2023 AIME I Problems/Problem 5
Contents
- 1 Problem
- 2 Solution 1 (Ptolemy's Theorem)
- 3 Solution 2 (Areas and Pythagorean Theorem)
- 4 Solution 3 (Similar Triangles)
- 5 Solution 4 (Heights and Half-Angle Formula)
- 6 Solution 5 (Analytic Geometry)
- 7 Solution 6 (Law of Cosines)
- 8 Solution 7 (Subtended Chords)
- 9 Solution 8 (Coordinates and Algebraic Manipulation)
- 10 Solution 9 (Law of Sines)
- 11 Solution 10 (Areas and Trigonometry)
- 12 Video Solution 1 by TheBeautyofMath
- 13 See also
Problem
Let be a point on the circle circumscribing square that satisfies and Find the area of
Solution 1 (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 ). ~mathboy100
Solution 2 (Areas and Pythagorean Theorem)
By the Inscribed Angle Theorem, we conclude that and are right triangles.
Let the brackets denote areas. We are given that Let be the center of the circle, be the foot of the perpendicular from to and be the foot of the perpendicular from to as shown below: Let be the diameter of It follows that Moreover, note that is a rectangle. By the Pythagorean Theorem, we have We rewrite this equation in terms of from which Therefore, we get ~MRENTHUSIASM
Solution 3 (Similar Triangles)
Let the center of the circle be , and the radius of the circle be . Since is a rhombus with diagonals and , its area is . Since and are diameters of the circle, and are right triangles. Let and be the foot of the altitudes to and , respectively. We have so . Similarly, so . Since But is a rectangle, so , and our equation becomes Cross multiplying and rearranging gives us , which rearranges to . Therefore .
~Cantalon
Solution 4 (Heights 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 the diameter of the circumcircle) and . Solving we get , , giving us a diagonal of length and area .
Solution 5 (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 6 (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 7 (Subtended Chords)
First draw a diagram. Let's say that the radius is . Then the area of the is Using the formula for the length of a chord subtended by an angle, we get Multiplying and simplifying these 2 equations gives Similarly and . Again, multiplying gives Dividing by gives , so . Pluging this back into one of the equations, gives If we imagine a -- right triangle, we see that if is opposite and is adjacent, . Now we see that ~Voldemort101
Solution 8 (Coordinates and Algebraic Manipulation)
Let on the upper quarter of the circle, and let be the side length of the square. Hence, we want to find . Let the center of the circle be . The two equations would thus become: Now, let , , , and . Our equations now change to and . Subtracting the first from the second, we have . Substituting back in and expanding, we have , so . We now have one of our terms we need (). Therefore, we only need to find to find . We now write the equation of the circle, which point satisfies: We can expand the second equation, yielding Now, with difference of squares, we get . We can add to this equation, which we can factor into . We realize that is the same as the equation of the circle, so we plug its equation in: . We can combine like terms to get , so . Since the answer is an integer, we know is a perfect square. Since it is even, it is divisible by , so we can factor . With some testing with approximations and last-digit methods, we can find that . Therefore, taking the square root, we find that , the area of square , is .
~wuwang2002
Solution 9 (Law of Sines)
WLOG, let be on minor arc Draw in , , , and let We can see, by the inscribed angle theorem, that , and Then, , , and Letting , we can use the law of sines on triangles and to get Making all the angles in the above equation acute gives
Note that we are looking for We are given that and This means that and However, and Therefore, and Therefore, by the Pythagorean Identity,
~pianoboy
Solution 10 (Areas and Trigonometry)
Similar to Solution 6, let be on minor arc , and be the radius and center of the circumcircle respectively, and . Since is a right triangle, equals the hypotenuse, , times its altitude, which can be represented as . Therefore, . Applying similar logic to , we get .
Dividing the two equations, we have Adding to both sides allows us to get rid of : Therefore, we have , and since the area of the square can be represented as , the answer is .
~phillipzeng
Video Solution 1 by TheBeautyofMath
~IceMatrix
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.