2011 USAJMO Problems/Problem 5

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

It's well known that $ABCD$ is a harmonic quadrilateral. Since $\overline{DE} \parallel \overline{AC}$ , we have that $AD=CE\rightarrow \angle ABD=\angle CBE$ . But $ABCD$ is a harmonic quadrilateral, thus $\overline{BD}$ is a symmedian of triangle $ABC$ , from which it follows that $\overline{BE}$ is a median of $\triangle ABC$ . $\blacksquare$

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