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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>
  
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When <math>\theta=\frac{\pi}{3}</math>, the limit of <math>x=\frac{1}{8}\implies y=\frac{3\sqrt{3}}{8}\implies \frac{7}{16}=OC^2, \boxed{023}</math>
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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

Problem

Let \(O=(0,0)\), \(A=\left(\tfrac{1}{2},0\right)\), and \(B=\left(0,\tfrac{\sqrt{3}}{2}\right)\) be points in the coordinate plane. Let \(\mathcal{F}\) be the family of segments \(\overline{PQ}\) of unit length lying in the first quadrant with \(P\) on the \(x\)-axis and \(Q\) on the \(y\)-axis. There is a unique point \(C\) on \(\overline{AB}\), distinct from \(A\) and \(B\), that does not belong to any segment from \(\mathcal{F}\) other than \(\overline{AB}\). Then \(OC^2=\tfrac{p}{q}\), where \(p\) and \(q\) are relatively prime positive integers. Find \(p+q\).

Solution 1

By Furaken [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); label("$A'$", A, S); label("$O$", O, SW); pair C=(1/8,3*sqrt(3)/8); dot(C); pair D=(1/8,0); dot(D); pair E=(0,3*sqrt(3)/8); dot(E); label("$C$", C, NE); label("$D$", D, S); label("$E$", E, W); draw(D--C--E); [/asy]

Let $C = (\tfrac18,\tfrac{3\sqrt3}8)$. this is sus, furaken randomly guessed C and proceeded to prove it works Draw a line through $C$ intersecting the $x$-axis at $A'$ and the $y$-axis at $B'$. We shall show that $A'B' \ge 1$, and that equality only holds when $A'=A$ and $B'=B$.

Let $\theta = \angle OA'C$. Draw $CD$ perpendicular to the $x$-axis and $CE$ perpendicular to the $y$-axis as shown in the diagram. Then \[8A'B' = 8CA' + 8CB' = \frac{3\sqrt3}{\sin\theta} + \frac{1}{\cos\theta}\] By some inequality (i forgot its name), \[\left(\frac{3\sqrt3}{\sin\theta} + \frac{1}{\cos\theta}\right) \cdot \left(\frac{3\sqrt3}{\sin\theta} + \frac{1}{\cos\theta}\right) \cdot (\sin^2\theta + \cos^2\theta) \ge (3+1)^3 = 64\] We know that $\sin^2\theta + \cos^2\theta = 1$. Thus $\tfrac{3\sqrt3}{\sin\theta} + \tfrac{1}{\cos\theta} \ge 8$. Equality holds if and only if \[\frac{3\sqrt3}{\sin\theta} : \frac{1}{\cos\theta} = \frac{3\sqrt3}{\sin\theta} : \frac{1}{\cos\theta} = \sin^2\theta : \cos^2\theta\] which occurs when $\theta=\tfrac\pi3$. Guess what, $\angle OAB$ happens to be $\tfrac\pi3$, thus $A'=A$ and $B'=B$. Thus, $AB$ is the only segment in $\mathcal{F}$ that passes through $C$. Finally, we calculate $OC^2 = \tfrac1{64} + \tfrac{27}{64} = \tfrac7{16}$, and the answer is $\boxed{023}$. ~Furaken

Solution 2

$y=-(\tan \theta) x+\sin \theta=-\sqrt{3}x+\frac{\sqrt{3}}{2}, x=\frac{\sqrt{3}-2\sin \theta}{2\sqrt{3}-2\tan \theta}$

Now, we want to find $\lim_{\theta\to\frac{\pi}{3}}\frac{\sqrt{3}-2\sin \theta}{2\sqrt{3}-2\tan \theta}$. By L'Hôpital's rule, we get $\lim_{\theta\to\frac{\pi}{3}}\frac{\sqrt{3}-2\sin \theta}{2\sqrt{3}-2\tan \theta}=\lim_{\theta\to\frac{\pi}{3}}cos^3{x}=\frac{1}{8}$. This means that $y=\frac{3\sqrt{3}}{8}\implies OC^2=\frac{7}{16}$, so we get $\boxed{023}$.

~Bluesoul

Solution 3

The equation of line $AB$ is \[ y = \frac{\sqrt{3}}{2} x - \sqrt{3} x. \hspace{1cm} (2) \]

The position of line $PQ$ can be characterized by $\angle QPO$, denoted as $\theta$. Thus, the equation of line $PQ$ is

\[ y = \sin \theta - \tan \theta \cdot x . \hspace{1cm} (2) \]

Solving (1) and (2), the $x$-coordinate of the intersecting point of lines $AB$ and $PQ$ 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 $f \left( \theta ; x \right)$.

We observe that $f \left( 60^\circ ; x \right) = 1$ for all $x$. Therefore, the point $C$ that this problem asks us to find can be equivalently stated in the following way:

We interpret Equation (1) as a parameterized equation that $x$ is a tuning parameter and $\theta$ is a variable that shall be solved and expressed in terms of $x$. In Equation (1), there exists a unique $x \in \left( 0, 1 \right)$, denoted as $x_C$ ($x$-coordinate of point $C$), such that the only solution is $\theta = 60^\circ$. For all other $x \in \left( 0, 1 \right) \backslash \{ x_C \}$, there are more than one solutions with one solution $\theta = 60^\circ$ and at least another solution.

Given that function $f \left( \theta ; x \right)$ is differentiable, the above condition is equivalent to the first-order-condition \[ \frac{\partial f \left( \theta ; x_C \right) }{\partial \theta} \bigg|_{\theta = 60^\circ} = 0 . \]

Calculating derivatives in this equation, we get \[ - \left( \frac{\sqrt{3}}{2} - \sqrt{3} x_C \right) \frac{\cos 60^\circ}{\sin^2 60^\circ} + x_C \frac{\sin 60^\circ}{\cos^2 60^\circ} = 0 . \]

By solving this equation, we get \[ x_C = \frac{1}{8} . \]

Plugging this into Equation (1), we get the $y$-coordinate of point $C$: \[ y_C = \frac{3 \sqrt{3}}{8} . \]

Therefore, \begin{align*} OC^2 & = x_C^2 + y_C^2 \\ & = \frac{7}{16} . \end{align*}


Therefore, the answer is $7 + 16 = \boxed{\textbf{(23) }}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution

https://youtu.be/QwLBBzHFPNE

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Query

[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 $C$ be a fixed point in the first quadrant. Let $A$ be a point on the positive $x$-axis and $B$ be a point on the positive $y$-axis such that $AB$ passes through $C$ and the length of $AB$ is minimal. Let $P$ be the point such that $OAPB$ is a rectangle. Prove that $PC \perp AB$. (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

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