Difference between revisions of "2006 USAMO Problems/Problem 6"
Ragnarok23 (talk | contribs) |
(→Solution 2) |
||
(10 intermediate revisions by 5 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | == Solution == | + | (''Zuming Feng, Zhonghao Ye'') Let <math>ABCD</math> be a quadrilateral, and let <math>E</math> and <math>F</math> be points on sides <math>AD</math> and <math>BC</math>, respectively, such that <math>AE/ED = BF/FC</math>. Ray <math>FE</math> meets rays <math>BA</math> and <math>CD</math> at <math>S</math> and <math>T</math> respectively. Prove that the circumcircles of triangles <math>SAE</math>, <math>SBF</math>, <math>TCF</math>, and <math>TDE</math> pass through a common point. |
− | == See | + | |
− | *[[ | + | == Solutions == |
+ | |||
+ | === Solution 1 === | ||
+ | Let the intersection of the circumcircles of <math>SAE</math> and <math>SBF</math> be <math>X</math>, and let the intersection of the circumcircles of <math>TCF</math> and <math>TDE</math> be <math>Y</math>. | ||
+ | |||
+ | <math>BXF=BSF=AXE</math> because <math>BSF</math> tends both arcs <math>AE</math> and <math>BF</math>. | ||
+ | <math>BFX=XSB=XEA</math> because <math>XSB</math> tends both arcs <math>XA</math> and <math>XB</math>. | ||
+ | Thus, <math>XAE\sim XBF</math> by AA similarity, and <math>X</math> is the center of spiral similarity for <math>A,E,B,</math> and <math>F</math>. | ||
+ | <math>FYC=FTC=EYD</math> because <math>FTC</math> tends both arcs <math>ED</math> and <math>FC</math>. | ||
+ | <math>FCY=FTY=EDY</math> because <math>FTY</math> tends both arcs <math>YF</math> and <math>YE</math>. | ||
+ | Thus, <math>YED\sim YFC</math> by AA similarity, and <math>Y</math> is the center of spiral similarity for <math>E,D,F,</math> and <math>C</math>. | ||
+ | |||
+ | From the similarity, we have that <math>XE/XF=AE/BF</math>. But we are given <math>ED/AE=CF/BF</math>, so multiplying the 2 equations together gets us <math>ED/FC=XE/XF</math>. <math>DEX,CFX</math> are the supplements of <math>AEX, BFX</math>, which are congruent, so <math>DEX=CFX</math>, and so <math>XED\sim XFC</math> by SAS similarity, and so <math>X</math> is also the center of spiral similarity for <math>E,D,F,</math> and <math>C</math>. Thus, <math>X</math> and <math>Y</math> are the same point, which all the circumcircles pass through, and so the statement is true. | ||
+ | |||
+ | === Solution 2 === | ||
+ | We will give a solution using complex coordinates. The first step is the following lemma. | ||
+ | |||
+ | '''Lemma.''' Suppose <math>s</math> and <math>t</math> are real numbers and <math>x</math>, <math>y</math> and <math>z</math> are complex. The circle in the complex plane passing through <math>x</math>, <math>x + ty</math> and <math>x + (s + t)z</math> also passes through the point <math>x + syz/(y - z)</math>, independent of <math>t</math>. | ||
+ | |||
+ | ''Proof.'' Four points <math>z_1</math>, <math>z_2</math>, <math>z_3</math> and <math>z_4</math> in the complex plane lie on a circle if and only if the cross-ratio | ||
+ | <cmath>cr(z_1, z_2, z_3, z_4) = \frac{(z_1 - z_3)(z_2 - z_4)}{(z_1 - z_4)(z_2 - z_3)}</cmath> | ||
+ | is real. Since we compute | ||
+ | <cmath>cr(x, x + ty, x + (s + t)z, x + syz/(y - z)) = \frac{s + t}{s}</cmath> | ||
+ | the given points are on a circle. <math>\blacksquare</math> | ||
+ | |||
+ | Lay down complex coordinates with <math>S = 0</math> and <math>E</math> and <math>F</math> on the positive real axis. Then there are real <math>r_1</math>, <math>r_2</math> and <math>R</math> with <math>B = r_1A</math>, <math>F = r_2E</math> and <math>D = E + R(A - E)</math> and hence <math>AE/ED = BF/FC</math> gives | ||
+ | <cmath>C = F + R(B - F) = r_2(1 - R)E + r_1RA.</cmath> | ||
+ | The line <math>CD</math> consists of all points of the form <math>sC + (1 - s)D</math> for real <math>s</math>. Since <math>T</math> lies on this line and has zero imaginary part, we see from <math>\text{Im}(sC + (1 - s)D) = (sr_1R + (1 - s)R)\text{Im}(A)</math> that it corresponds to <math>s = -1/(r_1 - 1)</math>. Thus | ||
+ | <cmath>T = \frac{r_1D - C}{r_1 - 1} = \frac{(r_2 - r_1)(R - 1)E}{r_1 - 1}.</cmath> | ||
+ | Apply the lemma with <math>x = E</math>, <math>y = A - E</math>, <math>z = (r_2 - r_1)E/(r_1 - 1)</math>, and <math>s = (r_2 - 1)(r_1 - r_2)</math>. Setting <math>t = 1</math> gives | ||
+ | <cmath>(x, x + y, x + (s + 1)z) = (E, A, S = 0)</cmath> | ||
+ | and setting <math>t = R</math> gives | ||
+ | <cmath>(x, x + Ry, x + (s + R)z) = (E, D, T).</cmath> | ||
+ | Therefore the circumcircles to <math>SAE</math> and <math>TDE</math> meet at | ||
+ | <cmath>x + \frac{syz}{y - z} = \frac{AE(r_1 - r_2)}{(1 - r_1)E - (1 - r_2)A} = \frac{AF - BE}{A + F - B - E}.</cmath> | ||
+ | This last expression is invariant under simultaneously interchanging <math>A</math> and <math>B</math> and interchanging <math>E</math> and <math>F</math>. Therefore it is also the intersection of the circumcircles of <math>SBF</math> and <math>TCF</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | Let <math>M</math> be the Miquel point of <math>ABCD</math>; then <math>M</math> is the center of the spiral similarity that takes <math>AD</math> to <math>BC</math>. Because <math>\frac{AE}{ED} = \frac{BF}{FC}</math>, the same spiral similarity also takes <math>E</math> to <math>F</math>, so <math>M</math> is the center of the spiral similarity that maps <math>AE</math> to <math>BF</math> and <math>ED</math> to <math>FC</math>. Then it is obvious that the circumcircles of <math>SAE</math>, <math>SBF</math>, <math>TCF</math>, and <math>TDE</math> pass through <math>M</math>. | ||
+ | |||
+ | {{alternate solutions}} | ||
+ | |||
+ | == See also == | ||
+ | * <url>viewtopic.php?t=84559 Discussion on AoPS/MathLinks</url> | ||
+ | |||
+ | {{USAMO newbox|year=2006|num-b=5|after=Last Question}} | ||
+ | |||
+ | [[Category:Olympiad Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:46, 18 May 2015
Problem
(Zuming Feng, Zhonghao Ye) Let be a quadrilateral, and let and be points on sides and , respectively, such that . Ray meets rays and at and respectively. Prove that the circumcircles of triangles , , , and pass through a common point.
Solutions
Solution 1
Let the intersection of the circumcircles of and be , and let the intersection of the circumcircles of and be .
because tends both arcs and . because tends both arcs and . Thus, by AA similarity, and is the center of spiral similarity for and . because tends both arcs and . because tends both arcs and . Thus, by AA similarity, and is the center of spiral similarity for and .
From the similarity, we have that . But we are given , so multiplying the 2 equations together gets us . are the supplements of , which are congruent, so , and so by SAS similarity, and so is also the center of spiral similarity for and . Thus, and are the same point, which all the circumcircles pass through, and so the statement is true.
Solution 2
We will give a solution using complex coordinates. The first step is the following lemma.
Lemma. Suppose and are real numbers and , and are complex. The circle in the complex plane passing through , and also passes through the point , independent of .
Proof. Four points , , and in the complex plane lie on a circle if and only if the cross-ratio is real. Since we compute the given points are on a circle.
Lay down complex coordinates with and and on the positive real axis. Then there are real , and with , and and hence gives The line consists of all points of the form for real . Since lies on this line and has zero imaginary part, we see from that it corresponds to . Thus Apply the lemma with , , , and . Setting gives and setting gives Therefore the circumcircles to and meet at This last expression is invariant under simultaneously interchanging and and interchanging and . Therefore it is also the intersection of the circumcircles of and .
Solution 3
Let be the Miquel point of ; then is the center of the spiral similarity that takes to . Because , the same spiral similarity also takes to , so is the center of the spiral similarity that maps to and to . Then it is obvious that the circumcircles of , , , and pass through .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See also
- <url>viewtopic.php?t=84559 Discussion on AoPS/MathLinks</url>
2006 USAMO (Problems • Resources) | ||
Preceded by Problem 5 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.