Difference between revisions of "1988 IMO Problems/Problem 1"
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+ | ==Problem== | ||
Consider 2 concentric circles with radii <math>R</math> and <math>r</math> (<math>R>r</math>) with center <math>O</math>. Fix <math>P</math> on the small circle and consider the variable chord <math>AP</math> of the small circle. Points <math>B</math> and <math>C</math> lie on the large circle; <math>B,P,C</math> are collinear and <math>BC</math> is perpendicular to <math>AP</math>. | Consider 2 concentric circles with radii <math>R</math> and <math>r</math> (<math>R>r</math>) with center <math>O</math>. Fix <math>P</math> on the small circle and consider the variable chord <math>AP</math> of the small circle. Points <math>B</math> and <math>C</math> lie on the large circle; <math>B,P,C</math> are collinear and <math>BC</math> is perpendicular to <math>AP</math>. | ||
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− | + | ==Solution== | |
'''i.)''' | '''i.)''' | ||
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We claim that the value <math>BC^2+CA^2+AB^2</math> stays constant as <math>\angle OPA</math> varies, and thus achieves its maximum at all value of <math>\angle OPA</math>. We have from the Pythagorean Theorem that <math>CA^2=AP^2+PC^2</math> and <math>AB^2=AP^2+PB^2</math> and so our expression becomes | We claim that the value <math>BC^2+CA^2+AB^2</math> stays constant as <math>\angle OPA</math> varies, and thus achieves its maximum at all value of <math>\angle OPA</math>. We have from the Pythagorean Theorem that <math>CA^2=AP^2+PC^2</math> and <math>AB^2=AP^2+PB^2</math> and so our expression becomes | ||
− | <math>BC^2+PB^2+PC^2+2AP^2=2BC^2+2AP^2-2PB\ | + | <math>BC^2+PB^2+PC^2+2AP^2=2BC^2+2AP^2-2PB\cdot PC</math> |
Since <math>PB\cdot PC</math> is the power of the point <math>P</math>, it stays constant as <math>A</math> varies. Thus, we are left to prove that the value <math>BC^2+AP^2</math> stays constant as <math>\angle OPA</math> varies. Let <math>G</math> be the midpoint of <math>AP</math> and let <math>H</math> be the midpoint of <math>BC</math>. Since <math>OG</math> is perpendicular to <math>AP</math>, we find that <math>PG=r\cos OPA</math>. Similarly, we find that <math>OH=r\sin OPC=r\cos OPA</math>. Thus, by the Pythagorean Theorem, we have | Since <math>PB\cdot PC</math> is the power of the point <math>P</math>, it stays constant as <math>A</math> varies. Thus, we are left to prove that the value <math>BC^2+AP^2</math> stays constant as <math>\angle OPA</math> varies. Let <math>G</math> be the midpoint of <math>AP</math> and let <math>H</math> be the midpoint of <math>BC</math>. Since <math>OG</math> is perpendicular to <math>AP</math>, we find that <math>PG=r\cos OPA</math>. Similarly, we find that <math>OH=r\sin OPC=r\cos OPA</math>. Thus, by the Pythagorean Theorem, we have |
Revision as of 20:00, 9 May 2011
Problem
Consider 2 concentric circles with radii and
(
) with center
. Fix
on the small circle and consider the variable chord
of the small circle. Points
and
lie on the large circle;
are collinear and
is perpendicular to
.
i.) For which values of is the sum
extremal?
ii.) What are the possible positions of the midpoints of
and
of
as
varies?
Solution
i.)
We claim that the value stays constant as
varies, and thus achieves its maximum at all value of
. We have from the Pythagorean Theorem that
and
and so our expression becomes
Since is the power of the point
, it stays constant as
varies. Thus, we are left to prove that the value
stays constant as
varies. Let
be the midpoint of
and let
be the midpoint of
. Since
is perpendicular to
, we find that
. Similarly, we find that
. Thus, by the Pythagorean Theorem, we have
Now it is obvious that is constant for all values of
.
ii.)
We claim that all points lie on a circle centered at the midpoint of
,
with radius
. Let
be the midpoint of
. Since
is the midpoint of
, it is clear that the projection of
onto
is the midpoint of
and
(the projection of
onto
). Thus, we have that
is perpendicular to
and thus the triangle
is iscoceles. We have
and
. Thus, from the Pythagorean Theorem we have
. Since we have shown already that
is constant, we have that
and the locus of points
is indeed a circle of radius
with center
.