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

m (Solution 4)
m (Solution 4)
(4 intermediate revisions by 2 users not shown)
Line 3: Line 3:
 
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 4 ==
+
== Solution 1 ==
 
Connect segment PO, and name the interaction of PO and the circle as point M.  
 
Connect segment PO, and name the interaction of PO and the circle as point M.  
  
Line 20: Line 20:
 
~ MVP Harry
 
~ MVP Harry
  
==Solution 1==
+
==Solution 2==
  
 
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>.
 
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>.
Line 32: Line 32:
 
~pandadude
 
~pandadude
  
==Solution 2==
+
==Solution 3==
 +
This is the solution from EGMO Problem 1.43 page 242
  
 
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>.
 
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>.
Line 80: Line 81:
 
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).
 
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).
  
==Solution 3==
+
==Solution 4==
 
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>,
 
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>,
 
<cmath>-1=(A,C;B,D)\stackrel{E}{=}(A,C;M,P_{\infty})</cmath>
 
<cmath>-1=(A,C;B,D)\stackrel{E}{=}(A,C;M,P_{\infty})</cmath>

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png