Difference between revisions of "2024 AIME II Problems/Problem 12"
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− | : | + | ==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== | ||
+ | |||
+ | <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} \\ | ||
+ | 2a\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 \\ | ||
+ | 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 <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) == | ||
+ | |||
+ | <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 <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>. | ||
+ | |||
+ | so <math>OC^2 = \frac{1}{64} + \frac{27}{64} = \frac{7}{16}</math>, the answer is <math>\boxed{023}</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 | ||
+ | |||
+ | ==See also== | ||
+ | {{AIME box|year=2024|num-b=11|num-a=13|n=II}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 00:31, 11 April 2024
Contents
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
Now, we want to find . By L'Hôpital's rule, we get . This means that , so we get .
~Bluesoul
Solution 3
The equation of line is
The position of line can be characterized by , denoted as . Thus, the equation of line is
Solving (1) and (2), the -coordinate of the intersecting point of lines and satisfies the following equation:
We denote the L.H.S. as .
We observe that for all . Therefore, the point that this problem asks us to find can be equivalently stated in the following way:
We interpret Equation (1) as a parameterized equation that is a tuning parameter and is a variable that shall be solved and expressed in terms of . In Equation (1), there exists a unique , denoted as (-coordinate of point ), such that the only solution is . For all other , there are more than one solutions with one solution and at least another solution.
Given that function is differentiable, the above condition is equivalent to the first-order-condition
Calculating derivatives in this equation, we get
By solving this equation, we get
Plugging this into Equation (1), we get the -coordinate of point :
Therefore, \begin{align*} OC^2 & = x_C^2 + y_C^2 \\ & = \frac{7}{16} . \end{align*}
Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 4 (coordinate bash)
Let be a segment in with x-intercept and y-intercept . We can write 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 lie on and no other segment in . We can find by solving and taking the limit as . Since has length , by the Pythagorean theorem. Solving this for , 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 , the equation for becomes
In , and . To find the x-coordinate of , we substitute these into the equation for 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 to get We substitute into the equation for to find the y-coordinate of : The problem asks for so .
Solution 5 (small perturb)
Let's move a little bit from to , then must move to to keep . intersects with at . Pick points and on and such that , , we have . Since is very small, , , so , , by similarity, . So the coordinates of is .
so , the answer is .
Video Solution
https://youtu.be/914687Yv6SY?si=tc6XfoOIHu0gu6AL
(no calculus)
~MathProblemSolvingSkills.com
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Query
Let be a fixed point in the first quadrant. Let be a point on the positive -axis and be a point on the positive -axis such that passes through and the length of is minimal. Let be the point such that is a rectangle. Prove that . (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 pass through while point is on the outside of line segment and point is in between and . We aim to show is longer than . Now since is the altitude of triangle yet just a cevian on the base of triangle (thus making the height shorter than ), it suffices to show the area of triangle is bigger than that of triangle . To do this, we compare these two triangles (let intersect at point ), and we just want to show . 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.