Difference between revisions of "2011 USAJMO Problems/Problem 5"

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Points <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, <math>E</math> lie on a circle <math>\omega</math> and point <math>P</math> lies outside the circle.  The given points are such that (i) lines <math>PB</math> and <math>PD</math> are tangent to <math>\omega</math>, (ii) <math>P</math>, <math>A</math>, <math>C</math> are collinear, and (iii) <math>\overline{DE} \parallel \overline{AC}</math>.  Prove that <math>\overline{BE}</math> bisects <math>\overline{AC}</math>.
 
Points <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math>, <math>E</math> lie on a circle <math>\omega</math> and point <math>P</math> lies outside the circle.  The given points are such that (i) lines <math>PB</math> and <math>PD</math> are tangent to <math>\omega</math>, (ii) <math>P</math>, <math>A</math>, <math>C</math> are collinear, and (iii) <math>\overline{DE} \parallel \overline{AC}</math>.  Prove that <math>\overline{BE}</math> bisects <math>\overline{AC}</math>.
  
== Solutions ==
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== Solution 4 ==
 
Connet segment PO, and name the interaction of PO and the circle as point M.  
 
Connet segment PO, and name the interaction of PO and the circle as point M.  
 
Since PB and PD are tangent to the circle, it's easy to see that M is the midpoint of arc BD.
 
Since PB and PD are tangent to the circle, it's easy to see that M is the midpoint of arc BD.
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Therefore, PBOM is cyclic, ∠ PFO = ∠ OBP = 90°, AF = AC (F is the interaction of BE and AC)
 
Therefore, PBOM is cyclic, ∠ PFO = ∠ OBP = 90°, AF = AC (F is the interaction of BE and AC)
 
BE bisects AC, proof completed!
 
BE bisects AC, proof completed!
 +
~ MVP Harry
  
 
==Solution 1==
 
==Solution 1==

Revision as of 07:39, 27 April 2019

Problem

Points $A$, $B$, $C$, $D$, $E$ lie on a circle $\omega$ and point $P$ lies outside the circle. The given points are such that (i) lines $PB$ and $PD$ are tangent to $\omega$, (ii) $P$, $A$, $C$ are collinear, and (iii) $\overline{DE} \parallel \overline{AC}$. Prove that $\overline{BE}$ bisects $\overline{AC}$.

Solution 4

Connet segment PO, and name the interaction of PO and the circle as point M. Since PB and PD are tangent to the circle, it's easy to see that M is the midpoint of arc BD. ∠ BOA = 1/2 arc AB + 1/2 arc CE Since AC // DE, arc AD = arc CE, thus, ∠ BOA = 1/2 arc AB + 1/2 arc AD = 1/2 arc BD = arc BM = ∠ BOM Therefore, PBOM is cyclic, ∠ PFO = ∠ OBP = 90°, AF = AC (F is the interaction of BE and AC) BE bisects AC, proof completed! ~ MVP Harry

Solution 1

Let $O$ be the center of the circle, and let $X$ be the intersection of $AC$ and $BE$. Let $\angle OPA$ be $x$ and $\angle OPD$ be $y$.

$\angle OPB = \angle OPD = y$, $\angle BED = \frac{\angle DOB}{2} = 90-y$, $\angle ODE = \angle PDE - 90 = 90-x-y$ $\angle OBE = \angle PBE - 90 = x = \angle OPA$

Thus $PBXO$ is a cyclic quadrilateral and $\angle OXP = \angle OBP = 90$ and so $X$ is the midpoint of chord $AC$.

~pandadude

Solution 2

Let $O$ be the center of the circle, and let $M$ be the midpoint of $AC$. Let $\theta$ denote the circle with diameter $OP$. Since $\angle OBP = \angle OMP = \angle ODP = 90^\circ$, $B$, $D$, and $M$ all lie on $\theta$.

[asy] import graph;  unitsize(2 cm);  pair A, B, C, D, E, M, O, P; path circ;  O = (0,0); circ = Circle(O,1); B = dir(100); D = dir(240); P = extension(B, B + rotate(90)*(B), D, D + rotate(90)*(D)); C = dir(-40); A = intersectionpoint((P--(P + 0.9*(C - P))),circ); E = intersectionpoint((D + 0.1*(C - A))--(D + C - A),circ); M = (A + C)/2;  draw(circ); draw(P--B); draw(P--D); draw(P--C); draw(B--E); draw(D--E); draw(O--B); draw(O--D); draw(O--M); draw(O--P); draw(Circle((O + P)/2, abs(O - P)/2),dashed); draw(D--M);  dot("$A$", A, NE); dot("$B$", B, NE); dot("$C$", C, SE); dot("$D$", D, S); dot("$E$", E, S); dot("$M$", M, NE); dot("$O$", O, dir(0)); dot("$P$", P, W); label("$\theta$", (O + P)/2 + abs(O - P)/2*dir(120), NW); [/asy]

Since quadrilateral $BOMP$ is cyclic, $\angle BMP = \angle BOP$. Triangles $BOP$ and $DOP$ are congruent, so $\angle BOP = \angle BOD/2 = \angle BED$, so $\angle BMP = \angle BED$. Because $AC$ and $DE$ are parallel, $M$ lies on $BE$ (using Euclid's Parallel Postulate).

Solution 3

Note that by Lemma 9.9 of EGMO, $(A,C;B,D)$ is a harmonic bundle. We project through $E$ onto $\overline{AC}$, \[-1=(A,C;B,D)\stackrel{E}{=}(A,C;M,P_{\infty})\] Where $P_{\infty}$ is the point at infinity for parallel lines $\overline{DE}$ and $\overline{AC}$. Thus, we get $\frac{MA}{MC}=-1$, and $M$ is the midpoint of $AC$. ~novus677

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