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Difference between revisions of "2024 AIME II Problems/Problem 12"
Line 42: Line 42:
  
 
~Bluesoul
 
~Bluesoul
 +
 +
==Solution 3==
 +
 +
The equation of line <math>AB</math> is
 +
\[
 +
y = \frac{\sqrt{3}}{2} x - \sqrt{3} x.  \hspace{1cm} (2)
 +
\]
 +
 +
The position of line  <math>PQ</math> can be characterized by <math>\angle QPO</math>, denoted as <math>\theta</math>.
 +
Thus, the equation of line <math>PQ</math> is
 +
\[
 +
y = \sin \theta - \tan \theta \cdot x . \hspace{1cm} (2)
 +
\]
 +
 +
Solving (1) and (2), the <math>x</math>-coordinate of the intersecting point of lines <math>AB</math> and <math>PQ</math> 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 <math>f \left( \theta ; x \right)</math>.
 +
 +
We observe that <math>f \left( 60^\circ ; x \right) = 1</math> for all <math>x</math>.
 +
Therefore, the point <math>C</math> that this problem asks us to find can be equivalently stated in the following way:
 +
 +
We interpret Equation (1) as a parameterized equation that <math>x</math> is a tuning parameter and <math>\theta</math> is a variable that shall be solved and expressed in terms of <math>x</math>.
 +
In Equation (1), there exists a unique <math>x \in \left( 0, 1 \right)</math>, denoted as <math>x_C</math> (<math>x</math>-coordinate of point <math>C</math>), such that the only solution is <math>\theta = 60^\circ</math>. For all other <math>x \in \left( 0, 1 \right) \backslash \{ x_C \}</math>, there are more than one solutions with one solution <math>\theta = 60^\circ</math> and at least another solution.
 +
 +
Given that function <math>f \left( \theta ; x \right)</math> is differentiable, the above condition is equivalent to the first-order-condition
 +
<cmath>
 +
\[
 +
\frac{\partial f \left( \theta ; x_C \right) }{\partial \theta} \bigg|_{\theta = 60^\circ} = 0 .
 +
\]
 +
</cmath>
 +
 +
Calculating derivatives in this equation, we get
 +
<cmath>
 +
\[
 +
- \left( \frac{\sqrt{3}}{2} - \sqrt{3} x_C \right) \frac{\cos 60^\circ}{\sin^2 60^\circ}
 +
+ x_C \frac{\sin 60^\circ}{\cos^2 60^\circ}
 +
= 0  .
 +
\]
 +
</cmath>
 +
 +
By solving this equation, we get
 +
<cmath>
 +
\[
 +
x_C = \frac{1}{8} .
 +
\]
 +
</cmath>
 +
 +
Plugging this into Equation (1), we get the <math>y</math>-coordinate of point <math>C</math>:
 +
<cmath>
 +
\[
 +
y_C = \frac{3 \sqrt{3}}{8} .
 +
\]
 +
</cmath>
 +
 +
Therefore,
 +
\begin{align*}
 +
OC^2 & = x_C^2 + y_C^2 \\
 +
& = \frac{7}{16} .
 +
\end{align*}
 +
 +
 +
Therefore, the answer is <math>7 + 16 = \boxed{\textbf{(23) }}</math>.
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 +
==Video Solution==
 +
 +
https://youtu.be/QwLBBzHFPNE
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Revision as of 17:19, 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)

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