Difference between revisions of "2014 USAMO Problems/Problem 5"
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Let <math>ABC</math> be a triangle with orthocenter <math>H</math> and let <math>P</math> be the second intersection of the circumcircle of triangle <math>AHC</math> with the internal bisector of the angle <math>\angle BAC</math>. Let <math>X</math> be the circumcenter of triangle <math>APB</math> and <math>Y</math> the orthocenter of triangle <math>APC</math>. Prove that the length of segment <math>XY</math> is equal to the circumradius of triangle <math>ABC</math>. | Let <math>ABC</math> be a triangle with orthocenter <math>H</math> and let <math>P</math> be the second intersection of the circumcircle of triangle <math>AHC</math> with the internal bisector of the angle <math>\angle BAC</math>. Let <math>X</math> be the circumcenter of triangle <math>APB</math> and <math>Y</math> the orthocenter of triangle <math>APC</math>. Prove that the length of segment <math>XY</math> is equal to the circumradius of triangle <math>ABC</math>. | ||
− | ==Solution== | + | ==Solution 1== |
Let <math>O_1</math> be the center of <math>(AHPC)</math>, <math>O</math> be the center of <math>(ABC)</math>. Note that <math>(O_1)</math> is the reflection of <math>(O)</math> across <math>AC</math>, so <math>AO=AO_1</math>. Additionally | Let <math>O_1</math> be the center of <math>(AHPC)</math>, <math>O</math> be the center of <math>(ABC)</math>. Note that <math>(O_1)</math> is the reflection of <math>(O)</math> across <math>AC</math>, so <math>AO=AO_1</math>. Additionally | ||
<cmath> | <cmath> | ||
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</cmath> | </cmath> | ||
But <math>YO=OA</math> as well, and <math>\angle YOX=\angle AOO_1</math>, so <math>\triangle OYX\cong \triangle OAO_1</math>. Thus <math>XY=AO_1=AO</math>. | But <math>YO=OA</math> as well, and <math>\angle YOX=\angle AOO_1</math>, so <math>\triangle OYX\cong \triangle OAO_1</math>. Thus <math>XY=AO_1=AO</math>. | ||
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+ | ==Solution 2== | ||
+ | Since <math>AHPC</math> is a cyclic quadrilateral, <math>\angle AHC = \angle APC</math>. <math>\angle AHC = 90^\circ + \angle ABC</math> and <math>\angle APC = 90^\circ + \angle AYC</math>, we find <math>\angle ABC = \angle AYC</math>. That is, <math>ABYC</math> is a cyclic quadrilateral. Let <math>D</math> be mid-point of <math>\overline{AB}</math>. <math>O, X, D</math> are collinear and <math>OX \perp AB</math>. Let <math>M</math> be second intersection of <math>AP</math> with circumcircle of the triangle <math>ABC</math>. Let <math>YP \cap AC = E</math>, <math>YM \cap AB = F</math>. Since <math>M</math> is mid-point of the arc <math>BC</math>, <math>OM\perp BC</math>. Since <math>AYMC</math> is a cyclic quadrilateral, <math>\angle CYM = \angle CAM = \angle BAC /2</math>. Since <math>Y</math> is the orthocenter of triangle <math>APC</math>, <math>\angle PYC = \angle CAP = \angle BAC /2</math>. Thus, <math>\angle PYM = \angle BAC</math> and <math>AEYF</math> is a cyclic quadrilateral. So, <math>YF \perp AB</math> and <math>OX \parallel MY</math>. We will prove that <math>XYMO</math> is a parallelogram. | ||
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+ | https://wiki-images.artofproblemsolving.com//7/7b/Usamo2014-5.png (figure link) | ||
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+ | We see that <math>YPM</math> is an isosceles triangle and <math>YM=YP</math>. Also <math>XB=XP</math> and <math>\angle BXP = 2\angle BAP = \angle BAC = \angle PYM</math>. Then, <math> BXP \sim MYP </math>. By spiral similarity, <math> BPM \sim XPY </math> and <math>\angle XYP = \angle BMP = \angle BCA</math>. Hence, <math>\angle XYP = \angle BCA</math>, <math>XY \perp BC</math>. Since <math>OM \perp BC</math>, we get <math>XYMO</math> is a parallelogram. As a result, <math>OM = XY</math>. | ||
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+ | (Lokman GÖKÇE) | ||
+ | |||
+ | ==See also== | ||
+ | {{USAMO newbox|year=2014|num-b=4|num-a=6}} |
Latest revision as of 11:08, 27 March 2022
Contents
Problem
Let be a triangle with orthocenter and let be the second intersection of the circumcircle of triangle with the internal bisector of the angle . Let be the circumcenter of triangle and the orthocenter of triangle . Prove that the length of segment is equal to the circumradius of triangle .
Solution 1
Let be the center of , be the center of . Note that is the reflection of across , so . Additionally so lies on . Now since are perpendicular to and their bisector, is isosceles with , and . Also But as well, and , so . Thus .
Solution 2
Since is a cyclic quadrilateral, . and , we find . That is, is a cyclic quadrilateral. Let be mid-point of . are collinear and . Let be second intersection of with circumcircle of the triangle . Let , . Since is mid-point of the arc , . Since is a cyclic quadrilateral, . Since is the orthocenter of triangle , . Thus, and is a cyclic quadrilateral. So, and . We will prove that is a parallelogram.
https://wiki-images.artofproblemsolving.com//7/7b/Usamo2014-5.png (figure link)
We see that is an isosceles triangle and . Also and . Then, . By spiral similarity, and . Hence, , . Since , we get is a parallelogram. As a result, .
(Lokman GÖKÇE)
See also
2014 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |