Difference between revisions of "2011 USAJMO Problems/Problem 5"
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Let <math>O</math> be the center of the circle, and let <math>X</math> be the intersection of <math>AC</math> and <math>BE</math>. Let <math>\angle OPA</math> be <math>x</math> and <math>\angle OPD</math> be <math>y</math>. | Let <math>O</math> be the center of the circle, and let <math>X</math> be the intersection of <math>AC</math> and <math>BE</math>. Let <math>\angle OPA</math> be <math>x</math> and <math>\angle OPD</math> be <math>y</math>. |
Revision as of 02:30, 24 July 2020
Problem
Points ,
,
,
,
lie on a circle
and point
lies outside the circle. The given points are such that (i) lines
and
are tangent to
, (ii)
,
,
are collinear, and (iii)
. Prove that
bisects
.
Solution 1
Connect segment PO, and name the interaction of PO and the circle as point M.
Since PB and PD are tangent to the circle, it's easy to see that M is the midpoint of arc BD.
∠ BOA = 1/2 arc AB + 1/2 arc CE
Since AC // DE, arc AD = arc CE,
thus, ∠ BOA = 1/2 arc AB + 1/2 arc AD = 1/2 arc BD = arc BM = ∠ BOM
Therefore, PBOM is cyclic, ∠ PFO = ∠ OBP = 90°, AF = AC (F is the interaction of BE and AC)
BE bisects AC, proof completed!
~ MVP Harry
Solution 2
Let be the center of the circle, and let
be the intersection of
and
. Let
be
and
be
.
,
,
Thus is a cyclic quadrilateral and
and so
is the midpoint of chord
.
~pandadude
Solution 2
This is the solution from EGMO Problem 1.43 page 242
Let be the center of the circle, and let
be the midpoint of
. Let
denote the circle with diameter
. Since
,
,
, and
all lie on
.
Since quadrilateral is cyclic,
. Triangles
and
are congruent, so
, so
. Because
and
are parallel,
lies on
(using Euclid's Parallel Postulate).
Solution 3
Note that by Lemma 9.9 of EGMO, is a harmonic bundle. We project through
onto
,
Where
is the point at infinity for parallel lines
and
. Thus, we get
, and
is the midpoint of
. ~novus677
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.