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 38: / Line 38: @@
 <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}}

2006 USAMO (Problems • Resources)
Preceded by Problem 5	Followed by Last Question
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