Difference between revisions of "2023 AIME I Problems/Problem 5"
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==Solution 2 (Trigonometry)== | ==Solution 2 (Trigonometry)== | ||
Drop a height from point P to line AC and BC. 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 and at the center of the circumcircle, call this intersection point O. | Drop a height from point P to line AC and BC. 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 and 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 the 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 | + | Since OXPY is a rectangle, OX is the distance from P to line BD. We know the 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 half angle formula for tangent, we get that tan(OCP) = -7/2 or 2/7. 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 sqrt(212) and area 106. | ||
==Solution 3 (Analytic geometry)== | ==Solution 3 (Analytic geometry)== |
Revision as of 12:43, 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
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 P to line AC and BC. 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 and 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 the 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 half angle formula for tangent, we get that tan(OCP) = -7/2 or 2/7. 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 sqrt(212) and area 106.
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