Difference between revisions of "1976 USAMO Problems/Problem 2"
Collinator (talk | contribs) |
|||
Line 41: | Line 41: | ||
Now note that at points <math>A</math> and <math>B</math>, this slope expression reduces to <math>y' = \frac{-b}{a}</math> and <math>y' = \frac{b}{a}</math> respectively, values which are identical to the slopes of lines <math>AO</math> and <math>BO</math>. Thus we conclude that the complete locus of intersection points is the circle tangent to lines <math>AO</math> and <math>BO</math> at points <math>A</math> and <math>B</math> respectively. | Now note that at points <math>A</math> and <math>B</math>, this slope expression reduces to <math>y' = \frac{-b}{a}</math> and <math>y' = \frac{b}{a}</math> respectively, values which are identical to the slopes of lines <math>AO</math> and <math>BO</math>. Thus we conclude that the complete locus of intersection points is the circle tangent to lines <math>AO</math> and <math>BO</math> at points <math>A</math> and <math>B</math> respectively. | ||
+ | ==Solution 2== | ||
+ | |||
+ | Notice that <math>m\angle AYP=\frac{1}{2}m\overarc{AB}</math> (Inscried angle theorem) and that <math>m\angle XAY=90</math> since <math>XY</math> is a diameter, and thus subtends an arc of <math>180</math>. This will hold for all <math>X</math> and all <math>Y</math>, and so by AA similarity, the angle <math>APY</math> will be constant for all P, thus implying that the points A, B, and all P will be concyclic. | ||
+ | If we assume that the center of this circle is <math>H</math>, we know that <math>m\angle AHB=2m\angle APB=2(90-\frac{1}{2}\overarc{AB})=180-\overarc{AB}</math>. We can assume that <math>\overline{AH}</math> and <math>\overline{BH}</math> intersect the original circle at points <math>M</math>, and <math>N</math> respectively. This will give us that <math>m\angle AHB=\frac{1}{2}|m\overarc{MN}-m\overarc{AB}|</math> (Since <math>H</math> lies on the perpendicular bisector of <math>AB</math>, we know that <math>M</math>, <math>N</math> will be on the same side of <math>AB</math>.) Now we also know that <math>m\overarc{MN}\leq m\overarc{AB}</math> or <math>m\overarc{MN}\leq 360-m\overarc{AB}</math>. The only case where <math>m\overarc{MN}</math> satisfies the measure of <math>\angle AHB</math>, is when <math>M=A</math> and <math>N=B</math>, implying that <math>AH</math> and <math>BH</math> are tangents, and so <math>H</math> is the intersection of the tangents from <math>A</math> and <math>B</math> to the original circle. | ||
Latest revision as of 23:41, 6 July 2020
Contents
Problem
If and
are fixed points on a given circle and
is a variable diameter of the same circle, determine the locus of the point of intersection of lines
and
. You may assume that
is not a diameter.
Solution
WLOG, assume that the circle is the unit circle centered at the origin. Then the points and
have coordinates
and
respectively and
and
have coordinates
and
. Note that these coordinates satisfy
and
since these points are on a unit circle. Now we can find equations for the lines:
Solving these simultaneous equations gives coordinates for
in terms of
and
:
. These coordinates can be parametrized in Cartesian variables as follows:
Now solve for
and
to get
and
. Then since
which reduces to
This equation defines a circle and is the locus of all intersection points
. In order to define this locus more generally, find the slope of this circle function using implicit differentiation:
Now note that at points
and
, this slope expression reduces to
and
respectively, values which are identical to the slopes of lines
and
. Thus we conclude that the complete locus of intersection points is the circle tangent to lines
and
at points
and
respectively.
Solution 2
Notice that (Inscried angle theorem) and that
since
is a diameter, and thus subtends an arc of
. This will hold for all
and all
, and so by AA similarity, the angle
will be constant for all P, thus implying that the points A, B, and all P will be concyclic.
If we assume that the center of this circle is
, we know that
. We can assume that
and
intersect the original circle at points
, and
respectively. This will give us that
(Since
lies on the perpendicular bisector of
, we know that
,
will be on the same side of
.) Now we also know that
or
. The only case where
satisfies the measure of
, is when
and
, implying that
and
are tangents, and so
is the intersection of the tangents from
and
to the original circle.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1976 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.