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 13:07, 20 April 2013
Contents
[hide]Problem
Let be a convex pentagon inscribed in a semicircle of diameter . Denote by the feet of the perpendiculars from onto lines , respectively. Prove that the acute angle formed by lines and is half the size of , where is the midpoint of segment .
Solution
Let , . Since is a chord of the circle with diameter , . From the chord , we conclude .
Triangles and are both right-triangles, and share the angle , therefore they are similar, and so the ratio . Now by Thales' theorem the angles are all right-angles. Also, , being the fourth angle in a quadrilateral with 3 right-angles is again a right-angle. Therefore and . Similarly, , and so .
Now is perpendicular to so the direction is counterclockwise from the vertical, and since we see that is clockwise from the vertical.
Similarly, is perpendicular to so the direction is clockwise from the vertical, and since is we see that is counterclockwise from the vertical.
Therefore the lines and intersect at an angle . Now by the central angle theorem and , and so , and we are done.
Footnote
We can prove a bit more. Namely, the extensions of the segments and meet at a point on the diameter that is vertically below the point .
Since and is inclined counterclockwise from the vertical, the point is horizontally to the right of .
Now , so is vertically above the diameter . Also, the segment is inclined clockwise from the vertical, so if we extend it down from towards the diameter it will meet the diameter at a point which is horizontally to the left of . This places the intersection point of and vertically below .
Similarly, and by symmetry the intersection point of and is directly below on , so the lines through and meet at a point on the diameter that is vertically below .
See Also
Problem
Let be a convex pentagon inscribed in a semicircle of diameter . Denote by the feet of the perpendiculars from onto lines , respectively. Prove that the acute angle formed by lines and is half the size of , where is the midpoint of segment .
Solution
Let , . Since is a chord of the circle with diameter , . From the chord , we conclude .
Triangles and are both right-triangles, and share the angle , therefore they are similar, and so the ratio . Now by Thales' theorem the angles are all right-angles. Also, , being the fourth angle in a quadrilateral with 3 right-angles is again a right-angle. Therefore and . Similarly, , and so .
Now is perpendicular to so the direction is counterclockwise from the vertical, and since we see that is clockwise from the vertical.
Similarly, is perpendicular to so the direction is clockwise from the vertical, and since is we see that is counterclockwise from the vertical.
Therefore the lines and intersect at an angle . Now by the central angle theorem and , and so , and we are done.
Footnote
We can prove a bit more. Namely, the extensions of the segments and meet at a point on the diameter that is vertically below the point .
Since and is inclined counterclockwise from the vertical, the point is horizontally to the right of .
Now , so is vertically above the diameter . Also, the segment is inclined clockwise from the vertical, so if we extend it down from towards the diameter it will meet the diameter at a point which is horizontally to the left of . This places the intersection point of and vertically below .
Similarly, and by symmetry the intersection point of and is directly below on , so the lines through and meet at a point on the diameter that is vertically below .
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 |