2024 AIME II Problems/Problem 12

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 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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2024 AIME II Problems/Problem 12

Contents

Problem

Solution 2

Solution 3

Solution 4 (coordinate bash)

Solution 5 (small perturb)

Video Solution

Video Solution

Query

See also