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Difference between revisions of "2004 IMO Shortlist Problems/G8"
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A cyclic quadrilateral <math>ABCD</math> is given. The lines <math>AD</math> and <math>BC</math> intersect at <math>E</math>, with <math>C</math> between <math>B</math> and <math>E</math>; the diagonals <math>AC</math> and <math>BD</math> intersect at <math>F</math>. Let <math>M</math> be the midpoint of the side <math>CD</math>, and let <math>N \neq M</math> be a point on the circumcircle of <math>\triangle ABM</math> such that <math>\frac{AN}{BN} = \frac{AM}{BM}</math>. Prove that <math>E, F, N</math> are collinear.
 
A cyclic quadrilateral <math>ABCD</math> is given. The lines <math>AD</math> and <math>BC</math> intersect at <math>E</math>, with <math>C</math> between <math>B</math> and <math>E</math>; the diagonals <math>AC</math> and <math>BD</math> intersect at <math>F</math>. Let <math>M</math> be the midpoint of the side <math>CD</math>, and let <math>N \neq M</math> be a point on the circumcircle of <math>\triangle ABM</math> such that <math>\frac{AN}{BN} = \frac{AM}{BM}</math>. Prove that <math>E, F, N</math> are collinear.
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== Solution ==
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Let <math>P = CD \cap EF</math>. Let <math>Q = AB \cap CD</math>. Let <math>R = AB \cap EF</math>. Let<math>(ABM)</math> denote the circumcircle of <math>\triangle ABM</math>. Let <math>N' = EF \cap (ABM)</math>. Note that <math>N' = PR \cap (ABM)</math>
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Claim: <math>P</math> is on <math>(ABM)</math>. Proof: <math>(C, D; P, Q) = -1</math> as complete quadrilaterals induce harmonic bundles. <math>QP \cdot QM = QC \cdot QD</math> by Lemma 9.17 on Euclidean Geometry in Maths Olympiad. By power of a point theorem, <math>QP \cdot QM = QA \cdot QB</math> and this is equivalent to our original claim.
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<math>(A, B; R, Q) = -1</math> as complete quadrilaterals induce harmonic bundles. By a projection through <math>P</math> from <math>AQ</math> onto <math>(ABM)</math>, <math>(A, B; N', M) = -1</math>. Since <math>\frac{AN}{BN} = \frac{AM}{BM}</math>, <math>M</math> and <math>N</math> are on the intersections of <math>(ABM)</math> and an Appollonian circle centered on AB, so N and M are on the opposite sides of AB. Therefore, <math>(A, B; N, M) = -\frac{AN}{BN}\div\frac{AM}{BM} = -1</math>. By uniqueness of harmonic conjugate, <math>N = N'</math>
  
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Latest revision as of 00:43, 6 July 2021

Problem

A cyclic quadrilateral $ABCD$ is given. The lines $AD$ and $BC$ intersect at $E$, with $C$ between $B$ and $E$; the diagonals $AC$ and $BD$ intersect at $F$. Let $M$ be the midpoint of the side $CD$, and let $N \neq M$ be a point on the circumcircle of $\triangle ABM$ such that $\frac{AN}{BN} = \frac{AM}{BM}$. Prove that $E, F, N$ are collinear.

Solution

Let $P = CD \cap EF$. Let $Q = AB \cap CD$. Let $R = AB \cap EF$. Let$(ABM)$ denote the circumcircle of $\triangle ABM$. Let $N' = EF \cap (ABM)$. Note that $N' = PR \cap (ABM)$


Claim: $P$ is on $(ABM)$. Proof: $(C, D; P, Q) = -1$ as complete quadrilaterals induce harmonic bundles. $QP \cdot QM = QC \cdot QD$ by Lemma 9.17 on Euclidean Geometry in Maths Olympiad. By power of a point theorem, $QP \cdot QM = QA \cdot QB$ and this is equivalent to our original claim.


$(A, B; R, Q) = -1$ as complete quadrilaterals induce harmonic bundles. By a projection through $P$ from $AQ$ onto $(ABM)$, $(A, B; N', M) = -1$. Since $\frac{AN}{BN} = \frac{AM}{BM}$, $M$ and $N$ are on the intersections of $(ABM)$ and an Appollonian circle centered on AB, so N and M are on the opposite sides of AB. Therefore, $(A, B; N, M) = -\frac{AN}{BN}\div\frac{AM}{BM} = -1$. By uniqueness of harmonic conjugate, $N = N'$

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