Difference between revisions of "2024 AIME II Problems/Problem 12"
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<math>y=-(\tan \theta) x+\sin \theta=-\sqrt{3}x+\frac{\sqrt{3}}{2}, x=\frac{\sqrt{3}-2\sin \theta}{2\sqrt{3}-2\tan \theta}</math> | <math>y=-(\tan \theta) x+\sin \theta=-\sqrt{3}x+\frac{\sqrt{3}}{2}, x=\frac{\sqrt{3}-2\sin \theta}{2\sqrt{3}-2\tan \theta}</math> | ||
− | + | Now, we want to find <math>\lim_{\theta\to\frac{\pi}{3}}\frac{\sqrt{3}-2\sin \theta}{2\sqrt{3}-2\tan \theta}</math>. By L'Hôpital's rule, we get <math>\lim_{\theta\to\frac{\pi}{3}}\frac{\sqrt{3}-2\sin \theta}{2\sqrt{3}-2\tan \theta}=\lim_{\theta\to\frac{\pi}{3}}cos^3{x}=\frac{1}{8}</math>. This means that <math>y=\frac{3\sqrt{3}}{8}\implies OC^2=\frac{7}{16}</math>, so we get <math>\boxed{023}</math>. | |
~Bluesoul | ~Bluesoul |
Revision as of 16:31, 11 February 2024
Contents
[hide]Problem
Let
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 forgot 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
Now, we want to find . By L'Hôpital's rule, we get . This means that , so we get .
~Bluesoul
Solution 3
The equation of line is
The position of line can be characterized by , denoted as . Thus, the equation of line is
Solving (1) and (2), the -coordinate of the intersecting point of lines and satisfies the following equation:
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)
Query
Let be a fixed point in the first quadrant. Let be a point on the positive -axis and be a point on the positive -axis such that passes through and the length of is minimal. Let be the point such that is a rectangle. Prove that . (One can solve this through algebra/calculus bash, but I'm trying to find a solution that mainly uses geometry. If you know such a solution, write it here on this wiki page.) ~Furaken
See also
2024 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.