Difference between revisions of "2024 AIME II Problems/Problem 12"
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</asy> | </asy> | ||
− | Let <math>C = (\tfrac18,\tfrac{3\sqrt3}8)</math>. <s> | + | Let <math>C = (\tfrac18,\tfrac{3\sqrt3}8)</math>. <s> This is sus, furaken randomly guessed C and proceeded to prove it works</s> Draw a line through <math>C</math> intersecting the <math>x</math>-axis at <math>A'</math> and the <math>y</math>-axis at <math>B'</math>. We shall show that <math>A'B' \ge 1</math>, and that equality only holds when <math>A'=A</math> and <math>B'=B</math>. |
Let <math>\theta = \angle OA'C</math>. Draw <math>CD</math> perpendicular to the <math>x</math>-axis and <math>CE</math> perpendicular to the <math>y</math>-axis as shown in the diagram. Then | Let <math>\theta = \angle OA'C</math>. Draw <math>CD</math> perpendicular to the <math>x</math>-axis and <math>CE</math> perpendicular to the <math>y</math>-axis as shown in the diagram. Then | ||
<cmath>8A'B' = 8CA' + 8CB' = \frac{3\sqrt3}{\sin\theta} + \frac{1}{\cos\theta}</cmath> | <cmath>8A'B' = 8CA' + 8CB' = \frac{3\sqrt3}{\sin\theta} + \frac{1}{\cos\theta}</cmath> | ||
− | By some inequality ( | + | By some inequality (I forgot its name), |
<cmath>\left(\frac{3\sqrt3}{\sin\theta} + \frac{1}{\cos\theta}\right) \cdot \left(\frac{3\sqrt3}{\sin\theta} + \frac{1}{\cos\theta}\right) \cdot (\sin^2\theta + \cos^2\theta) \ge (3+1)^3 = 64</cmath> | <cmath>\left(\frac{3\sqrt3}{\sin\theta} + \frac{1}{\cos\theta}\right) \cdot \left(\frac{3\sqrt3}{\sin\theta} + \frac{1}{\cos\theta}\right) \cdot (\sin^2\theta + \cos^2\theta) \ge (3+1)^3 = 64</cmath> | ||
We know that <math>\sin^2\theta + \cos^2\theta = 1</math>. Thus <math>\tfrac{3\sqrt3}{\sin\theta} + \tfrac{1}{\cos\theta} \ge 8</math>. Equality holds if and only if | We know that <math>\sin^2\theta + \cos^2\theta = 1</math>. Thus <math>\tfrac{3\sqrt3}{\sin\theta} + \tfrac{1}{\cos\theta} \ge 8</math>. Equality holds if and only if | ||
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</cmath> | </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>. |
We observe that <math>f \left( 60^\circ ; x \right) = 1</math> for all <math>x</math>. | We observe that <math>f \left( 60^\circ ; x \right) = 1</math> for all <math>x</math>. | ||
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- \left( \frac{\sqrt{3}}{2} - \sqrt{3} x_C \right) \frac{\cos 60^\circ}{\sin^2 60^\circ} | - \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} | + x_C \frac{\sin 60^\circ}{\cos^2 60^\circ} | ||
− | = 0 | + | = 0. |
\] | \] | ||
</cmath> | </cmath> |
Revision as of 04:14, 12 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.