Difference between revisions of "2023 AIME I Problems/Problem 5"
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==Solution 6 (Similar Triangles)== | ==Solution 6 (Similar Triangles)== | ||
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\draw (0,0) circle (4cm); | \draw (0,0) circle (4cm); | ||
\draw (2.8284, -2.8284) -- (2.8284, 2.8284) -- (-2.8284, 2.8284) -- (-2.8284, -2.8284) -- cycle; | \draw (2.8284, -2.8284) -- (2.8284, 2.8284) -- (-2.8284, 2.8284) -- (-2.8284, -2.8284) -- cycle; | ||
− | \draw (0, 0) node[anchor=north] { | + | \draw (0, 0) node[anchor=north] {$O$}; |
− | \draw (-2.8284, -2.8284) node[anchor=north east] { | + | \draw (-2.8284, -2.8284) node[anchor=north east] {$D$}; |
− | \draw (2.8284, -2.8284) node[anchor=north west] { | + | \draw (2.8284, -2.8284) node[anchor=north west] {$C$}; |
− | \draw (2.8284, 2.8284) node[anchor=south west] { | + | \draw (2.8284, 2.8284) node[anchor=south west] {$B$}; |
− | \draw (-2.8284, 2.8284) node[anchor=south east] { | + | \draw (-2.8284, 2.8284) node[anchor=south east] {$A$}; |
\draw (-0.531, 3.965) | \draw (-0.531, 3.965) | ||
− | node[anchor=south] { | + | node[anchor=south] {$P$}; |
\draw (-2.8284, -2.8284) -- (2.8284, 2.8284) -- (-0.531, 3.965) -- cycle; | \draw (-2.8284, -2.8284) -- (2.8284, 2.8284) -- (-0.531, 3.965) -- cycle; | ||
\draw (2.8284, -2.8284) -- (-2.8284, 2.8284) -- (-0.531, 3.965) -- cycle; | \draw (2.8284, -2.8284) -- (-2.8284, 2.8284) -- (-0.531, 3.965) -- cycle; | ||
\draw[dashed] (-0.531, 3.965) -- (1.717, 1.717); | \draw[dashed] (-0.531, 3.965) -- (1.717, 1.717); | ||
\draw[dashed] (-0.531, 3.965) -- (-2.248, 2.248); | \draw[dashed] (-0.531, 3.965) -- (-2.248, 2.248); | ||
− | \draw (-2.248, 2.248) node[anchor=north east] { | + | \draw (-2.248, 2.248) node[anchor=north east] {$X$}; |
− | \draw (1.717, 1.717) node[anchor=north west] { | + | \draw (1.717, 1.717) node[anchor=north west] {$Y$}; |
− | + | </asy> | |
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Let the center of the circle be <math>O</math>, and the radius of the circle be <math>r</math>. Since <math>ABCD</math> is a rhombus with diagonals <math>2r</math> and <math>2r</math>, its area is <math>\dfrac{1}{2}(2r)(2r) = 2r^2</math>. Since <math>AC</math> and <math>BD</math> are diameters of the circle, <math>\triangle APC</math> and <math>\triangle BPD</math> are right triangles. Let <math>X</math> and <math>Y</math> be the foot of the altitudes to <math>AC</math> and <math>BD</math>, respectively. We have | Let the center of the circle be <math>O</math>, and the radius of the circle be <math>r</math>. Since <math>ABCD</math> is a rhombus with diagonals <math>2r</math> and <math>2r</math>, its area is <math>\dfrac{1}{2}(2r)(2r) = 2r^2</math>. Since <math>AC</math> and <math>BD</math> are diameters of the circle, <math>\triangle APC</math> and <math>\triangle BPD</math> are right triangles. Let <math>X</math> and <math>Y</math> be the foot of the altitudes to <math>AC</math> and <math>BD</math>, respectively. We have |
Revision as of 14:57, 8 February 2023
Contents
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 ).
The answer is .
~mathboy100
Solution 2 (Circle Properties 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 also the diameter of the circumcircle) and . Solving we get , , 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
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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
Solution 6 (Similar Triangles)
\draw (0,0) circle (4cm); \draw (2.8284, -2.8284) -- (2.8284, 2.8284) -- (-2.8284, 2.8284) -- (-2.8284, -2.8284) -- cycle; \draw (0, 0) node[anchor=north] {$O$}; \draw (-2.8284, -2.8284) node[anchor=north east] {$D$}; \draw (2.8284, -2.8284) node[anchor=north west] {$C$}; \draw (2.8284, 2.8284) node[anchor=south west] {$B$}; \draw (-2.8284, 2.8284) node[anchor=south east] {$A$}; \draw (-0.531, 3.965) node[anchor=south] {$P$}; \draw (-2.8284, -2.8284) -- (2.8284, 2.8284) -- (-0.531, 3.965) -- cycle; \draw (2.8284, -2.8284) -- (-2.8284, 2.8284) -- (-0.531, 3.965) -- cycle; \draw[dashed] (-0.531, 3.965) -- (1.717, 1.717); \draw[dashed] (-0.531, 3.965) -- (-2.248, 2.248); \draw (-2.248, 2.248) node[anchor=north east] {$X$}; \draw (1.717, 1.717) node[anchor=north west] {$Y$}; (Error making remote request. Unknown error_msg)
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 similarity becomes Cross multiplying and rearranging gives us , which rearranges to . Therefore .
~Cantalon
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.