Difference between revisions of "1977 Canadian MO Problems/Problem 2"
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− | {{ | + | If <math>AB</math> is the chord perpendicular to <math>OX</math> through point <math>P</math>, then extend <math>AO</math> to meet the circle at point <math>C</math>. It is now evident that <math>O</math> is the midpoint of <math>AC</math>, <math>X</math> is the midpoint of <math>AB</math>, and hence <math>OX=\dfrac{BC}{2}</math>. |
+ | |||
+ | Similarly, let <math>P</math> be a point on arc <math>AB</math>. Extend <math>PO</math> to meet the circle at point <math>R</math>. Extend <math>PX</math> to meet the circle a second time at <math>Q</math>. | ||
+ | |||
+ | We now plot <math>S</math> on <math>XQ</math> such that <math>XS=XP</math>. Then, <math>OX=\dfrac{RS}{2}</math>. Since <math>\angle RQS=90</math>, <math>RS>RQ</math>. Hence, <math>RQ<\dfrac{OX}{2}</math>, and therefore, <math>\angle OPX=\angle OAX=\angle RPQ</math>. | ||
+ | |||
+ | Ergo, the points <math>P</math> such that <math>\angle OPA</math> is maximized are none other than points <math>A</math> and <math>B</math>. <math>\Box</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let <math>XY</math> be the chord perpendicular to <math>OA</math> through <math>A</math>. We claim that the choices of <math>P</math> that maximize <math>\angle{OPA}</math> are <math>X</math> and <math>Y</math>. | ||
+ | |||
+ | Let <math>\omega</math> be the circumcircle of <math>OPA</math>. Then clearly <math>\angle{OPA}</math> is maximized if the radius of <math>\omega</math> is minimized, which occurs if <math>\omega</math> is internally tangent to circle <math>O</math>. Hence <math>OP</math> is a diameter of <math>\omega</math> (whose center <math>OP</math> is collinear with), and so <math>\angle{OAP} = 90^\circ</math>. It follows that <math>P = X</math> or <math>Y</math>, as desired. | ||
{{Old CanadaMO box|num-b=1|num-a=3|year=1977}} | {{Old CanadaMO box|num-b=1|num-a=3|year=1977}} |
Latest revision as of 23:28, 21 September 2014
Let be the center of a circle and
be a fixed interior point of the circle different from
Determine all points
on the circumference of the circle such that the angle
is a maximum.
Solution
If is the chord perpendicular to
through point
, then extend
to meet the circle at point
. It is now evident that
is the midpoint of
,
is the midpoint of
, and hence
.
Similarly, let be a point on arc
. Extend
to meet the circle at point
. Extend
to meet the circle a second time at
.
We now plot on
such that
. Then,
. Since
,
. Hence,
, and therefore,
.
Ergo, the points such that
is maximized are none other than points
and
.
Solution 2
Let be the chord perpendicular to
through
. We claim that the choices of
that maximize
are
and
.
Let be the circumcircle of
. Then clearly
is maximized if the radius of
is minimized, which occurs if
is internally tangent to circle
. Hence
is a diameter of
(whose center
is collinear with), and so
. It follows that
or
, as desired.
1977 Canadian MO (Problems) | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • | Followed by Problem 3 |