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

Latest revision as of 11:34, 14 May 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 (completely no calculus required)

Begin by finding the equation of the line $\overline{AB}$ : $y= -\sqrt{3}x+\frac{\sqrt{3}}{2}$ Now, consider the general equation of all lines that belong to $\mathcal{F}$ . Let $P$ be located at $(a,0)$ and $Q$ be located at $(0,b)$ . With these assumptions, we may arrive at the equation $ay +bx =ab$ . However, a critical condition that must be satisfied by our parameters is that $a^2+b^2=1$ , since the length of $\overline{PQ}=1$ .

Here's the golden trick that resolves the problem: we wish to find some point $C$ along $\overline{AB}$ such that $\overline{PQ}$ passes through $C$ iff $a=\frac{1}{2}$ . It's not hard to convince oneself of this, since the property $a^2+b^2=1$ implies that if $a=\frac{1}{2}$ , then $\overline{PQ}=\overline{AB}$ .

We should now try to relate the point $C$ to some value of $a$ . This is accomplished by finding the intersection of two lines: $a(-\sqrt{3}x +\frac{\sqrt{3}}{2}) + x\sqrt{1-a^2} = a\sqrt{1-a^2}$

Where we have also used the fact that $b=\sqrt{1-a^2}$ , which follows nicely from $a^2+b^2 =1$ . $a(-\sqrt{3}x +\frac{\sqrt{3}}{2}) = (a-x)\sqrt{1-a^2}$

Square both sides and go through some algebraic manipulations to arrive at $-a^4 +2xa^3+(-4x^2+3x+\frac{1}{4})a^2-2xa+x^2=0$

Note how $a=\frac{1}{2}$ is a solution to this polynomial, and it is logically so. If we found the set of intersections consisting of line segment $\overline{AB}$ with an identical copy of itself, every single point on the line (all $x$ values) should satisfy the equation. Thus, we can perform polynomial division to eliminate the extraneous solution $a=\frac{1}{2}$ . $-a^3 + (2x-\frac{1}{2})a^2+(-4x^2+4x)a-2x^2=0$

Remember our original goal. It was to find an $x$ value such that $a=\frac{1}{2}$ is the only valid solution. Therefore, we can actually plug in $a=\frac{1}{2}$ back into the equation to look for values of $x$ such that the relation is satisfied, then eliminate undesirable answers. $16x^2-10x+1=0$ This is easily factored, allowing us to determine that $x=\frac{1}{8},\frac{1}{2}$ . The latter root is not our answer, since on line $\overline{AB}$ , $y(\frac{1}{2})=0$ , the horizontal line segment running from $(0,0)$ to $(1,0)$ covers that point. From this, we see that $x=\frac{1}{8}$ is the only possible candidate.

Going back to line $\overline{AB}, y= -\sqrt{3}x+\frac{\sqrt{3}}{2}$ , plugging in $x=\frac{1}{8}$ yields $y=\frac{3\sqrt{3}}{8}$ . The distance from the origin is then given by $\sqrt{\frac{1}{8^2}+(\frac{3\sqrt{3}}{8})^2} =\sqrt{\frac{7}{16}}$ . That number squared is $\frac{7}{16}$ , so the answer is $\boxed{023}$ .

~Installhelp_hex

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 $y = \frac{\sqrt{3}}{2} x - \sqrt{3} x. \hspace{1cm} (1)$

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)

Solution 4 (coordinate bash)

Let $s$ be a segment in $\mathcal{F}$ with x-intercept $a$ and y-intercept $b$ . We can write $s$ 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 $(x, y)$ lie on $s$ and no other segment in $\mathcal{F}$ . We can find $x$ by solving $b(1 - \frac{x}{a}) = (b + db)(1 - \frac{x}{a + da})$ and taking the limit as $da, db \to 0$ . Since $s$ has length $1$ , $a^2 + b^2 = 1^2$ by the Pythagorean theorem. Solving this for $db$ , 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 $db = -\frac{a}{b}da$ , the equation for $x$ becomes $b(1 - \frac{x}{a}) = (b -\frac{a}{b} da)(1 - \frac{x}{a + da}).$

In $\overline{AB}$ , $a = \frac{1}{2}$ and $b = \frac{\sqrt{3}}{2}$ . To find the x-coordinate of $C$ , we substitute these into the equation for $x$ 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 $da \to 0$ to get $x = \lim_{da \to 0} \frac{da + 2da^2}{8da} = \lim_{da \to 0} \frac{1 + 2da}{8} = \frac{1}{8}.$ We substitute $x = \frac{1}{8}$ into the equation for $\overline{AB}$ to find the y-coordinate of $C$ : $y = b(1 - \frac{x}{a}) = \frac{\sqrt{3}}{2}(1 - \frac{\frac{1}{8}}{\frac{1}{2}}) = \frac{3\sqrt{3}}{8}.$ The problem asks for $OC^2 = x^2 + y^2 = (\frac{1}{8})^2 + (\frac{3\sqrt{3}}{8})^2 = \frac{7}{16} = \frac{p}{q},$ so $p + q = 7 + 16 = \boxed{023}$ .

Solution 5 (small perturb)

$[asy] pair O=(0,0); pair X=(1,0); pair Y=(0,1); pair A=(0.5,0); pair B=(0,sin(pi/3)); pair A1=(0.6,0); pair B1=(0,0.8); pair A2=(0.575,0.04); pair B2=(0.03,0.816); dot(O); dot(X); dot(Y); dot(A); dot(B); dot(A1); dot(B1); dot(A2); dot(B2); draw(X--O--Y); draw(A--B); draw(A1--B1); draw(A--A2); draw(B1--B2); label("$B$", B, W); label("$A$", A, S); label("$B_1$", B1, SW); label("$A_1$", A1, S); label("$B_2$", B2, E); label("$A_2$", A2, NE); label("$O$", O, SW); pair C=(0.18,0.56); label("$C$", C, E); dot(C); [/asy]$

Let's move a little bit from $A$ to $A_1$ , then $B$ must move to $B_1$ to keep $A_1B_1 = 1$ . $AB$ intersects with $A_1B_1$ at $C$ . Pick points $A_2$ and $B_2$ on $CA_1$ and $CB$ such that $CA_2 = CA$ , $CB_2 = CB_1$ , we have $A_1A_2 = BB_2$ . Since $AA_1$ is very small, $\angle CA_1A \approx 60^\circ$ , $\angle CBB_1 \approx 30^\circ$ , so $AA_2\approx \sqrt{3}A_1A_2$ , $B_1B_2 \approx \frac{1}{\sqrt{3}}BB_2$ , by similarity, $\frac{CA}{CB} \approx \frac{CA}{CB_2} = \frac{AA_2}{B_1B_2} = \frac{\sqrt{3}A_1A_2}{\frac{1}{\sqrt{3}}BB_2} = 3$ . So the coordinates of $C$ is $\left(\frac{1}{8}, \frac{3\sqrt{3}}{8}\right)$ .

so $OC^2 = \frac{1}{64} + \frac{27}{64} = \frac{7}{16}$ , the answer is $\boxed{023}$ .

Video Solution

https://youtu.be/914687Yv6SY?si=tc6XfoOIHu0gu6AL

(no calculus)

~MathProblemSolvingSkills.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

I think there is such a geometry way: Let $DE$ pass through $C$ while point $D$ is on the outside of line segment $OA$ and point $E$ is in between $O$ and $B$ . We aim to show $DE$ is longer than $AB$ . Now since $PC$ is the altitude of triangle $PAB$ yet just a cevian on the base $DE$ of triangle $PDE$ (thus making the height shorter than $PC$ ), it suffices to show the area of triangle $PDE$ is bigger than that of triangle $PAB$ . To do this, we compare these two triangles (let $DE$ intersect $PA$ at point $F$ ), and we just want to show $PF*AD > AF*AO$ . This is trivial by similarity ratios. ~gougutheorem

Thanks! Now we know that it's possible to solve the AIME problem with only geometry. ~Furaken

2024 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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Difference between revisions of "2024 AIME II Problems/Problem 12"

Latest revision as of 11:34, 14 May 2024

Contents

Problem

Solution 1 (completely no calculus required)

Solution 2

Solution 3

Solution 4 (coordinate bash)

Solution 5 (small perturb)

Video Solution

Video Solution

Query

See also

@@ Line 1: / Line 1: @@
-Let <math>O(0,0),A(\tfrac{1}{2},0),</math> and <math>B(0,\tfrac{\sqrt{3}}{2})</math> be points in the coordinate plane. Let <math>\mathcal{F}</math> be the family of segments <math>\overline{PQ}</math> of unit length lying in the first quadrant with <math>P</math> on the <math>x</math>-axis and <math>Q</math> on the <math>y</math>-axis. There is a unique point <math>C</math> on <math>\overline{AB},</math> distinct from <math>A</math> and <math>B,</math> that does not belong to any segment from <math>\mathcal{F}</math> other than <math>\overline{AB}</math>. Then <math>OC^2=\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.
+==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 (completely no calculus required)==
+Begin by finding the equation of the line <math>\overline{AB}</math>: <math>y= -\sqrt{3}x+\frac{\sqrt{3}}{2}</math> Now, consider the general equation of all lines that belong to <math>\mathcal{F}</math>. Let <math>P</math> be located at <math>(a,0)</math> and <math>Q</math> be located at <math>(0,b)</math>. With these assumptions, we may arrive at the equation <math>ay +bx =ab</math>. However, a critical condition that must be satisfied by our parameters is that <math>a^2+b^2=1</math>, since the length of <math>\overline{PQ}=1</math>.
+Here's the golden trick that resolves the problem: we wish to find some point <math>C</math> along <math>\overline{AB}</math> such that <math>\overline{PQ}</math> passes through <math>C</math> iff <math>a=\frac{1}{2}</math>. It's not hard to convince oneself of this, since the property <math>a^2+b^2=1</math> implies that if <math>a=\frac{1}{2}</math>, then <math>\overline{PQ}=\overline{AB}</math>.
+We should now try to relate the point <math>C</math> to some value of <math>a</math>. This is accomplished by finding the intersection of two lines: <cmath>
+\[
+    a(-\sqrt{3}x +\frac{\sqrt{3}}{2}) + x\sqrt{1-a^2} = a\sqrt{1-a^2}
+\]
+</cmath>
+Where we have also used the fact that <math>b=\sqrt{1-a^2}</math>, which follows nicely from <math>a^2+b^2 =1</math>.  <cmath>
+\[
+    a(-\sqrt{3}x +\frac{\sqrt{3}}{2})   = (a-x)\sqrt{1-a^2}
+\]
+</cmath>
+Square both sides and go through some algebraic manipulations to arrive at
+<cmath>
+\[
+    -a^4 +2xa^3+(-4x^2+3x+\frac{1}{4})a^2-2xa+x^2=0
+\]
+</cmath>
+Note how <math>a=\frac{1}{2}</math> is a solution to this polynomial, and it is logically so. If we found the set of intersections consisting of line segment <math>\overline{AB}</math> with an identical copy of itself, every single point on the line (all <math>x</math> values) should satisfy the equation. Thus, we can perform polynomial division to eliminate the extraneous solution <math>a=\frac{1}{2}</math>. <cmath>
+\[
+    -a^3 + (2x-\frac{1}{2})a^2+(-4x^2+4x)a-2x^2=0
+\]
+</cmath>
+Remember our original goal. It was to find an <math>x</math> value such that <math>a=\frac{1}{2}</math> is the only valid solution. Therefore, we can actually plug in <math>a=\frac{1}{2}</math> back into the equation to look for values of <math>x</math> such that the relation is satisfied, then eliminate undesirable answers.
+<cmath>
+\[
+x^2-10x+1=0
+\]
+</cmath>
+This is easily factored, allowing us to determine that <math>x=\frac{1}{8},\frac{1}{2}</math>. The latter root is not our answer, since on line <math>\overline{AB}</math>, <math>y(\frac{1}{2})=0</math>, the horizontal line segment running from <math>(0,0)</math> to <math>(1,0)</math> covers that point. From this, we see that <math>x=\frac{1}{8}</math> is the only possible candidate.
+Going back to line <math>\overline{AB}, y= -\sqrt{3}x+\frac{\sqrt{3}}{2}</math>, plugging in <math>x=\frac{1}{8} </math> yields <math>y=\frac{3\sqrt{3}}{8}</math>. The distance from the origin is then given by <math>\sqrt{\frac{1}{8^2}+(\frac{3\sqrt{3}}{8})^2} =\sqrt{\frac{7}{16}}</math>. That number squared is <math>\frac{7}{16}</math>, so the answer is <math>\boxed{023}</math>.
+~Installhelp_hex
+==Solution 2==
+<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>
+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
+==Solution 3==
+The equation of line <math>AB</math> is <cmath>
+\[
+y = \frac{\sqrt{3}}{2} x - \sqrt{3} x.  \hspace{1cm} (1)
+\]
+</cmath>
+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
+<cmath>
+\[
+y = \sin \theta - \tan \theta \cdot x . \hspace{1cm} (2)
+\]
+</cmath>
+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:
+<cmath>
+\[
+\frac{\frac{\sqrt{3}}{2} - \sqrt{3} x}{\sin \theta}
++ \frac{x}{\cos \theta}
+= 1 . \hspace{1cm} (1)
+\]
+</cmath>
+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)
+== 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} \\
+a\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 \\
++ 6da - 6x - 12xda &= 3 + 4da - 6x - 4da^2 + 4xda \\
+da &= -4da^2 + 16xda \\
+xda &= 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 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 = x^2 + y^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>.
+== Solution 5 (small perturb) ==
-==Solution 1==
-By Furaken
 <asy>
 pair O=(0,0);
@@ Line 8: / Line 198: @@
 pair Y=(0,1);
 pair A=(0.5,0); pair B=(0,sin(pi/3));
+pair A1=(0.6,0); pair B1=(0,0.8);
+pair A2=(0.575,0.04); pair B2=(0.03,0.816);
 dot(O);
 dot(X);
 dot(Y); dot(A); dot(B);
+dot(A1); dot(B1);
+dot(A2); dot(B2);
 draw(X--O--Y);
 draw(A--B);
-label("$B'$", B, W);
+draw(A1--B1);
-label("$A'$", A, S);
+draw(A--A2);
+draw(B1--B2);
+label("$B$", B, W);
+label("$A$", A, S);
+label("$B_1$", B1, SW);
+label("$A_1$", A1, S);
+label("$B_2$", B2, E);
+label("$A_2$", A2, NE);
 label("$O$", O, SW);
-pair C=(1/8,3*sqrt(3)/8);
+pair C=(0.18,0.56);
+label("$C$", C, E);
 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 <math>C = (\tfrac18,\tfrac{3\sqrt3}8)</math>. <s>this is sus, furaken randomly guessed C and proceeded to prove it works</s> Draw a line through <math>C</math> intersecting the <math>x</math>-axis at <math>A'</math> and the <math>y</math>-axis at <math>B'</math>. We shall show that <math>A'B' \ge 1</math>, and that equality only holds when <math>A'=A</math> and <math>B'=B</math>.
+Let's move a little bit from <math>A</math> to <math>A_1</math>, then <math>B</math> must move to <math>B_1</math> to keep <math>A_1B_1 = 1</math>. <math>AB</math> intersects with <math>A_1B_1</math> at <math>C</math>. Pick points <math>A_2</math> and <math>B_2</math> on <math>CA_1</math> and <math>CB</math> such that <math>CA_2 = CA</math>, <math>CB_2 = CB_1</math>, we have <math>A_1A_2 = BB_2</math>. Since <math>AA_1</math> is very small, <math>\angle CA_1A \approx 60^\circ</math>, <math>\angle CBB_1 \approx 30^\circ</math>, so <math>AA_2\approx \sqrt{3}A_1A_2</math>, <math>B_1B_2 \approx \frac{1}{\sqrt{3}}BB_2</math>, by similarity, <math>\frac{CA}{CB} \approx \frac{CA}{CB_2} = \frac{AA_2}{B_1B_2} = \frac{\sqrt{3}A_1A_2}{\frac{1}{\sqrt{3}}BB_2} = 3</math>. So the coordinates of <math>C</math> is <math>\left(\frac{1}{8}, \frac{3\sqrt{3}}{8}\right)</math>.
-Let <math>\theta = \angle OA'C</math>. Draw <math>CD</math> perpendicular to the <math>x</math>-axis and <math>CE</math> perpendicular to the <math>y</math>-axis as shown in the diagram. Then
+so <math>OC^2 = \frac{1}{64} + \frac{27}{64} = \frac{7}{16}</math>, the answer is <math>\boxed{023}</math>.
-<cmath>8A'B' = 8CA' + 8CB' = \frac{3\sqrt3}{\sin\theta} + \frac{1}{\cos\theta}</cmath>
-By some inequality (i forgor its name),
-<cmath>\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</cmath>
-We know that <math>\sin^2\theta + \cos^2\theta = 1</math>. Thus <math>\tfrac{3\sqrt3}{\sin\theta} + \tfrac{1}{\cos\theta} \ge 8</math>. Equality holds if and only if
-<cmath>\frac{3\sqrt3}{\sin\theta} : \frac{1}{\cos\theta} = \frac{3\sqrt3}{\sin\theta} : \frac{1}{\cos\theta} = \sin^2\theta : \cos^2\theta</cmath>
-which occurs when <math>\theta=\tfrac\pi3</math>. Guess what, <math>\angle OAB</math> ''happens'' to be <math>\tfrac\pi3</math>, thus <math>A'=A</math> and <math>B'=B</math>. Thus, <math>AB</math> is the only segment in <math>\mathcal{F}</math> that passes through <math>C</math>. Finally, we calculate <math>OC^2 = \tfrac1{64} + \tfrac{27}{64} = \tfrac7{16}</math>, and the answer is <math>\boxed{023}</math>.
-~Furaken
-==Solution 2==
-<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>
+==Video Solution==
+https://youtu.be/914687Yv6SY?si=tc6XfoOIHu0gu6AL
+(no calculus)
+~MathProblemSolvingSkills.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 <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 such 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. If you know such a solution, write it here on this wiki page.) ~Furaken
+I think there is such a geometry way:
+Let <math>DE</math> pass through <math>C</math> while point <math>D</math> is on the outside of line segment <math>OA</math> and point <math>E</math> is in between <math>O</math> and <math>B</math>. We aim to show <math>DE</math> is longer than <math>AB</math>. Now since <math>PC</math> is the altitude of triangle <math>PAB</math> yet just a cevian on the base <math>DE</math> of triangle <math>PDE</math> (thus making the height shorter than <math>PC</math>), it suffices to show the area of triangle <math>PDE</math> is bigger than that of triangle <math>PAB</math>. To do this, we compare these two triangles (let <math>DE</math> intersect <math>PA</math> at point <math>F</math>), and we just want to show <math>PF*AD > AF*AO</math>. This is trivial by similarity ratios. ~gougutheorem
+Thanks! Now we know that it's possible to solve the AIME problem with only geometry. ~Furaken
-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>
+==See also==
+{{AIME box|year=2024|num-b=11|num-a=13|n=II}}
-~Bluesoul
+[[Category:Intermediate Geometry Problems]]
+{{MAA Notice}}