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.