Difference between revisions of "2023 AIME I Problems/Problem 5"
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− | + | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | |
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==Solution 4 (Law of Cosines)== | ==Solution 4 (Law of Cosines)== | ||
WLOG, let <math>P</math> be on minor arc <math>\overarc {AB}</math>. Let <math>r</math> and <math>O</math> be the radius and center of the circumcircle respectively, and let <math>\theta = \angle AOP</math>. | WLOG, let <math>P</math> be on minor arc <math>\overarc {AB}</math>. Let <math>r</math> and <math>O</math> be the radius and center of the circumcircle respectively, and let <math>\theta = \angle AOP</math>. |
Revision as of 14:38, 8 February 2023
Problem (not official; when the official problem statement comes out, please update this page; to ensure credibility until the official problem statement comes out, please add an O if you believe this is correct and add an X if you believe this is incorrect):
Let be a point on the circle circumscribing square
that satisfies
and
. Find the area of
.
Contents
Solution (Ptolemy's Theorem)
Ptolemy's theorem states that for cyclic quadrilateral ,
.
We may assume that is between
and
. Let
,
,
,
, and
. We have
, because
is a diameter of the circle. Similarly,
. Therefore,
. Similarly,
.
By Ptolemy's Theorem on ,
, and therefore
. By Ptolemy's on
,
, and therefore
. By squaring both equations, we obtain
Thus, , and
. Plugging these values into
, we obtain
, and
. Now, we can solve using
and
(though using
and
yields the same solution for
).
The answer is .
~mathboy100
Solution 2 (Circle Properties and Half-Angle Formula)
Drop a height from point to line
and line
. Call these two points to be
and
, respectively. Notice that the intersection of the diagonals of
meets at a right angle at the center of the circumcircle, call this intersection point
.
Since is a rectangle,
is the distance from
to line
. We know that
by triangle area and given information. Then, notice that the measure of
is half of
.
Using the half-angle formula for tangent,
Solving the equation above, we get that or
. Since this value must be positive, we pick
. Then,
(since
is a right triangle with line
also the diameter of the circumcircle) and
. Solving we get
,
, giving us a diagonal of length
and area
.
~Danielzh
Solution 3 (Analytic geometry)
Denote by the half length of each side of the square.
We put the square to the coordinate plane, with
,
,
,
.
The radius of the circumcircle of is
.
Denote by
the argument of point
on the circle.
Thus, the coordinates of
are
.
Thus, the equations and
can be written as
These equations can be reformulated as
These equations can be reformulated as
Taking , by solving the equation, we get
Plugging (3) into (1), we get
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 4 (Law of Cosines)
WLOG, let be on minor arc
. Let
and
be the radius and center of the circumcircle respectively, and let
.
By the Pythagorean Theorem, the area of the square is . We can use the Law of Cosines on isosceles triangles
to get
Taking the products of the first two and last two equations, respectively, and
Adding these equations,
so
~OrangeQuail9
Solution 5 (Double Angle)
Notice that and
are both