Difference between revisions of "2018 IMO Problems/Problem 6"
(Created page with "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>\angle XAB = \angle XC...") |
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<math>\angle XAB = \angle XCD</math> and <math>\angle XBC = \angle XDA.</math> | <math>\angle XAB = \angle XCD</math> and <math>\angle XBC = \angle XDA.</math> | ||
Prove that <math>\angle BXA + \angle DXC = 180^{\circ}</math> | Prove that <math>\angle BXA + \angle DXC = 180^{\circ}</math> | ||
− | . | + | |
+ | ==Solution== | ||
+ | <i><b>Special case</b></i> | ||
+ | |||
+ | We construct point <math>X_0</math> and prove that <math>X_0</math> coincides with the point <math>X.</math> | ||
+ | Let <math>AD = CD</math> and <math>AB = BC \implies AB \cdot CD = BC \cdot DA.</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 points <math>B</math> and <math>D</math> are symmetric with respect to the circle <math>\omega = EACF</math> <i><b>(Claim 1).</b></i> | ||
+ | The circle <math>\Omega = FBD</math> is orthogonal to the circle <math>\omega</math> <i><b>(Claim 2).</b></i> | ||
+ | Let <math>X_0</math> be the point of intersection of the circles <math>\omega</math> and <math>\Omega.</math> <math>\angle X_0AB = \angle X_0CD</math> (quadrilateral AX0CF is cyclic) and ∠X0BC = ∠X0DA (quadrangle DX0BF is cyclic) are equal to the property of opposite angles of cyclic quadrangles. This means that X0 coincides with the point X indicated in the condition. | ||
+ | The angle ∠FCX = ∠BCX subtend the arc XF of ω, ∠CBX = ∠XDA subtend the arc XF of Ω. The sum of these arcs is 180° (Claim 3 for orthogonal circles). Hence, the sum of the arcs XF is 180°, the sum of the angles ∠XСВ + ∠XВС = 90°, ∠СХВ = 90°. Similarly, ∠AXD = 90°, that is, ∠BXA + ∠DXC = 180°. |
Revision as of 13:47, 17 August 2022
A convex quadrilateral satisfies
Point
lies inside
so that
and
Prove that
Solution
Special case
We construct point and prove that
coincides with the point
Let
and
Let
and
be the intersection points of
and
and
and
respectively.
The points
and
are symmetric with respect to the circle
(Claim 1).
The circle
is orthogonal to the circle
(Claim 2).
Let
be the point of intersection of the circles
and
(quadrilateral AX0CF is cyclic) and ∠X0BC = ∠X0DA (quadrangle DX0BF is cyclic) are equal to the property of opposite angles of cyclic quadrangles. This means that X0 coincides with the point X indicated in the condition.
The angle ∠FCX = ∠BCX subtend the arc XF of ω, ∠CBX = ∠XDA subtend the arc XF of Ω. The sum of these arcs is 180° (Claim 3 for orthogonal circles). Hence, the sum of the arcs XF is 180°, the sum of the angles ∠XСВ + ∠XВС = 90°, ∠СХВ = 90°. Similarly, ∠AXD = 90°, that is, ∠BXA + ∠DXC = 180°.