Difference between revisions of "2010 USAMO Problems/Problem 1"
(Created page with '==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 fr…') |
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<asy> | <asy> | ||
import olympiad; | import olympiad; | ||
− | + | import markers; | |
− | |||
void langle(picture p=currentpicture, | 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; | picture p; | ||
+ | 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 A = r * plain.W; dot(p, A); label(p, "$A$", A, plain.W); | ||
Line 33: | Line 37: | ||
pair O = (0,0); dot(p, O); label(p, "$O$", O, plain.S); | pair O = (0,0); dot(p, O); label(p, "$O$", O, plain.S); | ||
− | real alpha = 22.5; | + | // Semi-circle with diameter A--B |
− | real beta = 15; | + | path C=arc(O, r, 0, 180)--cycle; draw(p, C); |
− | real delta = 30; | + | |
− | real gamma = 90 - alpha - beta - delta; // angle XBY | + | real alpha = 22.5; // angle BAZ |
+ | real beta = 15; // angble ABX | ||
+ | real delta = 30; // angle ZAY | ||
+ | real gamma = 90 - alpha - beta - delta; // angle XBY | ||
// Points X, Y, Z | // Points X, Y, Z | ||
Line 43: | Line 50: | ||
pair Z = r * dir(2*alpha); dot(p, Z); label(p, "$Z$", Z, plain.NE); | pair Z = r * dir(2*alpha); dot(p, Z); label(p, "$Z$", Z, plain.NE); | ||
− | langle(p, B, A, Z, "\alpha"); | + | // Angle labels |
− | langle(p, X, B, A, "\beta" | + | langle(p, B, A, Z, "\alpha" ); |
− | + | langle(p, X, B, A, "\beta", n=2); | |
− | langle(p, Y, | + | langle(p, Y, A, X, "\gamma", marks=1); |
− | langle(p, Y, | + | langle(p, Y, B, X, "\gamma", marks=1); |
− | langle(p, Z, A, Y, "\delta"); | + | langle(p, Z, A, Y, "\delta", marks=2); |
− | langle(p, Z, B, Y, "\delta" | + | langle(p, Z, B, Y, "\delta", marks=2); |
− | |||
− | |||
− | |||
+ | // Perpendiculars from Y | ||
pair Q = foot(Y, B, X); | pair Q = foot(Y, B, X); | ||
dot(p, Q); label(p, "$Q$", Q, plain.SW); | dot(p, Q); label(p, "$Q$", Q, plain.SW); | ||
Line 69: | Line 74: | ||
draw(p, Y--S); | draw(p, Y--S); | ||
draw(p, rightanglemark(Z, S, Y, 3)); | draw(p, rightanglemark(Z, S, Y, 3)); | ||
− | |||
pair R = foot(Y, B, Z); | pair R = foot(Y, B, Z); | ||
Line 77: | Line 81: | ||
draw(p, rightanglemark(B, R, Y, 3)); | draw(p, rightanglemark(B, R, Y, 3)); | ||
− | // | + | // 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, A--Y); draw(p, A--Z); | ||
draw(p, B--X); draw(p, B--Y); | draw(p, B--X); draw(p, B--Y); | ||
Line 88: | Line 96: | ||
dot(p, T); label(p, "$T$", T, plain.S); | dot(p, T); label(p, "$T$", T, plain.S); | ||
− | + | // Label for sought angle "chi" | |
− | + | langle(p, R, T, P, "\chi", r=15); | |
− | draw(p, P--T); | + | |
− | draw(p, R--T) | + | // Ligher lines for PT, RT |
− | + | draw(p, P--T, linewidth(0.2)); draw(p, R--T, linewidth(0.2)); | |
− | |||
add(p); | add(p); |
Revision as of 01:29, 8 May 2010
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
anti-clockwise 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
anti-clockwise 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
anti-clockwise
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
.