Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎See Also)
Line 116: Line 116:
  
 
== See Also ==
 
== See Also ==
 +
==Problem==
 +
Let <math>AXYZB</math> be a convex pentagon inscribed in a semicircle of diameter
 +
<math>AB</math>. Denote by <math>P, Q, R, S</math> the feet of the perpendiculars from <math>Y</math> onto
 +
lines <math>AX, BX, AZ, BZ</math>, respectively. Prove that the acute angle
 +
formed by lines <math>PQ</math> and <math>RS</math> is half the size of <math>\angle XOZ</math>, where
 +
<math>O</math> is the midpoint of segment <math>AB</math>.
 +
 +
==Solution==
 +
Let <math>\alpha = \angle BAZ</math>, <math>\beta = \angle ABX</math>.
 +
Since <math>XY</math> is a chord of the circle with diameter <math>AB</math>,
 +
<math>\angle XAY = \angle XBY = \gamma</math>. From the chord <math>YZ</math>,
 +
we conclude <math>\angle YAZ = \angle YBZ = \delta</math>.
 +
<center>
 +
<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("$S$", R, unit(R-Y)); dot(R);
 +
pair S = foot(Y, A, Z); dot(S); label("$R$", 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>
 +
</center>
 +
 +
Triangles <math>BQY</math> and <math>APY</math> are both right-triangles, and share the
 +
angle <math>\gamma</math>, therefore they are similar, and so the ratio <math>PY :
 +
YQ = AY : YB</math>. Now by [[Thales' theorem]] the angles <math>\angle AXB =
 +
\angle AYB = \angle AZB</math> are all right-angles. Also, <math>\angle PYQ</math>,
 +
being the fourth angle in a quadrilateral with 3 right-angles is
 +
again a right-angle.  Therefore <math>\triangle PYQ \sim \triangle AYB</math> and
 +
<math>\angle YQP = \angle YBA = \gamma + \beta</math>.
 +
Similarly, <math>RY : YS = BY : YA</math>, and so <math>\angle YSR = \angle YAB = \alpha + \delta</math>.
 +
 +
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.
 +
 +
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.
 +
 +
Therefore the lines <math>PQ</math> and <math>RS</math> intersect at an angle <math>\chi = \gamma
 +
+ \delta</math>. Now by the central angle theorem <math>2\gamma = \angle XOY</math>
 +
and <math>2\delta = \angle YOZ</math>, and so <math>2(\gamma + \delta) = \angle XOZ</math>,
 +
and we are done.
 +
 +
===Footnote===
 +
We can prove a bit more. Namely, the extensions of the segments
 +
<math>RS</math> and <math>PQ</math> meet at a point on the diameter <math>AB</math> that is vertically
 +
below the point <math>Y</math>.
 +
 +
Since <math>YS = AY \sin(\delta)</math> and is inclined <math>\alpha</math> counterclockwise
 +
from the vertical, the point <math>S</math> is <math>AY \sin(\delta) \sin(\alpha)</math>
 +
horizontally to the right of <math>Y</math>.
 +
 +
Now <math>AS = AY \cos(\delta)</math>, so <math>S</math> is <math>AS \sin(\alpha) = AY
 +
\cos(\delta)\sin(\alpha)</math> vertically above the diameter <math>AB</math>. Also,
 +
the segment <math>SR</math> is inclined <math>\delta</math> clockwise from the vertical,
 +
so if we extend it down from <math>S</math> towards the diameter <math>AB</math> it will
 +
meet the diameter at a point which is
 +
<math>AY \cos(\delta)\sin(\alpha)\tan(\delta) = AY \sin(\delta)\sin(\alpha)</math>
 +
horizontally to the left of <math>S</math>. This places the intersection point
 +
of <math>RS</math> and <math>AB</math> vertically below <math>Y</math>.
 +
 +
Similarly, and by symmetry the intersection point of <math>PQ</math> and <math>AB</math>
 +
is directly below <math>Y</math> on <math>AB</math>, so the lines through <math>PQ</math> and <math>RS</math>
 +
meet at a point <math>T</math> on the diameter that is vertically below <math>Y</math>.
 +
 +
== See Also ==
 +
{{USAMO newbox|year=2010|before=First problem|num-a=2}}
 +
{{USAJMO newbox|year=2010|num-b=2|num-a=4}}
 +
 +
[[Category:Olympiad Number Theory Problems]]
 +
 
{{USAJMO newbox|year=2010|num-b=2|num-a=4}}
 
{{USAJMO newbox|year=2010|num-b=2|num-a=4}}
  
 
[[Category:Olympiad Number Theory Problems]]
 
[[Category:Olympiad Number Theory Problems]]

Revision as of 14:07, 20 April 2013

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("$S$", R, unit(R-Y)); dot(R); pair S = foot(Y, A, Z); dot(S); label("$R$", 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 $RY$ is perpendicular to $AZ$ so the direction $RY$ is $\alpha$ counterclockwise from the vertical, and since $\angle YRS = \alpha + \delta$ 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 $\gamma$ counterclockwise 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 $\alpha$ counterclockwise 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 $Y$.

See Also

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("$S$", R, unit(R-Y)); dot(R); pair S = foot(Y, A, Z); dot(S); label("$R$", 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 $RY$ is perpendicular to $AZ$ so the direction $RY$ is $\alpha$ counterclockwise from the vertical, and since $\angle YRS = \alpha + \delta$ 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 $\gamma$ counterclockwise 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 $\alpha$ counterclockwise 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 $Y$.

See Also

2010 USAMO (Problems • Resources)
Preceded by
First problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions
2010 USAJMO (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions
2010 USAJMO (Problems • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_USAMO_Problems/Problem_1&oldid=52574"