Difference between revisions of "2018 IMO Problems/Problem 6"
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+ | ==Problem== | ||
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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> | |
− | + | Let <math>AD = CD</math> and <math>AB = BC \implies AB \cdot CD = BC \cdot DA.</math> | |
− | The | + | 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> | |
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− | <math>\hspace{10mm} \angle FCX = \angle BCX = \frac { | ||
− | |||
− | <math>\hspace{10mm} \angle CBX = \angle XDA = \frac { | ||
− | |||
− | |||
+ | <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> | <math>\angle XCB + \angle XBC = 90^\circ \implies \angle CXB = 90^\circ.</math> | ||
− | Similarly, <math>\angle AXD = 90^\circ \implies \angle BXA + \angle DXC = 180^\circ.</ | + | 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> | + | <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> | |
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− | |||
− | |||
− | < | ||
− | + | 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 points <math>B</math> and <math>D</math> are symmetric with respect to | + | 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> | 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> | Circles <math>ACF</math> and <math>ACE</math> are symmetric with respect to the circle <math>\omega</math> <i><b>(Lemma).</b></i> | ||
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Denote by <math>Y</math> the point of intersection of circles <math>BDF</math> and <math>ACF.</math> | Denote by <math>Y</math> the point of intersection of circles <math>BDF</math> and <math>ACF.</math> | ||
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Quadrangle <math>BYDF</math> is cyclic <math>\implies \angle CBY = \angle ADY.</math> | Quadrangle <math>BYDF</math> is cyclic <math>\implies \angle CBY = \angle ADY.</math> | ||
− | Quadrangle <math> | + | Quadrangle <math>AYCF</math> is cyclic <math>\implies \angle YAD = \angle YCB.</math> |
− | The triangles <math>\triangle YAD \sim \triangle | + | 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> |
− | 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 | + | 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 | + | 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>\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> | <cmath>\frac {OB}{OD} = \frac {OB}{OC} \cdot \frac {OC}{OD} = \frac{BC^2}{CD^2} = \frac{BC}{CD} \cdot \frac {AB}{AD}.</cmath> | ||
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<cmath>\frac {BX}{DY} = \frac {R^2}{OD \cdot OY} ,\frac {DX}{BY} = \frac{R^2}{OB \cdot OY}.</cmath> | <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> | <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> | Denote <math>\angle XAB = \angle XCD = \alpha, \angle BXA = \varphi, \angle DXC = \psi.</math> | ||
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By the law of sines for <math>\triangle CDX,</math> we obtain <math>\frac {CD}{\sin \psi} = \frac {DX}{\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 01:47, 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 |