Difference between revisions of "2024 AIME II Problems/Problem 12"

Revision as of 18:02, 9 February 2024

Let $O(0,0),A(\tfrac{1}{2},0),$ and $B(0,\tfrac{\sqrt{3}}{2})$ 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 forgor 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

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

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

~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)

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 $C$ 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 so 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.)

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