Difference between revisions of "2011 USAJMO Problems/Problem 5"
m (→Solution 2) |
m (→Solution 4) |
||
(10 intermediate revisions by 6 users not shown) | |||
Line 3: | Line 3: | ||
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 1 == |
+ | Connect segment PO, and name the interaction of PO and the circle as point M. | ||
− | ==Solution | + | 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 32: | ||
~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>. | ||
Line 63: | Line 79: | ||
</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). |
− | ==Solution | + | ==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>, | 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> | <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 | 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}} |
Revision as of 02:31, 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 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).
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.