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Difference between revisions of "2013 APMO Problems/Problem 5"

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Now by some simple angle chasing, we get <math>\angle DEB = 90^{\circ} </math>, <math>\angle DEQ = 135^{\circ} </math> and <math>\angle REQ = 45^{\circ} </math>.  
 
Now by some simple angle chasing, we get <math>\angle DEB = 90^{\circ} </math>, <math>\angle DEQ = 135^{\circ} </math> and <math>\angle REQ = 45^{\circ} </math>.  
  
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<math>\implies </math> <math>\angle DER = 90^{\circ} </math>, which proves that <math>B </math>, <math>E </math>, <math>R </math> are collinear, finishing our proof.
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<math>\implies </math> <math>\angle DER = 90^{\circ} </math>, which proves that <math>B </math>, <math>E </math>, <math>R </math> are collinear, finishing our proof.  
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<math>\implies </math>
 
  
 
http://www.artofproblemsolving.com/community/c6h532682p3046946
 
http://www.artofproblemsolving.com/community/c6h532682p3046946

Latest revision as of 08:46, 25 November 2024

Problem

Let $ABCD$ be a quadrilateral inscribed in a circle $\omega$, and let $P$ be a point on the extension of $AC$ such that $PB$ and $PD$ are tangent to $\omega$. The tangent at $C$ intersects $PD$ at $Q$ and the line $AD$ at $R$. Let $E$ be the second point of intersection between $AQ$ and $\omega$. Prove that $B$, $E$, $R$ are collinear.

Solution

Solution 1

Let $R'=AD \cap BE$. Note that the tangents at $B$ and $D$ concur on $AC$ at $P$, so $ABCD$ is harmonic, hence the tangents at $C$ and $A$ concur on $BD$ at $X$, say.

Now apply Pascal's Theorem to hexagon $AAEBDD$ to find that $X=AA \cap BD$, $Q=AE \cap DD$ and $R'=BE \cap AD$ are collinear. Now note that $X$ and $Q$ both lie on the tangent at $C$, hence $R'$ also lies on the tangent at $C$. It follows that $R'=CC \cap AD \cap BE=CC \cap AD=R$. So $R'$ and $R$ are in fact the same point. Since $R'=R$ lies on $BE$ by definition, it follows that $B, E, R$, are indeed collinear, and thus the problem is solved.

Solution 2

We use complex numbers. Let $\omega$ be the unit circle, and let the lowercase letter of a point be its complex coordinate.

Since $P$ lies on the intersection of the tangents to $\omega$ at $B$ and $D$, we have $p = \frac{2bd}{b+d}$. In addition, $P$ lies on chord $AC$, so $p = a+c-ac\bar{p}$. This implies that $\frac{2bd}{b+d} = a+c - \frac{2ac}{b+d}$, or $ab+bc+cd+da=2ac+2bd$.

$Q$ lies on the tangent at $C$, and lies on $PD$, so $q = \frac{2cd}{c+d}$.

$R$ lies on chord $AD$ and on the tangent at $C$. Therefore we have $r = 2c - c^2\bar{r}$ and $r = a+d-ad\bar{r}$. Solving for $r$ yields \[r = a + d - ad \frac{2c-r}{c^2} \implies r = \frac{ac^2+c^2d-2acd}{c^2-ad}\] $A, Q, E$ are collinear, so we have $\frac{a-q}{\bar{a}-\bar{q}} = \frac{a-e}{\bar{a}-\bar{e}}$, or \[\frac{a-\frac{2cd}{c+d}}{\frac{1}{a}-\frac{2}{c+d}} = -ae \implies e = \frac{ac + ad - 2cd}{2a-c-d}\] We must prove that $B, E, R$ are collinear, or that \[\frac{b-e}{\bar{b}-\bar{e}} = \frac{b-r}{\bar{b}-\bar{r}} \implies -be = \frac{b-r}{\frac{1}{b} - \bar{r}}\] or \[\frac{abc+abd-2bcd}{c+d-2a} = \frac{b-\frac{ac^2+c^2d-2acd}{c^2-ad}}{\frac{1}{b}- \frac{\frac{1}{ac^2}+\frac{1}{c^2d}-\frac{2}{acd}}{\frac{1}{c^2}-\frac{1}{ad}}} = \frac{b - \frac{ac^2+c^2d-2acd}{c^2-ad}}{\frac{1}{b} - \frac{d+a-2c}{ad-c^2}} = \frac{b(c^2-ad)-(ac^2+c^2d-2acd)}{\frac{c^2-ad}{b} + (d+a-2c)}\] Cross-multiplying, we have \[(ac+ad-2cd)(c^2-ad+bd+ab-2bc) = (c+d-2a)(bc^2-abd - ac^2 - c^2d + 2acd)\] \[\implies (a-c)(c-d)(ab+bc+cd+da-2ac-2bd) = 0\] which is true.

Solution 3

Set $T = \overline{BD} \cap \overline{CR}$, $K = \overline{AC} \cap \overline{BD}$, $Z = \overline{AB} \cap \overline{CR}$ and $E' = \overline{BR} \cap \omega$, where $E' \neq A$. Note that since $ABCD$ is harmonic, we have $T,K,B,D$ collinear and with \[-1 = (T,K;B,D) \stackrel{A}{=} (T,C; Z,R) \stackrel{B}{=} (D,C; A,E').\] But $DACE$ is harmonic; therefore $E = E'$. $\blacksquare$

Solution 4

Use barycentrics on $CBD$, in that order. It is easy to derive that $P=(-a^2:b^2:c^2)$ and $A=(-a^2:2b^2:2c^2)$. Clearly, line $PD$ has equation $\frac{x}{y}=\frac{-a^2}{b^2}$, and the tangent from $C$ has equation $c^2y+b^2z=0$, so we get that $Q=(-a^2:b^2:c^2)$. Line $AD$ has equation $\frac{x}{y}=\frac{-a^2}{2b^2}$, so we also get that $R=(-a^2:2b^2:-2c^2)$. Line $AQ$ has equation $4b^2c^2x+3a^2c^2y-a^2b^2z=0$, and line $BR$ has equation $\frac{x}{z}=\frac{a^2}{2c^2}$, so we quickly derive that $E$ has coordinates $(3a^2:-2b^2:6c^2)$, and it is easy to verify that this lies on the circumcircle ${\sum_{\text{cyc}}a^2yz=0}$, so we're done.

Solution 5

First off, let's state the constructions:

Let $AE \cap BD \cong K$; let the tangent from $R$ to $(ABCD)$ not containing $C$ be $l$, and let $l \cap (ABCD) \cong F$.

Now we make a couple of slick observations;

Since $ACED$ is harmonic, $AE$ is the $A-$ symmedian in $\triangle ACD$, and since $ABCD$ is harmonic, $DB$ is the $B-$ symmedian in $\triangle ACD$. Then $K$ is the symmedian point of $\triangle ACD$.

Now by the definition of $F$, $ACDF$ is harmonic. Hence $CF$ is a symmedian of $\triangle ACD$ as well! Hence $K$ lies on $CF$.

Now since $R$ is the pole of $\overline{CF}$, $R$ lies on the polar of point $K$ (w.r.t. $(ABCD)$, obviously). But by Brocard's theorem on quadrilateral $ABED$, $BE \cap AD$ lies on the polar of $K$.

Hence $R$ lies on $\overleftrightarrow{BE}$ and we are done.

Solution 6

Note that this problem is purely projective! So we can take a projective transformation fixing $(ABCD)$ and taking $\overline{AC} \cap \overline{BD}$ to the center of $(ABCD)$. This implies $ABCD$ is a rectangle. But it is also harmonic; hence it is a square.

Now by some simple angle chasing, we get $\angle DEB = 90^{\circ}$, $\angle DEQ = 135^{\circ}$ and $\angle REQ = 45^{\circ}$.

$\implies$ $\angle DER = 90^{\circ}$, which proves that $B$, $E$, $R$ are collinear, finishing our proof.

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