Difference between revisions of "2006 USAMO Problems/Problem 6"

Latest revision as of 21:46, 18 May 2015

Problem

(Zuming Feng, Zhonghao Ye) Let $ABCD$ be a quadrilateral, and let $E$ and $F$ be points on sides $AD$ and $BC$ , respectively, such that $AE/ED = BF/FC$ . Ray $FE$ meets rays $BA$ and $CD$ at $S$ and $T$ respectively. Prove that the circumcircles of triangles $SAE$ , $SBF$ , $TCF$ , and $TDE$ pass through a common point.

Solutions

Solution 1

Let the intersection of the circumcircles of $SAE$ and $SBF$ be $X$ , and let the intersection of the circumcircles of $TCF$ and $TDE$ be $Y$ .

$BXF=BSF=AXE$ because $BSF$ tends both arcs $AE$ and $BF$ . $BFX=XSB=XEA$ because $XSB$ tends both arcs $XA$ and $XB$ . Thus, $XAE\sim XBF$ by AA similarity, and $X$ is the center of spiral similarity for $A,E,B,$ and $F$ . $FYC=FTC=EYD$ because $FTC$ tends both arcs $ED$ and $FC$ . $FCY=FTY=EDY$ because $FTY$ tends both arcs $YF$ and $YE$ . Thus, $YED\sim YFC$ by AA similarity, and $Y$ is the center of spiral similarity for $E,D,F,$ and $C$ .

From the similarity, we have that $XE/XF=AE/BF$ . But we are given $ED/AE=CF/BF$ , so multiplying the 2 equations together gets us $ED/FC=XE/XF$ . $DEX,CFX$ are the supplements of $AEX, BFX$ , which are congruent, so $DEX=CFX$ , and so $XED\sim XFC$ by SAS similarity, and so $X$ is also the center of spiral similarity for $E,D,F,$ and $C$ . Thus, $X$ and $Y$ 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 $s$ and $t$ are real numbers and $x$ , $y$ and $z$ are complex. The circle in the complex plane passing through $x$ , $x + ty$ and $x + (s + t)z$ also passes through the point $x + syz/(y - z)$ , independent of $t$ .

Proof. Four points $z_1$ , $z_2$ , $z_3$ and $z_4$ in the complex plane lie on a circle if and only if the cross-ratio $cr(z_1, z_2, z_3, z_4) = \frac{(z_1 - z_3)(z_2 - z_4)}{(z_1 - z_4)(z_2 - z_3)}$ is real. Since we compute $cr(x, x + ty, x + (s + t)z, x + syz/(y - z)) = \frac{s + t}{s}$ the given points are on a circle. $\blacksquare$

Lay down complex coordinates with $S = 0$ and $E$ and $F$ on the positive real axis. Then there are real $r_1$ , $r_2$ and $R$ with $B = r_1A$ , $F = r_2E$ and $D = E + R(A - E)$ and hence $AE/ED = BF/FC$ gives $C = F + R(B - F) = r_2(1 - R)E + r_1RA.$ The line $CD$ consists of all points of the form $sC + (1 - s)D$ for real $s$ . Since $T$ lies on this line and has zero imaginary part, we see from $\text{Im}(sC + (1 - s)D) = (sr_1R + (1 - s)R)\text{Im}(A)$ that it corresponds to $s = -1/(r_1 - 1)$ . Thus $T = \frac{r_1D - C}{r_1 - 1} = \frac{(r_2 - r_1)(R - 1)E}{r_1 - 1}.$ Apply the lemma with $x = E$ , $y = A - E$ , $z = (r_2 - r_1)E/(r_1 - 1)$ , and $s = (r_2 - 1)(r_1 - r_2)$ . Setting $t = 1$ gives $(x, x + y, x + (s + 1)z) = (E, A, S = 0)$ and setting $t = R$ gives $(x, x + Ry, x + (s + R)z) = (E, D, T).$ Therefore the circumcircles to $SAE$ and $TDE$ meet at $x + \frac{syz}{y - z} = \frac{AE(r_1 - r_2)}{(1 - r_1)E - (1 - r_2)A} = \frac{AF - BE}{A + F - B - E}.$ This last expression is invariant under simultaneously interchanging $A$ and $B$ and interchanging $E$ and $F$ . Therefore it is also the intersection of the circumcircles of $SBF$ and $TCF$ .

Solution 3

Let $M$ be the Miquel point of $ABCD$ ; then $M$ is the center of the spiral similarity that takes $AD$ to $BC$ . Because $\frac{AE}{ED} = \frac{BF}{FC}$ , the same spiral similarity also takes $E$ to $F$ , so $M$ is the center of the spiral similarity that maps $AE$ to $BF$ and $ED$ to $FC$ . Then it is obvious that the circumcircles of $SAE$ , $SBF$ , $TCF$ , and $TDE$ pass through $M$ .

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

@@ Line 1: / Line 1: @@
 == Problem ==
+(''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.
-Let <math> \displaystyle ABCD </math> be a quadrilateral, and let <math> \displaystyle  E </math> and <math> \displaystyle F </math> be points on sides <math> \displaystyle AD </math> and <math> \displaystyle BC </math>, respectively, such that <math> \displaystyle AE/ED = BF/FC </math>.  Ray <math> \displaystyle FE </math> meets rays <math> \displaystyle BA </math> and <math> \displaystyle CD </math> at <math> \displaystyle S </math> and <math> \displaystyle T </math> respectively. Prove that the circumcircles of triangles  <math> \displaystyle SAE, SBF, TCF, </math> and <math> \displaystyle TDE </math> pass through a common point.
+== Solutions ==
-== Solution ==
+=== 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>.
-{{solution}}
+<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>.
-== See Also ==
+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.
-* [[2006 USAMO Problems]]
+=== Solution 2 ===
-* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=490691#p490691 Discussion on AoPS/MathLinks]
+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}}

2006 USAMO (Problems • Resources)
Preceded by Problem 5	Followed by Last Question
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

Page

Toolbox

Search