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Difference between revisions of "2010 USAMO Problems/Problem 1"

(→‎Solution)
(Cleaner asy code...)
Line 14: Line 14:
 
<asy>
 
<asy>
 
import olympiad;
 
import olympiad;
−
import markers;
 
−
 
−
void langle(picture p=currentpicture,
 
−
    pair A, pair B, pair C, string l="", real r=40,
 
−
    int n=1, int marks = 0)
 
−
{
 
−
marker m;
 
−
string sl = "$\scriptstyle{" + l + "}$";
 
−
if (marks == 0) {
 
−
m = nomarker;
 
−
} else {
 
−
m = marker(markinterval(stickframe(n=marks, 2mm),true));
 
−
}
 
−
markangle(p, Label(sl), radius=r, n=n, A, B, C, m);
 
−
}
 
  
−
picture p;
+
// Scale
 
unitsize(1inch);
 
unitsize(1inch);
 
real r = 1.75;
 
real r = 1.75;
−
pair A = r * plain.W; dot(p, A); label(p, "$A$", A, plain.W);
 
−
pair B = r * plain.E; dot(p, B); label(p, "$B$", B, plain.E);
 
−
pair O = (0,0); dot(p, O); label(p, "$O$", O, plain.S);
 
  
−
// Semi-circle with diameter A--B
+
// Semi-circle: centre O, radius r, diameter A--B.
−
path C=arc(O, r, 0, 180)--cycle; draw(p, C);
+
pair O = (0,0); dot(O); label("$O$", O, plain.S);
 +
pair A = r * plain.W; dot(A); label("$A$", A, unit(A));
 +
pair B = r * plain.E; dot(B); label("$B$", B, unit(B));
 +
draw(arc(O, r, 0, 180)--cycle);
  
−
real alpha = 22.5; // angle BAZ
+
// points X, Y, Z
−
real beta = 15; // angble ABX
+
real alpha = 22.5;
−
real delta = 30; // angle ZAY
+
real beta = 15;
−
real gamma = 90 - alpha - beta - delta; // angle XBY
+
real delta = 30;
 +
pair X = r * dir(180 - 2*beta);      dot(X); label("$X$", X, unit(X));
 +
pair Y = r * dir(2*(alpha + delta)); dot(Y); label("$Y$", Y, unit(Y));
 +
pair Z = r * dir(2*alpha);          dot(Z); label("$Z$", Z, unit(Z));
  
−
// Points X, Y, Z
+
// Feet of perpendiculars from Y
−
pair X = r * dir(180-2*beta); dot(p, X); label(p, "$X$", X, plain.WNW);
+
pair P = foot(Y, A, X); dot(P); label("$P$", P, unit(P-Y)); dot(P);
−
pair Y = r * dir(2*(alpha + delta)); dot(p, Y); label(p, "$Y$", Y, plain.NNW);
+
pair Q = foot(Y, B, X); dot(P); label("$Q$", Q, unit(A-Q)); dot(Q);
−
pair Z = r * dir(2*alpha); dot(p, Z); label(p, "$Z$", Z, plain.NE);
+
pair R = foot(Y, B, Z); dot(R); label("$R$", R, unit(R-Y)); dot(R);
 +
pair S = foot(Y, A, Z); dot(S); label("$S$", S, unit(B-S)); dot(S);
 +
pair T = foot(Y, A, B); dot(T); label("$T$", T, unit(T-Y)); dot(T);
  
−
// Angle labels
+
// Segments
−
langle(p, B, A, Z, "\alpha" );
+
draw(B--X); draw(B--Y); draw(B--R);
−
langle(p, X, B, A, "\beta", n=2);
+
draw(A--Z); draw(A--Y); draw(A--P);
−
langle(p, Y, A, X, "\gamma", marks=1);
+
draw(Y--P); draw(Y--Q); draw(Y--R); draw(Y--S);
−
langle(p, Y, B, X, "\gamma", marks=1);
+
draw(R--T); draw(P--T);
−
langle(p, Z, A, Y, "\delta", marks=2);
 
−
langle(p, Z, B, Y, "\delta", marks=2);
 
  
−
// Perpendiculars from Y
+
// Right angles
−
pair Q = foot(Y, B, X);
+
draw(rightanglemark(A, X, B, 3));
−
dot(p, Q); label(p, "$Q$", Q, plain.SW);
+
draw(rightanglemark(A, Y, B, 3));
−
draw(p, Y--Q);
+
draw(rightanglemark(A, Z, B, 3));
−
draw(p, rightanglemark(Y, Q, X, 3));
+
draw(rightanglemark(A, P, Y, 3));
 +
draw(rightanglemark(Y, R, B, 3));
 +
draw(rightanglemark(Y, S, A, 3));
 +
draw(rightanglemark(B, Q, Y, 3));
  
−
pair P = foot(Y, A, X);
+
// Acute angles
−
dot(p, P); label(p, "$P$", P, plain.NW);
+
import markers;
−
draw(p, A--P);
+
void langle(pair A, pair B, pair C, string l="", real r=40, int n=1, int nm = 0)
−
draw(p, Y--P);
+
{
−
draw(p, rightanglemark(Y, P, A, 3));
+
  string sl = "$\scriptstyle{" + l + "}$";
−
 
+
  marker m = (nm > 0) ? marker(markinterval(stickframe(n=nm, 2mm), true)) : nomarker;
−
pair S = foot(Y, A, Z);
+
  markangle(Label(sl), radius=r, n=n, A, B, C, m);
−
dot(p, S); label(p, "$S$", S, plain.SE);
+
}
−
draw(p, Y--S);
+
langle(B, A, Z, "\alpha" );
−
draw(p, rightanglemark(Z, S, Y, 3));
+
langle(X, B, A, "\beta", n=2);
−
 
+
langle(Y, A, X, "\gamma", nm=1);
−
pair R = foot(Y, B, Z);
+
langle(Y, B, X, "\gamma", nm=1);
−
dot(p, R); label(p, "$R$", R, plain.NE);
+
langle(Z, A, Y, "\delta", nm=2);
−
draw(p, B--R);
+
langle(Z, B, Y, "\delta", nm=2);
−
draw(p, Y--R);
+
langle(R, S, Y, "\alpha+\delta", r=23);
−
draw(p, rightanglemark(B, R, Y, 3));
+
langle(Y, Q, P, "\beta+\gamma", r=23);
−
 
+
langle(R, T, P, "\chi", r=15);
−
// Angle labels
 
−
langle(p, R, S, Y, "\alpha+\delta", r=23);
 
−
langle(p, Y, Q, P, "\beta+\gamma", r=23);
 
−
 
 
−
// Right triangles AB{X,Y,Z}
 
−
draw(p, A--Y); draw(p, A--Z);
 
−
draw(p, B--X); draw(p, B--Y);
 
−
draw(p, rightanglemark(B, Y, A, 3));
 
−
draw(p, rightanglemark(B, Z, A, 3));
 
−
draw(p, rightanglemark(B, X, A, 3));
 
−
 
 
−
// Y projection on AB
 
−
pair T = foot(Y, A, B);
 
−
dot(p, T); label(p, "$T$", T, plain.S);
 
−
 
 
−
// Label for sought angle "chi"
 
−
langle(p, R, T, P, "\chi", r=15);
 
−
 
 
−
// Ligher lines for PT, RT
 
−
draw(p, P--T, linewidth(0.2)); draw(p, R--T, linewidth(0.2));
 
−
 
 
−
add(p);
 
 
</asy>
 
</asy>
 
</center>
 
</center>

Revision as of 14:22, 9 May 2010

Problem

Let $AXYZB$ be a convex pentagon inscribed in a semicircle of diameter $AB$. Denote by $P, Q, R, S$ the feet of the perpendiculars from $Y$ onto lines $AX, BX, AZ, BZ$, respectively. Prove that the acute angle formed by lines $PQ$ and $RS$ is half the size of $\angle XOZ$, where $O$ is the midpoint of segment $AB$.

Solution

Let $\alpha = \angle BAZ$, $\beta = \angle ABX$. Since $XY$ is a chord of the circle with diameter $AB$, $\angle XAY = \angle XBY = \gamma$. From the chord $YZ$, we conclude $\angle YAZ = \angle YBZ = \delta$.

[asy] import olympiad; // Scale unitsize(1inch); real r = 1.75; // Semi-circle: centre O, radius r, diameter A--B. pair O = (0,0); dot(O); label("$O$", O, plain.S); pair A = r * plain.W; dot(A); label("$A$", A, unit(A)); pair B = r * plain.E; dot(B); label("$B$", B, unit(B)); draw(arc(O, r, 0, 180)--cycle); // points X, Y, Z real alpha = 22.5; real beta = 15; real delta = 30; pair X = r * dir(180 - 2*beta); dot(X); label("$X$", X, unit(X)); pair Y = r * dir(2*(alpha + delta)); dot(Y); label("$Y$", Y, unit(Y)); pair Z = r * dir(2*alpha); dot(Z); label("$Z$", Z, unit(Z)); // Feet of perpendiculars from Y pair P = foot(Y, A, X); dot(P); label("$P$", P, unit(P-Y)); dot(P); pair Q = foot(Y, B, X); dot(P); label("$Q$", Q, unit(A-Q)); dot(Q); pair R = foot(Y, B, Z); dot(R); label("$R$", R, unit(R-Y)); dot(R); pair S = foot(Y, A, Z); dot(S); label("$S$", S, unit(B-S)); dot(S); pair T = foot(Y, A, B); dot(T); label("$T$", T, unit(T-Y)); dot(T); // Segments draw(B--X); draw(B--Y); draw(B--R); draw(A--Z); draw(A--Y); draw(A--P); draw(Y--P); draw(Y--Q); draw(Y--R); draw(Y--S); draw(R--T); draw(P--T); // Right angles draw(rightanglemark(A, X, B, 3)); draw(rightanglemark(A, Y, B, 3)); draw(rightanglemark(A, Z, B, 3)); draw(rightanglemark(A, P, Y, 3)); draw(rightanglemark(Y, R, B, 3)); draw(rightanglemark(Y, S, A, 3)); draw(rightanglemark(B, Q, Y, 3)); // Acute angles import markers; void langle(pair A, pair B, pair C, string l="", real r=40, int n=1, int nm = 0) { string sl = "$\scriptstyle{" + l + "}$"; marker m = (nm > 0) ? marker(markinterval(stickframe(n=nm, 2mm), true)) : nomarker; markangle(Label(sl), radius=r, n=n, A, B, C, m); } langle(B, A, Z, "\alpha" ); langle(X, B, A, "\beta", n=2); langle(Y, A, X, "\gamma", nm=1); langle(Y, B, X, "\gamma", nm=1); langle(Z, A, Y, "\delta", nm=2); langle(Z, B, Y, "\delta", nm=2); langle(R, S, Y, "\alpha+\delta", r=23); langle(Y, Q, P, "\beta+\gamma", r=23); langle(R, T, P, "\chi", r=15); [/asy]

Triangles $BQY$ and $APY$ are both right-triangles, and share the angle $\gamma$, therefore they are similar, and so the ratio $PY : YQ = AY : YB$. Now by Thales' theorem the angles $\angle AXB = \angle AYB = \angle AZB$ are all right-angles. Also, $\angle PYQ$, being the fourth angle in a quadrilateral with 3 right-angles is again a right-angle. Therefore $\triangle PYQ \sim \triangle AYB$ and $\angle YQP = \angle YBA = \gamma + \beta$. Similarly, $RY : YS = BY : YA$, and so $\angle YSR = \angle YAB = \alpha + \delta$.

Now $SY$ is perpendicular to $AZ$ so the direction $SY$ is $\alpha$ anti-clockwise from the vertical, and since $\angle YSR = \alpha + \delta$ we see that $SR$ is $\delta$ clockwise from the vertical.

Similarly, $QY$ is perpendicular to $BX$ so the direction $QY$ is $\beta$ clockwise from the vertical, and since $\angle YQP$ is $\gamma + \beta$ we see that $QY$ is $\gamma$ anti-clockwise from the vertical.

Therefore the lines $PQ$ and $RS$ intersect at an angle $\chi = \gamma + \delta$. Now by the central angle theorem $2\gamma = \angle XOY$ and $2\delta = \angle YOZ$, and so $2(\gamma + \delta) = \angle XOZ$, and we are done.

Footnote

We can prove a bit more. Namely, the extensions of the segments $RS$ and $PQ$ meet at a point on the diameter $AB$ that is vertically below the point $Y$.

Since $YS = AY \sin(\delta)$ and is inclined $\alpha$ anti-clockwise from the vertical, the point $S$ is $AY \sin(\delta) \sin(\alpha)$ horizontally to the right of $Y$.

Now $AS = AY \cos(\delta)$, so $S$ is $AS \sin(\alpha) = AY \cos(\delta)\sin(\alpha)$ vertically above the diameter $AB$. Also, the segment $SR$ is inclined $\delta$ clockwise from the vertical, so if we extend it down from $S$ towards the diameter $AB$ it will meet the diameter at a point which is $AY \cos(\delta)\sin(\alpha)\tan(\delta) = AY \sin(\delta)\sin(\alpha)$ horizontally to the left of $S$. This places the intersection point of $RS$ and $AB$ vertically below $Y$.

Similarly, and by symmetry the intersection point of $PQ$ and $AB$ is directly below $Y$ on $AB$, so the lines through $PQ$ and $RS$ meet at a point $T$ on the diameter that is vertically below $Y$.

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