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>. | ||
− | == | + | == The Power of Angle Chasing == |
+ | https://www.youtube.com/watch?v=Dn1IIx9Cnqw&list=PLqgsN351HEtHgi3Ax2oXJk_bEiROCvNF5 | ||
− | ==Solution 1== | + | ==Solution== |
+ | Since <math>\overline{DE} \parallel \overline{AC}</math>, we have that <math>AD=CE\rightarrow \angle ABD=\angle CBE</math>. But it is well known that <math>ABCD</math> is a harmonic quadrilateral, thus <math>\overline{BD}</math> is a symmedian of triangle <math>ABC</math>, from which it follows that <math>\overline{BE}</math> is a median of <math>\triangle ABC</math>. <math>\blacksquare</math> | ||
+ | |||
+ | == 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 <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>. | ||
Line 17: | Line 38: | ||
~pandadude | ~pandadude | ||
− | ==Solution | + | ==Solution 3== |
+ | This is the solution from EGMO Problem 1.43 page 242 | ||
Let <math>O</math> be the center of the circle, and let <math>M</math> be the midpoint of <math>AC</math>. Let <math>\theta</math> denote the circle with diameter <math>OP</math>. Since <math>\angle OBP = \angle OMP = \angle ODP = 90^\circ</math>, <math>B</math>, <math>D</math>, and <math>M</math> all lie on <math>\theta</math>. | Let <math>O</math> be the center of the circle, and let <math>M</math> be the midpoint of <math>AC</math>. Let <math>\theta</math> denote the circle with diameter <math>OP</math>. Since <math>\angle OBP = \angle OMP = \angle ODP = 90^\circ</math>, <math>B</math>, <math>D</math>, and <math>M</math> all lie on <math>\theta</math>. | ||
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</asy> | </asy> | ||
− | Since quadrilateral <math>BOMP</math> is cyclic, <math>\angle BMP = \angle BOP</math>. Triangles <math>BOP</math> and <math>DOP</math> are congruent, so <math>\angle BOP = \angle BOD/2 = \angle BED</math>, so <math>\angle BMP = \angle BED</math>. Because <math>AC</math> and <math>DE</math> are parallel, <math>M</math> lies on <math>BE</math> (using | + | Since quadrilateral <math>BOMP</math> is cyclic, <math>\angle BMP = \angle BOP</math>. Triangles <math>BOP</math> and <math>DOP</math> are congruent, so <math>\angle BOP = \angle BOD/2 = \angle BED</math>, so <math>\angle BMP = \angle BED</math>. Because <math>AC</math> and <math>DE</math> are parallel, <math>M</math> lies on <math>BE</math> (using Euclid's Parallel Postulate). |
+ | |||
+ | -Evan Chen (vEnhance) | ||
+ | |||
+ | ==Solution 4== | ||
+ | Note that by Lemma 9.9 of EGMO, <math>(A,C;B,D)</math> is a harmonic bundle. We project through <math>E</math> onto <math>\overline{AC}</math>, | ||
+ | <cmath>-1=(A,C;B,D)\stackrel{E}{=}(A,C;M,P_{\infty})</cmath> | ||
+ | Where <math>P_{\infty}</math> is the point at infinity for parallel lines <math>\overline{DE}</math> and <math>\overline{AC}</math>. Thus, we get <math>\frac{MA}{MC}=-1</math>, and <math>M</math> is the midpoint of <math>AC</math>. ~novus677 | ||
+ | |||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 17:00, 14 April 2024
Contents
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
.
The Power of Angle Chasing
https://www.youtube.com/watch?v=Dn1IIx9Cnqw&list=PLqgsN351HEtHgi3Ax2oXJk_bEiROCvNF5
Solution
Since , we have that
. But it is well known that
is a harmonic quadrilateral, thus
is a symmedian of triangle
, from which it follows that
is a median of
.
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 3
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).
-Evan Chen (vEnhance)
Solution 4
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.