2024 AIME II Problems/Problem 12


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 $\boxed{023}$.


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


(no calculus)


Video Solution


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


[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

See also

2024 AIME II (ProblemsAnswer KeyResources)
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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