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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}} |
| | | | |
| | [[Category:Olympiad Number Theory Problems]] | | [[Category:Olympiad Number Theory Problems]] |
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 
.
![[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]](//latex.artofproblemsolving.com/3/b/8/3b8595caea22f3bdf34ff3334c724f17808ca631.png)
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.
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
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 
, 
.
Since 
is a chord of the circle with diameter 
. From the chord 
,
we conclude 
.
![[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]](http://latex.artofproblemsolving.com/3/b/8/3b8595caea22f3bdf34ff3334c724f17808ca631.png)
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 
.
is perpendicular to 
so the direction 
counterclockwise from the vertical, and since 
we see that 
is 
clockwise from the vertical.
is perpendicular to 
so the direction 
clockwise from the vertical, and since 
is 
we see that 
. Now by the central angle theorem 
and 
, and so 
,
and we are done.
and is inclined 
is 
horizontally to the right of 
, so 
vertically above the diameter 
horizontally to the left of 
on the diameter that is vertically below 
