Difference between revisions of "2011 USAJMO Problems/Problem 5"
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Points <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, <math>E</math> lie on a circle <math>\omega</math> and point <math>P</math> lies outside the circle. The given points are such that (i) lines <math>PB</math> and <math>PD</math> are tangent to <math>\omega</math>, (ii) <math>P</math>, <math>A</math>, <math>C</math> are collinear, and (iii) <math>\overline{DE} \parallel \overline{AC}</math>. Prove that <math>\overline{BE}</math> bisects <math>\overline{AC}</math>. | Points <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, <math>E</math> lie on a circle <math>\omega</math> and point <math>P</math> lies outside the circle. The given points are such that (i) lines <math>PB</math> and <math>PD</math> are tangent to <math>\omega</math>, (ii) <math>P</math>, <math>A</math>, <math>C</math> are collinear, and (iii) <math>\overline{DE} \parallel \overline{AC}</math>. Prove that <math>\overline{BE}</math> bisects <math>\overline{AC}</math>. | ||
− | == | + | == Solution 4 == |
Connet segment PO, and name the interaction of PO and the circle as point M. | Connet 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. | Since PB and PD are tangent to the circle, it's easy to see that M is the midpoint of arc BD. | ||
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Therefore, PBOM is cyclic, ∠ PFO = ∠ OBP = 90°, AF = AC (F is the interaction of BE and AC) | Therefore, PBOM is cyclic, ∠ PFO = ∠ OBP = 90°, AF = AC (F is the interaction of BE and AC) | ||
BE bisects AC, proof completed! | BE bisects AC, proof completed! | ||
+ | ~ MVP Harry | ||
==Solution 1== | ==Solution 1== |
Revision as of 07:39, 27 April 2019
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 4
Connet 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 1
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
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.