Difference between revisions of "2023 AIME I Problems/Problem 5"
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~mathboy100 | ~mathboy100 | ||
− | ==Solution 2 (Trigonometry)== | + | ==Solution 2 (Trigonometry and Half-Angle Formula)== |
Drop a height from point <math>P</math> to line <math>\overline{AC}</math> and line <math>\overline{BC}</math>. Call these two points to be X and Y, respectively. Notice that the intersection of the diagonals of square ABCD meets at a right angle at the center of the circumcircle, call this intersection point O. | Drop a height from point <math>P</math> to line <math>\overline{AC}</math> and line <math>\overline{BC}</math>. Call these two points to be X and Y, respectively. Notice that the intersection of the diagonals of square ABCD meets at a right angle at the center of the circumcircle, call this intersection point O. | ||
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</cmath> | </cmath> | ||
− | we get that <math>\tan{\angle{OCP}} = -7/2</math> or <math>2/7</math>. Since this value must be positive, we pick 2/ | + | we get that <math>\tan{\angle{OCP}} = -7/2</math> or <math>2/7</math>. Since this value must be positive, we pick <math>\frac{2}{7}</math>. Then, PA/PC = 2/7 (since triangle CAP is a right triangle with AC also the diameter of the circumcircle) and PA * PC = 56. Solving we get PA = 4, PC = 14, giving us a diagonal of length <math>\sqrt{212}</math> and area <math>\boxed{106}</math>. |
~Danielzh | ~Danielzh |
Revision as of 14:20, 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 there be a circle circumscribing a square ABCD, and let P be a point on the circle. PA*PC = 56, PB*PD = 90. What is the area of the square?
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 (Trigonometry and Half-Angle Formula)
Drop a height from point to line
and line
. Call these two points to be X and Y, respectively. Notice that the intersection of the diagonals of square ABCD meets at a right angle at the center of the circumcircle, call this intersection point O.
Since OXPY is a rectangle, OX is the distance from P to line BD. We know that tan(YOX) = PX/XO = 28/45 by triangle area and given information. Then, notice that the measure of angle OCP is half of the angle of angle XOY.
Using the half-angle formula for tangent,
we get that or
. Since this value must be positive, we pick
. Then, PA/PC = 2/7 (since triangle CAP is a right triangle with AC also the diameter of the circumcircle) and PA * PC = 56. Solving we get PA = 4, PC = 14, 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
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