Difference between revisions of "2018 IMO Problems/Problem 6"
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+ | ==Problem== | ||
+ | |||
A convex quadrilateral <math>ABCD</math> satisfies <math>AB\cdot CD=BC \cdot DA.</math> Point <math>X</math> lies inside | A convex quadrilateral <math>ABCD</math> satisfies <math>AB\cdot CD=BC \cdot DA.</math> Point <math>X</math> lies inside | ||
<math>ABCD</math> so that | <math>ABCD</math> so that | ||
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==Solution== | ==Solution== | ||
− | [[File:2018 IMO 6.png| | + | [[File:2018 IMO 6bb.png|470px|right]] |
− | [[File:2018 IMO | + | [[File:2018 IMO 6.png|470px|right]] |
+ | [[File:2018 IMO 6e.png|470px|right]] | ||
+ | We want to find the point <math>X.</math> Let <math>E</math> and <math>F</math> be the intersection points of <math>AB</math> and <math>CD,</math> and <math>BC</math> and <math>DA,</math> respectively. | ||
+ | The poinx <math>X</math> is inside <math>ABCD,</math> so points <math>E,A,X,C</math> follow in this order. | ||
+ | |||
+ | <math>\angle XAB = \angle XCD \implies \angle XAE + \angle XCE = 180^\circ </math> | ||
+ | <math>\implies AXCE</math> is cyclic <math>\implies X</math> lie on circle <math>ACE.</math> | ||
+ | |||
+ | Similarly, <math>X</math> lie on circle <math>BDF.</math> | ||
+ | |||
+ | Point <math>X</math> is the point of intersection of circles <math>ACE</math> and <math>\Omega = BDF.</math> | ||
<i><b>Special case</b></i> | <i><b>Special case</b></i> | ||
− | + | Let <math>AD = CD</math> and <math>AB = BC \implies AB \cdot CD = BC \cdot DA.</math> | |
+ | |||
+ | The points <math>B</math> and <math>D</math> are symmetric with respect to the circle <math>\theta = ACEF</math> <i><b>(Claim 1).</b></i> | ||
+ | |||
+ | The circle <math>BDF</math> is orthogonal to the circle <math>\theta</math> <i><b>(Claim 2).</b></i> | ||
− | + | <math>\hspace{10mm} \angle FCX = \angle BCX = \frac {\overset{\Large\frown} {XAF}}{2}</math> of <math>\theta.</math> <math>\hspace{10mm} \angle CBX = \angle XDA = \frac {\overset{\Large\frown} {XBF}}{2}</math> of <math>\Omega.</math> | |
+ | |||
+ | <math>\overset{\Large\frown} {XAF} + \overset{\Large\frown} {XBF} = 180^\circ</math> <i><b>(Claim 3)</b></i> <math>\implies</math> | ||
+ | <math>\angle XCB + \angle XBC = 90^\circ \implies \angle CXB = 90^\circ.</math> | ||
+ | |||
+ | Similarly, <math>\angle AXD = 90^\circ \implies</math> | ||
+ | <cmath>\angle BXA + \angle DXC = 360^\circ -\angle AXD -\angle CXB = 180^\circ.</cmath> | ||
+ | |||
+ | [[File:2018 IMO 6c.png|470px|right]] | ||
+ | [[File:2018 IMO 6d.png|470px|right]] | ||
+ | |||
+ | <i><b>Common case </b></i> | ||
− | + | Denote by <math>O</math> the intersection point of <math>BD</math> and the perpendicular bisector of <math>AC.</math> Let <math>\omega</math> be a circle (red) with center <math>O</math> and radius <math>OA = R.</math> | |
− | + | We will prove <math>\sin\angle BXA =\sin \angle DXC</math> using point <math>Y</math> symmetric to <math>X</math> with respect to <math>\omega.</math> | |
− | The | + | The points <math>B</math> and <math>D</math> are symmetric with respect to <math>\omega</math> <i><b>(Claim 1).</b></i> |
− | + | The circles <math>BDF</math> and <math>BDE</math> are orthogonal to the circle <math>\omega</math> <i><b>(Claim 2).</b></i> | |
− | |||
− | + | Circles <math>ACF</math> and <math>ACE</math> are symmetric with respect to the circle <math>\omega</math> <i><b>(Lemma).</b></i> | |
− | + | Denote by <math>Y</math> the point of intersection of circles <math>BDF</math> and <math>ACF.</math> | |
− | <math> | + | Quadrangle <math>BYDF</math> is cyclic <math>\implies \angle CBY = \angle ADY.</math> |
− | <math> | + | Quadrangle <math>AYCF</math> is cyclic <math>\implies \angle YAD = \angle YCB.</math> |
− | The | + | The triangles <math>\triangle YAD \sim \triangle YCB</math> by two angles, so <cmath>\frac {BC}{AD} = \frac {CY}{AY} = \frac {BY} {DY} \hspace{10mm} (1).</cmath> |
− | <math>\ | + | The points <math>X</math> and <math>Y</math> are symmetric with respect to the circle <math>\omega</math>, since they lie on the intersection of the circles <math>ACF</math> and <math>ACE</math> symmetric with respect to <math>\omega</math> and the circle <math>BDF</math> orthogonal to <math>\omega.</math> |
− | + | The point <math>B</math> is symmetric to <math>D</math> with respect to <math>\omega \implies</math> | |
+ | <cmath>\triangle OBC \sim \triangle OCD \implies \frac {OB}{OC} = \frac {BC}{CD} = \frac {OC}{OD},</cmath> | ||
+ | <cmath>\frac {OB}{OD} = \frac {OB}{OC} \cdot \frac {OC}{OD} = \frac{BC^2}{CD^2} = \frac{BC}{CD} \cdot \frac {AB}{AD}.</cmath> | ||
+ | The point <math>B</math> is symmetric to <math>D</math> and the point <math>X</math> is symmetric to <math>Y</math> with respect to <math>\omega,</math> hence | ||
+ | <cmath>\frac {BX}{DY} = \frac {R^2}{OD \cdot OY} ,\frac {DX}{BY} = \frac{R^2}{OB \cdot OY}.</cmath> | ||
+ | <cmath>\frac{BX}{DX} =\frac{DY}{BY} \cdot \frac {OB}{OD} = \frac{AD}{BC} \cdot \frac{BC}{CD} \cdot \frac{AB}{AD} = \frac{AB}{CD}.</cmath> | ||
+ | [[File:2018 IMO 6 angles.png|390px|right]] | ||
+ | [[File:2018 IMO 6 Claim 3.png|390px|right]] | ||
+ | [[File:2018 IMO 6a.png|390px|right]] | ||
− | + | Denote <math>\angle XAB = \angle XCD = \alpha, \angle BXA = \varphi, \angle DXC = \psi.</math> | |
+ | |||
+ | By the law of sines for <math>\triangle ABX,</math> we obtain <math>\frac {AB}{\sin \varphi} = \frac{BX}{\sin \alpha}.</math> | ||
− | + | By the law of sines for <math>\triangle CDX,</math> we obtain <math>\frac {CD}{\sin \psi} = \frac {DX}{\sin \alpha}.</math> | |
− | + | Hence we get <math>\frac{\sin \psi} {\sin \varphi}= \frac {CD}{DX} \cdot \frac{BX}{AB} = 1.</math> | |
− | |||
− | + | If <math>\varphi = \psi,</math> then <math>\triangle XAB \sim \triangle XCD \implies \frac {CD}{AB} = \frac {BX}{DX} = \frac{AX}{CX} = \frac {AD}{BC}.</math> | |
− | + | <math>CD \cdot BC = AB \cdot AD \implies AD = CD, AB = BC.</math> This is a special case. | |
− | |||
− | < | ||
− | + | In all other cases, the equality of the sines follows <math>\varphi + \psi = 180^\circ .</math> | |
− | + | <i><b>Claim 1</b></i> Let <math>A, C,</math> and <math>E</math> be arbitrary points on a circle <math>\omega, l</math> be the perpendicular bisector to the segment <math>AC.</math> Then the straight lines <math>AE</math> and <math>CE</math> intersect <math>l</math> at the points <math>B</math> and <math>D,</math> symmetric with respect to <math>\omega.</math> | |
− | + | <i><b>Claim 2</b></i> Let points <math>B</math> and <math>D</math> be symmetric with respect to the circle <math>\omega.</math> Then any circle <math>\Omega</math> passing through these points is orthogonal to <math>\omega.</math> | |
− | + | <i><b>Claim 3</b></i> The sum of the arcs between the points of intersection of two perpendicular circles is <math>180^\circ.</math> | |
+ | In the figure they are a blue and red arcs <math>\overset{\Large\frown} {CD}, \alpha + \beta = 180^\circ.</math> | ||
− | + | <i><b>Lemma</b></i> The opposite sides of the quadrilateral <math>ABCD</math> intersect at points <math>E</math> and <math>F</math> (<math>E</math> lies on <math>AB</math>). The circle <math>\omega</math> centered at the point <math>O</math> contains the ends of the diagonal <math>AC.</math> The points <math>B</math> and <math>D</math> are symmetric with respect to the circle <math>\omega</math> (in other words, the inversion with respect to <math>\omega</math> maps <math>B</math> into <math>D).</math> Then the circles <math>ACE</math> and <math>ACF</math> are symmetric with respect to <math>\omega.</math> | |
− | |||
− | |||
− | + | <i><b>Proof</b></i> We will prove that the point <math>G,</math> symmetric to the point <math>E</math> with respect to <math>\omega,</math> belongs to the circle <math>ACF</math> becouse <math>\angle AGC = \angle AFC.</math> | |
− | + | A circle <math>BDE</math> containing points <math>B</math> and <math>D</math> symmetric with respect to <math>\omega,</math> is orthogonal to <math>\omega</math> <i><b>(Claim 2)</b></i> and maps into itself under inversion with respect to the circle <math>\omega.</math> Hence, the point <math>E</math> under this inversion passes to some point <math>G,</math> of the same circle <math>BDE.</math> | |
− | + | A straight line <math>ABE</math> containing the point <math>A</math> of the circle <math>\omega,</math> under inversion with respect to <math>\omega,</math> maps into the circle <math>OADG.</math> Hence, the inscribed angles of this circle are equal <math>\angle ADB = \angle AGE.</math> | |
+ | <math>\angle OCE = \angle CGE (CE</math> maps into <math>CG)</math> and <math>\angle OCE = \angle BCD (BC</math> maps into <math>DC).</math> | ||
+ | Consequently, the angles <math>\angle AFC = \angle ADB – \angle FBD = \angle AGE - \angle CGE = \angle AGC.</math> | ||
+ | These angles subtend the <math>\overset{\Large\frown} {AC}</math> of the <math>ACF</math> circle, that is, the point <math>G,</math> symmetric to the point <math>E</math> with respect to <math>\omega,</math> belongs to the circle <math>ACF.</math> | ||
− | + | '''vladimir.shelomovskii@gmail.com, vvsss''' | |
− | |||
− | + | ==See Also== | |
− | + | {{IMO box|year=2018|num-b=5|after=Last Problem}} |
Latest revision as of 00:47, 19 November 2023
Problem
A convex quadrilateral satisfies Point lies inside so that and Prove that
Solution
We want to find the point Let and be the intersection points of and and and respectively. The poinx is inside so points follow in this order.
is cyclic lie on circle
Similarly, lie on circle
Point is the point of intersection of circles and
Special case
Let and
The points and are symmetric with respect to the circle (Claim 1).
The circle is orthogonal to the circle (Claim 2).
of of
(Claim 3)
Similarly,
Common case
Denote by the intersection point of and the perpendicular bisector of Let be a circle (red) with center and radius
We will prove using point symmetric to with respect to
The points and are symmetric with respect to (Claim 1).
The circles and are orthogonal to the circle (Claim 2).
Circles and are symmetric with respect to the circle (Lemma).
Denote by the point of intersection of circles and
Quadrangle is cyclic
Quadrangle is cyclic
The triangles by two angles, so
The points and are symmetric with respect to the circle , since they lie on the intersection of the circles and symmetric with respect to and the circle orthogonal to
The point is symmetric to with respect to The point is symmetric to and the point is symmetric to with respect to hence
Denote
By the law of sines for we obtain
By the law of sines for we obtain
Hence we get
If then This is a special case.
In all other cases, the equality of the sines follows
Claim 1 Let and be arbitrary points on a circle be the perpendicular bisector to the segment Then the straight lines and intersect at the points and symmetric with respect to
Claim 2 Let points and be symmetric with respect to the circle Then any circle passing through these points is orthogonal to
Claim 3 The sum of the arcs between the points of intersection of two perpendicular circles is In the figure they are a blue and red arcs
Lemma The opposite sides of the quadrilateral intersect at points and ( lies on ). The circle centered at the point contains the ends of the diagonal The points and are symmetric with respect to the circle (in other words, the inversion with respect to maps into Then the circles and are symmetric with respect to
Proof We will prove that the point symmetric to the point with respect to belongs to the circle becouse
A circle containing points and symmetric with respect to is orthogonal to (Claim 2) and maps into itself under inversion with respect to the circle Hence, the point under this inversion passes to some point of the same circle
A straight line containing the point of the circle under inversion with respect to maps into the circle Hence, the inscribed angles of this circle are equal maps into and maps into Consequently, the angles These angles subtend the of the circle, that is, the point symmetric to the point with respect to belongs to the circle
vladimir.shelomovskii@gmail.com, vvsss
See Also
2018 IMO (Problems) • Resources | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Problem |
All IMO Problems and Solutions |