Difference between revisions of "2023 AIME I Problems/Problem 5"
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Notice that <math>\angle APC</math> and <math>\angle BPD</math> are both <math>90^\circ</math> | Notice that <math>\angle APC</math> and <math>\angle BPD</math> are both <math>90^\circ</math> | ||
+ | |||
+ | ==Solution 6 (Similar Triangles)== | ||
+ | |||
+ | \begin{center} | ||
+ | \begin{tikzpicture} | ||
+ | \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] {<math>O</math>}; | ||
+ | \draw (-2.8284, -2.8284) node[anchor=north east] {<math>D</math>}; | ||
+ | \draw (2.8284, -2.8284) node[anchor=north west] {<math>C</math>}; | ||
+ | \draw (2.8284, 2.8284) node[anchor=south west] {<math>B</math>}; | ||
+ | \draw (-2.8284, 2.8284) node[anchor=south east] {<math>A</math>}; | ||
+ | \draw (-0.531, 3.965) | ||
+ | node[anchor=south] {<math>P</math>}; | ||
+ | \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] {<math>X</math>}; | ||
+ | \draw (1.717, 1.717) node[anchor=north west] {<math>Y</math>}; | ||
+ | \end{tikzpicture} | ||
+ | \end{center} | ||
+ | |||
+ | 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 ACP</math> and <math>\triangle BDP</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 | ||
+ | <cmath>[\triangle APC] = \frac{1}{2}(PA)(PC) = \frac{1}{2}(PX)(AC),</cmath> | ||
+ | so <math>PX = \dfrac{(PA)(PC)}{AC} = \dfrac{28}{r}</math>. Similarly, | ||
+ | <cmath>[\triangle BPD] = \frac{1}{2}(PB)(PD) = \frac{1}{2}(PY)(PB),</cmath> | ||
+ | so <math>PY = \dfrac{(PB)(PD)}{BD} = \dfrac{45}{r}</math>. Since <math>\triangle APX \sim \triangle PCX,</math> | ||
+ | <cmath>\frac{AX}{PX} = \frac{PX}{XC}.</cmath> | ||
+ | But <math>PXOY</math> is a rectangle, so <math>PY</math> = <math>XO</math>, and our similarity becomes | ||
+ | <cmath>\frac{AO - XO}{PX} = \frac{PX}{OC + XO}</cmath> | ||
+ | <cmath>\frac{r - PY}{PX} = \frac{PX}{r + PY}.</cmath> | ||
+ | Cross multiplying and rearranging gives us <math>r^2 = PX^2 + PY^2 = \left(\dfrac{28}{r}\right)^2 + \left(\dfrac{45}{r}\right)^2</math>, which rearranges to <math>r^4 = 2809</math>. Therefore <math>[ABCD] = 2r^2 = \boxed{106}</math>. | ||
+ | |||
+ | ~Cantalon |
Revision as of 14:16, 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 be a point on the circle circumscribing square that satisfies and . Find the area of .
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 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)
\begin{center}
\begin{tikzpicture} \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] {}; \draw (-2.8284, -2.8284) node[anchor=north east] {}; \draw (2.8284, -2.8284) node[anchor=north west] {}; \draw (2.8284, 2.8284) node[anchor=south west] {}; \draw (-2.8284, 2.8284) node[anchor=south east] {}; \draw (-0.531, 3.965) node[anchor=south] {}; \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] {}; \draw (1.717, 1.717) node[anchor=north west] {}; \end{tikzpicture}
\end{center}
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