2023 AIME I Problems/Problem 5
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 diagonal. 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)
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,
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we get that or . Since this value must be positive, we pick 2/7. 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
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