Difference between revisions of "2024 AIME II Problems/Problem 12"
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Point <math>D = C</math> is a unique point on <math>AB</math> if the ellipse is tangent to the line <math>AB.</math> | Point <math>D = C</math> is a unique point on <math>AB</math> if the ellipse is tangent to the line <math>AB.</math> | ||
− | In this case in point <math> | + | In this case in point <math>C</math> we get <math>\frac{dy}{dx} = \frac{dy}{d \alpha} : \frac{dx}{d \alpha} = - k \cot \alpha.</math> |
− | The tangent of line <math>AB</math> is <math>- tan \alpha \implies k = \tan^2 \alpha.</math> | + | The tangent of the line <math>AB</math> is <math>- \tan \alpha \implies k = \tan^2 \alpha.</math> |
+ | |||
+ | For point <math>D</math> we get <math>k = \tan^2 \theta.</math> | ||
+ | |||
+ | <cmath>\tan \angle DOX = \tan \varphi = \frac {D_y}{D_x} = \frac{k \sin \theta}{k+1} : \frac{\cos \theta}{k+1} = k \cdot \tan \theta = \tan^3 \theta.</cmath> | ||
+ | |||
+ | For the line <math>AB \tan \theta = \frac {B_y}{A_x} = \sqrt{3} \implies k = 3 \implies</math> | ||
+ | <math>C = (\frac{\cos \alpha}{k+1}, \frac{k \sin \alpha}{k+1} = (\frac{1}{\sqrt{1+ \tan^2 \alpha} \cdot (k+1)}, \frac{k \tan \alpha}{\sqrt{1+ \tan^2 \alpha} \cdot (k+1)}) =</math> | ||
+ | <math> (\frac{1}{\sqrt{1+ k} \cdot (k+1)}, \frac{k \cdot k}{\sqrt{1+ k} \cdot (k+1)}) = \frac{1}{8}, \frac{3 \sqrt{3}}{8}.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==Video Solution== | ==Video Solution== |
Revision as of 09:41, 1 September 2024
Contents
[hide]Problem
Let
Solution 1 (completely no calculus required)
Begin by finding the equation of the line : Now, consider the general equation of all lines that belong to . Let be located at and be located at . With these assumptions, we may arrive at the equation . However, a critical condition that must be satisfied by our parameters is that , since the length of .
Here's the golden trick that resolves the problem: we wish to find some point along such that passes through if . It's not hard to convince oneself of this, since the property implies that if , then .
We should now try to relate the point to some value of . This is accomplished by finding the intersection of two lines:
Where we have also used the fact that , which follows nicely from .
Square both sides and go through some algebraic manipulations to arrive at
Note how is a solution to this polynomial, and it is logically so. If we found the set of intersections consisting of line segment with an identical copy of itself, every single point on the line (all values) should satisfy the equation. Thus, we can perform polynomial division to eliminate the extraneous solution .
Remember our original goal. It was to find an value such that is the only valid solution. Therefore, we can actually plug in back into the equation to look for values of such that the relation is satisfied, then eliminate undesirable answers. This is easily factored, allowing us to determine that . The latter root is not our answer, since on line , , the horizontal line segment running from to covers that point. From this, we see that is the only possible candidate.
Going back to line , plugging in yields . The distance from the origin is then given by . That number squared is , so the answer is .
~Installhelp_hex
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 .
Solution 6(trig identities and questionable rigidity)
Let's try to find the general form of a line that is in based on what angle it makes with the x-axis, , and so its slope is and due to us knowing its y-intercept we know that our line has form
Now we can analyze the system of equations made by and , this gives us that
We can proceed to simplify our expression further:
Seeing that there are only valid solutions when is acute(all that is allowed anyways) and when since one of the expressions in our simplified solution will equal . Since there is only one intersection point for every and vice versa in the appropriate domain and range(we can easily prove this by contradiction), we know that the missing element of the range(the points) must correspond with the excluded value. The x-coordinate of which which can be evaluated by taking the limit of our expression as goes to which is regardless of the direction we approach from. The corresponding is and using the distance formula gives us as our answer.
While this solution may seem long all of these steps come naturally.
~SailS
Solution 7
Denote Then
Let be the point with property
So locus of points is the ellipse with semiaxes and
Point is a unique point on if the ellipse is tangent to the line
In this case in point we get
The tangent of the line is
For point we get
For the line
vladimir.shelomovskii@gmail.com, vvsss
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.