Difference between revisions of "2024 AIME II Problems/Problem 12"
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so <math>OC^2 = \frac{1}{64} + \frac{27}{64} = \frac{7}{16}</math>, the answer is <math>\boxed{023}</math>. | so <math>OC^2 = \frac{1}{64} + \frac{27}{64} = \frac{7}{16}</math>, the answer is <math>\boxed{023}</math>. | ||
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+ | ==Video Solution== | ||
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+ | https://youtu.be/914687Yv6SY?si=tc6XfoOIHu0gu6AL | ||
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+ | (no calculus) | ||
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+ | ~MathProblemSolvingSkills.com | ||
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==Video Solution== | ==Video Solution== |
Revision as of 16:06, 8 April 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)
Solution 4 (coordinate bash)
Let be a segment in with x-intercept and y-intercept . We can write as
In , and . To find the x-coordinate of , we substitute these into the equation for and get
Solution 5 (small perturb)
Let's move a little bit from to , then must move to to keep . intersects with at . Pick points and on and such that , , we have . Since is very small, , , so , , by similarity, . So the coordinates of is .
so , the answer is .
Video Solution
https://youtu.be/914687Yv6SY?si=tc6XfoOIHu0gu6AL
(no calculus)
~MathProblemSolvingSkills.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
I think there is such a geometry way: Let pass through while point is on the outside of line segment and point is in between and . We aim to show is longer than . Now since is the altitude of triangle yet just a cevian on the base of triangle (thus making the height shorter than ), it suffices to show the area of triangle is bigger than that of triangle . To do this, we compare these two triangles (let intersect at point ), and we just want to show . This is trivial by similarity ratios. ~gougutheorem
Thanks! Now we know that it's possible to solve the AIME problem with only geometry. ~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.