Difference between revisions of "2024 AIME II Problems/Problem 12"
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Question== | ||
+ | <asy> | ||
+ | pair O=(0,0); | ||
+ | pair X=(1,0); | ||
+ | pair Y=(0,1); | ||
+ | pair A=(0.5,0); pair B=(0,sin(pi/3)); | ||
+ | dot(O); | ||
+ | dot(X); | ||
+ | dot(Y); dot(A); dot(B); | ||
+ | draw(X--O--Y); | ||
+ | draw(A--B); | ||
+ | label("$B$", B, W); | ||
+ | pair P=(0.5, sin(pi/3)); | ||
+ | dot(P); | ||
+ | draw(A--P--B); | ||
+ | label("$A$", A, S); | ||
+ | label("$O$", O, SW); | ||
+ | pair C=(1/8,3*sqrt(3)/8); | ||
+ | dot(C); | ||
+ | label("$C$", C, SW); | ||
+ | draw(C--P); label("$P$", P, NE); | ||
+ | </asy> | ||
+ | Let <math>C</math> be a fixed point in the first quadrant. Let <math>A</math> be a point on the positive <math>x</math>-axis and <math>B</math> be a point on the positive <math>y</math>-axis so that <math>AB</math> passes through <math>C</math> and the length of <math>AB</math> is minimal. Let <math>P</math> be the point such that <math>OAPB</math> is a rectangle. Prove that <math>PC \perp AB</math>. (One can solve this through algebra/calculus bash, but I'm trying to find a solution that mainly uses geometry.) |
Revision as of 18:02, 9 February 2024
Let and be points in the coordinate plane. Let be the family of segments of unit length lying in the first quadrant with on the -axis and on the -axis. There is a unique point on distinct from and that does not belong to any segment from other than . Then , where and are relatively prime positive integers. Find .
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 forgor 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
When , the limit of
~Bluesoul
Solution 3
The equation of line is \[ y = \frac{\sqrt{3}}{2} x - \sqrt{3} x. \hspace{1cm} (2) \]
The position of line can be characterized by , denoted as . Thus, the equation of line is \[ y = \sin \theta - \tan \theta \cdot x . \hspace{1cm} (2) \]
Solving (1) and (2), the -coordinate of the intersecting point of lines and satisfies the following equation: \[ \frac{\frac{\sqrt{3}}{2} - \sqrt{3} x}{\sin \theta} + \frac{x}{\cos \theta} = 1 . \hspace{1cm} (1) \] 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)
Question
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 so 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.)