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

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so <math>OC^2 = \frac{1}{64} + \frac{27}{64} = \frac{7}{16}</math>, the answer is <math>\boxed{023}</math>.
 
so <math>OC^2 = \frac{1}{64} + \frac{27}{64} = \frac{7}{16}</math>, the answer is <math>\boxed{023}</math>.
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==Video Solution==
 +
 +
https://youtu.be/914687Yv6SY?si=tc6XfoOIHu0gu6AL
 +
 +
(no calculus)
 +
 +
~MathProblemSolvingSkills.com
 +
  
 
==Video Solution==
 
==Video Solution==

Revision as of 17:06, 8 April 2024

Problem

Let O=(0,0), A=(12,0), and B=(0,32) be points in the coordinate plane. Let F be the family of segments 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 AB―, distinct from A and B, that does not belong to any segment from F other than AB―. Then OC2=pq, 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 forgot 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

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

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, OC2=xC2+yC2=716.

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 xa+yb=1y=b(1−xa). 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 a2+b2=1b2=1−a2db2da=d(1−a2)da2adbda=−2adb=−abda. 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 32(1−x12)=(32−1232da)(1−x12+da)32(1−2x)=(32−da3)(1−x1+2da2)32−3x=3−2da23(1−2x1+2da)32−3x=3−2da23⋅1+2da−2x1+2da32−3x=3+6da−6x−2da−4da2+4xda23+43da(32−3x)(23+43da)=3+6da−6x−2da−4da2+4xda3+6da−6x−12xda=3+4da−6x−4da2+4xda2da=−4da2+16xda16xda=2da+4da2x=da+2da28da. 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

See also

2024 AIME II (Problems • Answer Key • Resources)
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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