AN}{BN\cdot CM}=\frac {bc}{a^2}, where M\in AC\cap BP, N\in AB\cap CP. If (x:y:z) are the barycentric coordinates of P with respect to \triangle ABC, then locus of P is the conic \mathcal{K} \equiv bcx^2-a^2yz=0. \mathcal{K} passes through the incenter I, A-excenter I_a and is tangent to AB,AC through B,C. If AB=AC (the proposed problem), then \mathcal{K} becomes the circumcircle of \triangle IBC.
20.08.2010 04:10
Thank you, Luis! It's nice result. Maybe somebody presents a draw. I'm learning just now geogebra. Thanks.
20.08.2010 06:43
Virgil Nicula wrote: Maybe somebody presents a draw. I'm learning just now geogebra. Thanks. I also use Geogebra, since it's quite easy to manage. Note that Geogebra can only draw conic sections passing through 5 points. In this case, we need to draw \mathcal{K} given 4 points B,C,I,I_a and the tangent line AC through C. We can define a homology \mathcal{U} with center C mapping \mathcal{K} into an arbitrary circle \omega tangent to AC through C, in order to find as many points on \mathcal{K} as we want. The following sketch shows the construction of a fith point N on \mathcal{K}.
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