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− | ==Problem==
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− | Let <math>AXYZB</math> be a convex pentagon inscribed in a semicircle of diameter
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− | <math>AB</math>. Denote by <math>P, Q, R, S</math> the feet of the perpendiculars from <math>Y</math> onto
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− | lines <math>AX, BX, AZ, BZ</math>, respectively. Prove that the acute angle
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− | formed by lines <math>PQ</math> and <math>RS</math> is half the size of <math>\angle XOZ</math>, where
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− | <math>O</math> is the midpoint of segment <math>AB</math>.
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− |
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− | ==Solution==
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− | Let <math>\alpha = \angle BAZ</math>, <math>\beta = \angle ABX</math>.
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− | Since <math>XY</math> is a chord of the circle with diameter <math>AB</math>,
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− | <math>\angle XAY = \angle XBY = \gamma</math>. From the chord <math>YZ</math>,
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− | we conclude <math>\angle YAZ = \angle YBZ = \delta</math>.
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− | <center>
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− | <asy>
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− | import olympiad;
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− |
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− | // Scale
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− | unitsize(1inch);
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− | real r = 1.75;
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− |
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− | // Semi-circle: centre O, radius r, diameter A--B.
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− | pair O = (0,0); dot(O); label("$O$", O, plain.S);
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− | pair A = r * plain.W; dot(A); label("$A$", A, unit(A));
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− | pair B = r * plain.E; dot(B); label("$B$", B, unit(B));
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− | draw(arc(O, r, 0, 180)--cycle);
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− |
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− | // points X, Y, Z
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− | real alpha = 22.5;
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− | real beta = 15;
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− | real delta = 30;
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− | pair X = r * dir(180 - 2*beta); dot(X); label("$X$", X, unit(X));
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− | pair Y = r * dir(2*(alpha + delta)); dot(Y); label("$Y$", Y, unit(Y));
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− | pair Z = r * dir(2*alpha); dot(Z); label("$Z$", Z, unit(Z));
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− |
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− | // Feet of perpendiculars from Y
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− | pair P = foot(Y, A, X); dot(P); label("$P$", P, unit(P-Y)); dot(P);
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− | pair Q = foot(Y, B, X); dot(P); label("$Q$", Q, unit(A-Q)); dot(Q);
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− | pair R = foot(Y, B, Z); dot(R); label("$S$", R, unit(R-Y)); dot(R);
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− | pair S = foot(Y, A, Z); dot(S); label("$R$", S, unit(B-S)); dot(S);
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− | pair T = foot(Y, A, B); dot(T); label("$T$", T, unit(T-Y)); dot(T);
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− |
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− | // Segments
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− | draw(B--X); draw(B--Y); draw(B--R);
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− | draw(A--Z); draw(A--Y); draw(A--P);
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− | draw(Y--P); draw(Y--Q); draw(Y--R); draw(Y--S);
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− | draw(R--T); draw(P--T);
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− |
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− | // Right angles
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− | draw(rightanglemark(A, X, B, 3));
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− | draw(rightanglemark(A, Y, B, 3));
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− | draw(rightanglemark(A, Z, B, 3));
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− | draw(rightanglemark(A, P, Y, 3));
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− | draw(rightanglemark(Y, R, B, 3));
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− | draw(rightanglemark(Y, S, A, 3));
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− | draw(rightanglemark(B, Q, Y, 3));
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− |
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− | // Acute angles
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− | import markers;
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− | void langle(pair A, pair B, pair C, string l="", real r=40, int n=1, int nm = 0)
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− | {
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− | string sl = "$\scriptstyle{" + l + "}$";
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− | marker m = (nm > 0) ? marker(markinterval(stickframe(n=nm, 2mm), true)) : nomarker;
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− | markangle(Label(sl), radius=r, n=n, A, B, C, m);
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− | }
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− | langle(B, A, Z, "\alpha" );
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− | langle(X, B, A, "\beta", n=2);
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− | langle(Y, A, X, "\gamma", nm=1);
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− | langle(Y, B, X, "\gamma", nm=1);
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− | langle(Z, A, Y, "\delta", nm=2);
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− | langle(Z, B, Y, "\delta", nm=2);
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− | langle(R, S, Y, "\alpha+\delta", r=23);
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− | langle(Y, Q, P, "\beta+\gamma", r=23);
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− | langle(R, T, P, "\chi", r=15);
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− | </asy>
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− | </center>
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− |
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− | Triangles <math>BQY</math> and <math>APY</math> are both right-triangles, and share the
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− | angle <math>\gamma</math>, therefore they are similar, and so the ratio <math>PY :
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− | YQ = AY : YB</math>. Now by [[Thales' theorem]] the angles <math>\angle AXB =
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− | \angle AYB = \angle AZB</math> are all right-angles. Also, <math>\angle PYQ</math>,
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− | being the fourth angle in a quadrilateral with 3 right-angles is
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− | again a right-angle. Therefore <math>\triangle PYQ \sim \triangle AYB</math> and
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− | <math>\angle YQP = \angle YBA = \gamma + \beta</math>.
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− | Similarly, <math>RY : YS = BY : YA</math>, and so <math>\angle YSR = \angle YAB = \alpha + \delta</math>.
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− |
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− | Now <math>RY</math> is perpendicular to <math>AZ</math> so the direction <math>RY</math> is <math>\alpha</math> counterclockwise from the vertical, and since <math>\angle YRS = \alpha + \delta</math> we see that <math>SR</math> is <math>\delta</math> clockwise from the vertical.
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− |
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− | Similarly, <math>QY</math> is perpendicular to <math>BX</math> so the direction <math>QY</math> is <math>\beta</math> clockwise from the vertical, and since <math>\angle YQP</math> is <math>\gamma + \beta</math> we see that <math>QY</math> is <math>\gamma</math> counterclockwise from the vertical.
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− |
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− | Therefore the lines <math>PQ</math> and <math>RS</math> intersect at an angle <math>\chi = \gamma
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− | + \delta</math>. Now by the central angle theorem <math>2\gamma = \angle XOY</math>
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− | and <math>2\delta = \angle YOZ</math>, and so <math>2(\gamma + \delta) = \angle XOZ</math>,
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− | and we are done.
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− |
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− | ===Footnote===
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− | We can prove a bit more. Namely, the extensions of the segments
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− | <math>RS</math> and <math>PQ</math> meet at a point on the diameter <math>AB</math> that is vertically
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− | below the point <math>Y</math>.
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− |
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− | Since <math>YS = AY \sin(\delta)</math> and is inclined <math>\alpha</math> counterclockwise
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− | from the vertical, the point <math>S</math> is <math>AY \sin(\delta) \sin(\alpha)</math>
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− | horizontally to the right of <math>Y</math>.
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− |
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− | Now <math>AS = AY \cos(\delta)</math>, so <math>S</math> is <math>AS \sin(\alpha) = AY
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− | \cos(\delta)\sin(\alpha)</math> vertically above the diameter <math>AB</math>. Also,
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− | the segment <math>SR</math> is inclined <math>\delta</math> clockwise from the vertical,
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− | so if we extend it down from <math>S</math> towards the diameter <math>AB</math> it will
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− | meet the diameter at a point which is
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− | <math>AY \cos(\delta)\sin(\alpha)\tan(\delta) = AY \sin(\delta)\sin(\alpha)</math>
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− | horizontally to the left of <math>S</math>. This places the intersection point
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− | of <math>RS</math> and <math>AB</math> vertically below <math>Y</math>.
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− |
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− | Similarly, and by symmetry the intersection point of <math>PQ</math> and <math>AB</math>
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− | is directly below <math>Y</math> on <math>AB</math>, so the lines through <math>PQ</math> and <math>RS</math>
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− | meet at a point <math>T</math> on the diameter that is vertically below <math>Y</math>.
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− |
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− | == See Also ==
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| ==Problem== | | ==Problem== |
| Let <math>AXYZB</math> be a convex pentagon inscribed in a semicircle of diameter | | Let <math>AXYZB</math> be a convex pentagon inscribed in a semicircle of diameter |
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| == See Also == | | == See Also == |
| {{USAMO newbox|year=2010|before=First problem|num-a=2}} | | {{USAMO newbox|year=2010|before=First problem|num-a=2}} |
− | {{USAJMO newbox|year=2010|num-b=2|num-a=4}}
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− |
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− | [[Category:Olympiad Number Theory Problems]]
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− |
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| {{USAJMO newbox|year=2010|num-b=2|num-a=4}} | | {{USAJMO newbox|year=2010|num-b=2|num-a=4}} |
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| [[Category:Olympiad Number Theory Problems]] | | [[Category:Olympiad Number Theory Problems]] |