Difference between revisions of "2024 AIME II Problems/Problem 12"
(→Solution 3) |
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==Solution 3== | ==Solution 3== | ||
− | The equation of line <math>AB</math> is | + | The equation of line <math>AB</math> is <cmath> |
\[ | \[ | ||
y = \frac{\sqrt{3}}{2} x - \sqrt{3} x. \hspace{1cm} (2) | y = \frac{\sqrt{3}}{2} x - \sqrt{3} x. \hspace{1cm} (2) | ||
\] | \] | ||
+ | |||
+ | </cmath> | ||
The position of line <math>PQ</math> can be characterized by <math>\angle QPO</math>, denoted as <math>\theta</math>. | 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 | Thus, the equation of line <math>PQ</math> is | ||
+ | |||
+ | <cmath> | ||
\[ | \[ | ||
y = \sin \theta - \tan \theta \cdot x . \hspace{1cm} (2) | y = \sin \theta - \tan \theta \cdot x . \hspace{1cm} (2) | ||
\] | \] | ||
+ | </cmath> | ||
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: | 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: | ||
+ | |||
+ | <cmath> | ||
\[ | \[ | ||
\frac{\frac{\sqrt{3}}{2} - \sqrt{3} x}{\sin \theta} | \frac{\frac{\sqrt{3}}{2} - \sqrt{3} x}{\sin \theta} | ||
Line 63: | Line 70: | ||
= 1 . \hspace{1cm} (1) | = 1 . \hspace{1cm} (1) | ||
\] | \] | ||
+ | </cmath> | ||
+ | |||
We denote the L.H.S. as <math>f \left( \theta ; x \right)</math>. | We denote the L.H.S. as <math>f \left( \theta ; x \right)</math>. | ||
Revision as of 03:19, 10 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 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
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.