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

(Created page with 'Let <math>O</math> be the center of the circle, and let <math>M</math> be the midpoint of <math>AC</math>. Let <math>\omega</math> denote the circle with diameter <math>OP</math…')
 
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== Problem ==
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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>.
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== Solution ==
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Let <math>O</math> be the center of the circle, and let <math>M</math> be the midpoint of <math>AC</math>.  Let <math>\omega</math> denote the circle with diameter <math>OP</math>.  Since <math>\angle OBP = \angle OMP = \angle ODP = 90^\circ</math>, <math>B</math>, <math>D</math>, and <math>M</math> all lie on <math>\omega</math>.
 
Let <math>O</math> be the center of the circle, and let <math>M</math> be the midpoint of <math>AC</math>.  Let <math>\omega</math> denote the circle with diameter <math>OP</math>.  Since <math>\angle OBP = \angle OMP = \angle ODP = 90^\circ</math>, <math>B</math>, <math>D</math>, and <math>M</math> all lie on <math>\omega</math>.
  

Revision as of 18:31, 28 April 2011

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

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

[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("$\omega$", (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 $\angle BMP = \angle BED$. Since $AC$ and $DE$ are parallel, $M$ lies on $BE$.

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