Difference between revisions of "2024 AIME II Problems/Problem 12"
(grammar) |
(add a solution) |
||
Line 115: | Line 115: | ||
& = \frac{7}{16} . | & = \frac{7}{16} . | ||
\end{align*} | \end{align*} | ||
− | |||
Therefore, the answer is <math>7 + 16 = \boxed{\textbf{(23) }}</math>. | Therefore, the answer is <math>7 + 16 = \boxed{\textbf{(23) }}</math>. | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | == Solution 4 (coordinate bash) == | ||
+ | |||
+ | Let <math>s</math> be a segment in <math>\mathcal{F}</math> with x-intercept <math>a</math> and y-intercept <math>b</math>. We can write <math>s</math> as | ||
+ | \begin{align*} | ||
+ | \frac{x}{a} + \frac{y}{b} &= 1 \ | ||
+ | y &= b(1 - \frac{x}{a}). | ||
+ | \end{align*} | ||
+ | Let the unique point in the first quadrant <math>(x, y)</math> lie on <math>s</math> and no other segment in <math>\mathcal{F}</math>. We can find <math>x</math> by solving | ||
+ | <cmath> | ||
+ | b(1 - \frac{x}{a}) = (b + db)(1 - \frac{x}{a + da}) | ||
+ | </cmath> | ||
+ | and taking the limit as <math>da, db \to 0</math>. Since <math>s</math> has length <math>1</math>, <math>a^2 + b^2 = 1^2</math> by the Pythagorean theorem. Solving this for <math>db</math>, we get | ||
+ | \begin{align*} | ||
+ | a^2 + b^2 &= 1 \ | ||
+ | b^2 &= 1 - a^2 \ | ||
+ | \frac{db^2}{da} &= \frac{d(1 - a^2)}{da} \ | ||
+ | 2a\frac{db}{da} &= -2a \ | ||
+ | db &= -\frac{a}{b}da. | ||
+ | \end{align*} | ||
+ | After we substitute <math>db = -\frac{a}{b}da</math>, the equation for <math>x</math> becomes | ||
+ | <cmath> | ||
+ | b(1 - \frac{x}{a}) = (b -\frac{a}{b} da)(1 - \frac{x}{a + da}). | ||
+ | </cmath> | ||
+ | |||
+ | In <math>\overline{AB}</math>, <math>a = \frac{1}{2}</math> and <math>b = \frac{\sqrt{3}}{2}</math>. To find the x-coordinate of <math>C</math>, we substitute these into the equation for <math>x</math> and get | ||
+ | \begin{align*} | ||
+ | \frac{\sqrt{3}}{2}(1 - \frac{x}{\frac{1}{2}}) &= (\frac{\sqrt{3}}{2} - \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} da)(1 - \frac{x}{\frac{1}{2} + da}) \ | ||
+ | \frac{\sqrt{3}}{2}(1 - 2x) &= (\frac{\sqrt{3}}{2} - \frac{da}{\sqrt{3}})(1 - \frac{x}{\frac{1 + 2da}{2}}) \ | ||
+ | \frac{\sqrt{3}}{2} - \sqrt{3}x &= \frac{3 - 2da}{2\sqrt{3}}(1 - \frac{2x}{1 + 2da}) \ | ||
+ | \frac{\sqrt{3}}{2} - \sqrt{3}x &= \frac{3 - 2da}{2\sqrt{3}} \cdot \frac{1 + 2da - 2x}{1 + 2da} \ | ||
+ | \frac{\sqrt{3}}{2} - \sqrt{3}x &= \frac{3 + 6da - 6x - 2da - 4da^2 + 4xda}{2\sqrt{3} + 4\sqrt{3}da} \ | ||
+ | (\frac{\sqrt{3}}{2} - \sqrt{3}x)(2\sqrt{3} + 4\sqrt{3}da) &= 3 + 6da - 6x - 2da - 4da^2 + 4xda \ | ||
+ | 3 + 6da - 6x - 12xda &= 3 + 4da - 6x - 4da^2 + 4xda \ | ||
+ | 2da &= -4da^2 + 16xda \ | ||
+ | 16xda &= 2da + 4da^2 \ | ||
+ | x &= \frac{da + 2da^2}{8da}. | ||
+ | \end{align*} | ||
+ | We take the limit as <math>da \to 0</math> to get | ||
+ | <cmath> | ||
+ | x = \lim_{da \to 0} \frac{da + 2da^2}{8da} = \lim_{da \to 0} \frac{1 + 2da}{8} = \frac{1}{8}. | ||
+ | </cmath> | ||
+ | We substitute <math>x = \frac{1}{8}</math> into the equation for <math>\overline{AB}</math> to find the y-coordinate <math>y</math> of <math>C</math>: | ||
+ | <cmath> | ||
+ | y = b(1 - \frac{x}{a}) = \frac{\sqrt{3}}{2}(1 - \frac{\frac{1}{8}}{\frac{1}{2}}) = \frac{3\sqrt{3}}{8}. | ||
+ | </cmath> | ||
+ | The problem asks for | ||
+ | <cmath> | ||
+ | OC^2 = (\frac{1}{8})^2 + (\frac{3\sqrt{3}}{8})^2 = \frac{7}{16} = \frac{p}{q}, | ||
+ | </cmath> | ||
+ | so <math>p + q = 7 + 16 = \boxed{023}</math>. | ||
==Video Solution== | ==Video Solution== |
Revision as of 04:22, 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)
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
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.