Difference between revisions of "2013 IMO Problems/Problem 4"
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''Lemma 1:'' <math>T</math> is on <math>XY</math>. | ''Lemma 1:'' <math>T</math> is on <math>XY</math>. | ||
− | Proof: We have <math> | + | Proof: We have <math>\angle{XTW} = \angle{YTW} = 90^\circ</math> because they intercept semicircles. Hence, <math>\angle{XTY} = \angle{XTW} + \angle{YTW} = 180^\circ</math>, so <math>XTY</math> is a straight line. |
''Lemma 2:'' <math>T</math> is on <math>AW</math>. | ''Lemma 2:'' <math>T</math> is on <math>AW</math>. | ||
− | Proof: Let the circumcircles of <math>NBW</math> and <math>MWC</math> be <math>\omega_1</math> and <math>\omega_2</math>, respectively, and, as <math>BNMC</math> is cyclic (from congruent <math> | + | Proof: Let the circumcircles of <math>NBW</math> and <math>MWC</math> be <math>\omega_1</math> and <math>\omega_2</math>, respectively, and, as <math>BNMC</math> is cyclic (from congruent <math>\angle{BNC} = \angle{BMC} = 90^\circ</math>), let its circumcircle be <math>\omega_3</math>. Then each pair of circles' radical axises, <math>BN, TW,</math> and <math>MC</math>, must concur at the intersection of <math>BN</math> and <math>MC</math>, which is <math>A</math>. |
''Lemma 3:'' <math>YT</math> is perpendicular to <math>AW</math>. | ''Lemma 3:'' <math>YT</math> is perpendicular to <math>AW</math>. | ||
− | Proof: This is immediate from <math> | + | Proof: This is immediate from <math>\angle{YTW} = 90^\circ</math>. |
− | Let <math>AH</math> meet <math>BC</math> at <math>L</math>, which is also the foot of the altitude to that side. Hence, <math> | + | Let <math>AH</math> meet <math>BC</math> at <math>L</math>, which is also the foot of the altitude to that side. Hence, <math>\angle{ALB} = 90^\circ.</math> |
''Lemma 4:'' Quadrilateral <math>THLW</math> is cyclic. | ''Lemma 4:'' Quadrilateral <math>THLW</math> is cyclic. | ||
− | Proof: We know that <math>NHLB</math> is cyclic because <math> | + | Proof: We know that <math>NHLB</math> is cyclic because <math>\angle{BNH}</math> and <math>\angle{BLH}</math>, opposite and right angles, sum to <math>180^\circ</math>. Furthermore, we are given that <math>NTWB</math> is cyclic. Hence, by Power of a Point, |
<cmath>AT * AW = AN * AB = AH * AL.</cmath> | <cmath>AT * AW = AN * AB = AH * AL.</cmath> | ||
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The converse of Power of a Point then proves <math>THLW</math> cyclic. | The converse of Power of a Point then proves <math>THLW</math> cyclic. | ||
− | Hence, <math> | + | Hence, <math>\angle{WTH} = 180^\circ - \angle{WLH} = 90^\circ</math>, and so <math>HT</math> is perpendicular to <math>AW</math> as well. Combining this with Lemma 3's statement, we deduce that <math>T, H, Y</math> are collinear. But, as <math>X</math> is on <math>YT</math> (from Lemma 1), <math>X, Y, H</math> are collinear. This completes the proof. |
<math>\blacksquare</math> | <math>\blacksquare</math> | ||
--[[User:Suli|Suli]] 13:51, 25 August 2014 (EDT) | --[[User:Suli|Suli]] 13:51, 25 August 2014 (EDT) |
Revision as of 20:49, 28 February 2016
Problem
Let be an acute triangle with orthocenter , and let be a point on the side , lying strictly between and . The points and are the feet of the altitudes from and , respectively. Denote by is [sic] the circumcircle of , and let be the point on such that is a diameter of . Analogoously, denote by the circumcircle of triangle , and let be the point such that is a diameter of . Prove that and are collinear.
Hint
Draw a good diagram, or use the one below. What do you notice? (In particular, what do you want to be true? How do you prove it true?)
Solution
Let be the intersection of and other than .
Lemma 1: is on .
Proof: We have because they intercept semicircles. Hence, , so is a straight line.
Lemma 2: is on .
Proof: Let the circumcircles of and be and , respectively, and, as is cyclic (from congruent ), let its circumcircle be . Then each pair of circles' radical axises, and , must concur at the intersection of and , which is .
Lemma 3: is perpendicular to .
Proof: This is immediate from .
Let meet at , which is also the foot of the altitude to that side. Hence,
Lemma 4: Quadrilateral is cyclic.
Proof: We know that is cyclic because and , opposite and right angles, sum to . Furthermore, we are given that is cyclic. Hence, by Power of a Point,
The converse of Power of a Point then proves cyclic.
Hence, , and so is perpendicular to as well. Combining this with Lemma 3's statement, we deduce that are collinear. But, as is on (from Lemma 1), are collinear. This completes the proof.
--Suli 13:51, 25 August 2014 (EDT)