Difference between revisions of "2023 AIME I Problems/Problem 5"
(Added LaTeX to problem statement; added second solution.) |
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\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
+ | |||
+ | |||
+ | ==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>. | ||
+ | |||
+ | By the [[Pythagorean Theorem]], the area of the square is <math>2r^2</math>. We can use the [[Law of Cosines]] on isosceles triangles <math>\triangle AOP, \, \triangle COP, \, \triangle BOP, \, \triangle DOP</math> to get | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | PA^2 &= 2r^2(1 - \cos \theta), \ | ||
+ | PC^2 &= 2r^2(1 - \cos (180 - \theta)) = 2r^2(1 + \cos \theta), \ | ||
+ | PB^2 &= 2r^2(1 - \cos (90 - \theta)) = 2r^2(1 - \sin \theta), \ | ||
+ | PD^2 &= 2r^2(1 - \cos (90 + \theta)) = 2r^2(1 + \sin \theta). | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Taking the products of the first two and last two equations, respectively, <cmath>56^2 = (PA \cdot PC)^2 = 4r^4(1 - \cos \theta)(1 + \cos \theta) = 4r^4(1 - \cos^2 \theta) = 4r^4 \sin^2 \theta,</cmath> and <cmath>90^2 = (PB \cdot PD)^2 = 4r^4(1 - \sin \theta)(1 + \sin \theta) = 4r^4(1 - \sin^2 \theta) = 4r^4 \cos^2 \theta.</cmath> | ||
+ | Adding these equations, <cmath>56^2 + 90^2 = 4r^4,</cmath> so | ||
+ | <cmath>2r^2 = \sqrt{56^2+90^2} = 2\sqrt{28^2+45^2} = 2\sqrt{2809} = 2 \cdot 53 = \boxed{106}.</cmath> | ||
+ | ~OrangeQuail9 |
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
[hide]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 using half angle formula,
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