Difference between revisions of "2024 AIME II Problems/Problem 12"
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~Bluesoul | ~Bluesoul | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | The equation of line <math>AB</math> is | ||
+ | \[ | ||
+ | y = \frac{\sqrt{3}}{2} x - \sqrt{3} x. \hspace{1cm} (2) | ||
+ | \] | ||
+ | |||
+ | The position of line <math>PQ</math> can be characterized by <math>\angle QPO</math>, denoted as <math>\theta</math>. | ||
+ | Thus, the equation of line <math>PQ</math> is | ||
+ | \[ | ||
+ | y = \sin \theta - \tan \theta \cdot x . \hspace{1cm} (2) | ||
+ | \] | ||
+ | |||
+ | Solving (1) and (2), the <math>x</math>-coordinate of the intersecting point of lines <math>AB</math> and <math>PQ</math> satisfies the following equation: | ||
+ | \[ | ||
+ | \frac{\frac{\sqrt{3}}{2} - \sqrt{3} x}{\sin \theta} | ||
+ | + \frac{x}{\cos \theta} | ||
+ | = 1 . \hspace{1cm} (1) | ||
+ | \] | ||
+ | We denote the L.H.S. as <math>f \left( \theta ; x \right)</math>. | ||
+ | |||
+ | We observe that <math>f \left( 60^\circ ; x \right) = 1</math> for all <math>x</math>. | ||
+ | Therefore, the point <math>C</math> that this problem asks us to find can be equivalently stated in the following way: | ||
+ | |||
+ | We interpret Equation (1) as a parameterized equation that <math>x</math> is a tuning parameter and <math>\theta</math> is a variable that shall be solved and expressed in terms of <math>x</math>. | ||
+ | In Equation (1), there exists a unique <math>x \in \left( 0, 1 \right)</math>, denoted as <math>x_C</math> (<math>x</math>-coordinate of point <math>C</math>), such that the only solution is <math>\theta = 60^\circ</math>. For all other <math>x \in \left( 0, 1 \right) \backslash \{ x_C \}</math>, there are more than one solutions with one solution <math>\theta = 60^\circ</math> and at least another solution. | ||
+ | |||
+ | Given that function <math>f \left( \theta ; x \right)</math> is differentiable, the above condition is equivalent to the first-order-condition | ||
+ | <cmath> | ||
+ | \[ | ||
+ | \frac{\partial f \left( \theta ; x_C \right) }{\partial \theta} \bigg|_{\theta = 60^\circ} = 0 . | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | Calculating derivatives in this equation, we get | ||
+ | <cmath> | ||
+ | \[ | ||
+ | - \left( \frac{\sqrt{3}}{2} - \sqrt{3} x_C \right) \frac{\cos 60^\circ}{\sin^2 60^\circ} | ||
+ | + x_C \frac{\sin 60^\circ}{\cos^2 60^\circ} | ||
+ | = 0 . | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | By solving this equation, we get | ||
+ | <cmath> | ||
+ | \[ | ||
+ | x_C = \frac{1}{8} . | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | Plugging this into Equation (1), we get the <math>y</math>-coordinate of point <math>C</math>: | ||
+ | <cmath> | ||
+ | \[ | ||
+ | y_C = \frac{3 \sqrt{3}}{8} . | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | Therefore, | ||
+ | \begin{align*} | ||
+ | OC^2 & = x_C^2 + y_C^2 \ | ||
+ | & = \frac{7}{16} . | ||
+ | \end{align*} | ||
+ | |||
+ | |||
+ | Therefore, the answer is <math>7 + 16 = \boxed{\textbf{(23) }}</math>. | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/QwLBBzHFPNE | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) |
Revision as of 16:19, 9 February 2024
Let and be points in the coordinate plane. Let be the family of segments of unit length lying in the first quadrant with on the -axis and on the -axis. There is a unique point on distinct from and that does not belong to any segment from other than . Then , where and are relatively prime positive integers. Find .
Contents
[hide]Solution 1
By Furaken
Let . this is sus, furaken randomly guessed C and proceeded to prove it works Draw a line through intersecting the -axis at and the -axis at . We shall show that , and that equality only holds when and .
Let . Draw perpendicular to the -axis and perpendicular to the -axis as shown in the diagram. Then By some inequality (i forgor its name), We know that . Thus . Equality holds if and only if which occurs when . Guess what, happens to be , thus and . Thus, is the only segment in that passes through . Finally, we calculate , and the answer is . ~Furaken
Solution 2
When , the limit of
~Bluesoul
Solution 3
The equation of line is \[ y = \frac{\sqrt{3}}{2} x - \sqrt{3} x. \hspace{1cm} (2) \]
The position of line can be characterized by , denoted as . Thus, the equation of line is \[ y = \sin \theta - \tan \theta \cdot x . \hspace{1cm} (2) \]
Solving (1) and (2), the -coordinate of the intersecting point of lines and satisfies the following equation: \[ \frac{\frac{\sqrt{3}}{2} - \sqrt{3} x}{\sin \theta} + \frac{x}{\cos \theta} = 1 . \hspace{1cm} (1) \] We denote the L.H.S. as .
We observe that for all . Therefore, the point that this problem asks us to find can be equivalently stated in the following way:
We interpret Equation (1) as a parameterized equation that is a tuning parameter and is a variable that shall be solved and expressed in terms of . In Equation (1), there exists a unique , denoted as (-coordinate of point ), such that the only solution is . For all other , there are more than one solutions with one solution and at least another solution.
Given that function is differentiable, the above condition is equivalent to the first-order-condition
Calculating derivatives in this equation, we get
By solving this equation, we get
Plugging this into Equation (1), we get the -coordinate of point :
Therefore,
Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)