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>.
  
== Solution ==
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== Solution 1 ==
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Connect segment PO, and name the interaction of PO and the circle as point M.
  
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>.
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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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∠ BOA = 1/2 arc AB + 1/2 arc CE
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Since AC // DE, arc AD = arc CE,
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thus, ∠ BOA = 1/2 arc AB + 1/2 arc AD = 1/2 arc BD = arc BM = ∠ BOM
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Therefore, PBOM is cyclic, ∠ PFO = ∠ OBP = 90°, AF = AC (F is the interaction of BE and AC)
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BE bisects AC, proof completed!
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~ MVP Harry
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==Solution 2==
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Let <math>O</math> be the center of the circle, and let <math>X</math> be the intersection of <math>AC</math> and <math>BE</math>. Let <math>\angle OPA</math> be <math>x</math> and <math>\angle OPD</math> be <math>y</math>.
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<math> \angle OPB = \angle OPD = y </math>, <math> \angle BED = \frac{\angle DOB}{2} = 90-y </math>,
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<math> \angle ODE = \angle PDE - 90 = 90-x-y </math>
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<math> \angle OBE = \angle PBE - 90 = x = \angle OPA </math>
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Thus <math>PBXO</math> is a cyclic quadrilateral and <math>\angle OXP = \angle OBP = 90</math> and so <math>X</math> is the midpoint of chord <math>AC</math>.
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~pandadude
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==Solution 3==
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This is the solution from EGMO Problem 1.43 page 242
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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>\theta</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>\theta</math>.
  
 
<asy>
 
<asy>
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dot("$O$", O, dir(0));
 
dot("$O$", O, dir(0));
 
dot("$P$", P, W);
 
dot("$P$", P, W);
label("$\omega$", (O + P)/2 + abs(O - P)/2*dir(120), NW);
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label("$\theta$", (O + P)/2 + abs(O - P)/2*dir(120), NW);
 
</asy>
 
</asy>
  
Since quadrilateral <math>BOMP</math> is cyclic, <math>\angle BMP = \angle BOP</math>.  Triangles <math>BOP</math> and <math>DOP</math> are congruent, so <math>\angle BOP = \angle BOD/2 = \angle BED</math>, so <math>\angle BMP = \angle BED</math>. Since <math>AC</math> and <math>DE</math> are parallel, <math>M</math> lies on <math>BE</math>.
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Since quadrilateral <math>BOMP</math> is cyclic, <math>\angle BMP = \angle BOP</math>.  Triangles <math>BOP</math> and <math>DOP</math> are congruent, so <math>\angle BOP = \angle BOD/2 = \angle BED</math>, so <math>\angle BMP = \angle BED</math>. Because <math>AC</math> and <math>DE</math> are parallel, <math>M</math> lies on <math>BE</math> (using Euclid's Parallel Postulate).
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==Solution 4==
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Note that by Lemma 9.9 of EGMO, <math>(A,C;B,D)</math> is a harmonic bundle. We project through <math>E</math> onto <math>\overline{AC}</math>,
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<cmath>-1=(A,C;B,D)\stackrel{E}{=}(A,C;M,P_{\infty})</cmath>
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Where <math>P_{\infty}</math> is the point at infinity for parallel lines <math>\overline{DE}</math> and <math>\overline{AC}</math>. Thus, we get <math>\frac{MA}{MC}=-1</math>, and <math>M</math> is the midpoint of <math>AC</math>. ~novus677
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{{MAA Notice}}

Revision as of 02:31, 24 July 2020

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 1

Connect 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 2

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 3

This is the solution from EGMO Problem 1.43 page 242

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 4

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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